summarize  this for power point presentation please: AP Photo/Ed Bailey
167
ATM USERS ARE DISCRETE
AS WELL AS DISCREET
Yes, we know automatic teller machine (ATM) customers use discretion as to
when and how they use the ATM. For example, many zealously shield the
keyboard when entering their personal ID number into the machine.
But ATM users are also discrete: They are members of a discrete probability
distribution. For x 5 the number of persons arriving at the machine over a period
of time, x will be a discrete variable—0, 1, 2, 3, and so on.
The next time you’re in the vicinity of your local ATM, relax and enjoy a soft
drink for 5 minutes while you count how many people arrive at the machine during
that time. Do this on 9 more occasions. Now figure out what percent of
the time x was equal to 0, what percent
of the time x was equal to 1, and so
on. Congratulations, you’ve just created
a discrete probability distribution, a
type of distribution that is the topic of
this chapter.
More specifically, the phenomenon
you’ve been observing tends to be well
represented by a theoretical discrete
probability distribution called the Poisson
distribution, discussed later in the chapter.
This distribution is useful in both describing
and predicting the number of people arriving
at an ATM, a tollbooth, an emergency room,
and many other kinds of service facilities over a
given period of time.
Stuck in the Friday afternoon discrete probability
distribution . . .
C H A P T E R 6
DiscDriestteriPbruotbioanbsility• Understand both the concept and the applications of a probability
distribution for a random variable.
• Determine whether the random variable in a probability distribution is
of the discrete type or the continuous type.
• Differentiate among the binomial, hypergeometric, and Poisson
discrete probability distributions and their applications.
• Understand what is meant by a Bernoulli process and how this applies
to the consecutive trials associated with the binomial distribution.
• Use the appropriate probability distribution in determining the
probability that a discrete random variable will have a given value or a
value in a given range.
LEARNING
OBJECTIVES
After reading this chapter,
you should be able to:
168
INTRODUCTION
Chapter 2 introduced the relative frequency distribution as a way of describing
the proportion of actual observations that either had one of several possible
values or fell into one of a number of different ranges of values. In Chapter 5,
we discussed classical probability as a probability that could be theoretically
determined without engaging in any experiments [e.g., P(heads) 5 0.5 for a coin
flip]. In this chapter, we’ll see how a probability distribution is really a very
special combination of these two concepts.
In general, we can define a probability distribution as the relative frequency
distribution that should theoretically occur for observations from a given
population. In business and other contexts, it can be helpful to proceed from
(1) a basic understanding of how a natural process seems to operate in generating
events to (2) identifying the probability that a given event may occur. By using a
probability distribution as a model that represents the possible events and their
respective likelihoods of occurrence, we can make more effective decisions and
preparations in dealing with the events that the process is generating.
Random Variables: Discrete versus Continuous
A random variable is a variable that can take on different values according to the
outcome of an experiment. Random variables can be either discrete or continuous,
a distinction discussed briefly in Chapter 1:
(6.1)
DISCRETE RANDOM VARIABLE A random variable that can take on only certain
values along an interval, with the possible values having gaps between them.
For example, in a given group of five children, the number who got at least one
electronic toy for Christmas would be 0, 1, 2, 3, 4, or 5. It could not be a number
between any of these values, such as 2.338.
CONTINUOUS RANDOM VARIABLE A random variable that can take on a value at any
point along an interval. The exact temperature outside as you’re reading this book
could be 13.568, 78.352, or 83.815 degrees Fahrenheit, or any of an infinity of other
values in the range of temperatures where colleges and universities are located.
(The temperature examples have been expressed to just three decimal places here,
but there would be countless possibilities along the temperature continuum.)Chapter 6: Discrete Probability Distributions 169
A random variable is described as random because we don’t know ahead
of time exactly what value it will have following the experiment. For example,
when we toss a coin, we don’t know for sure whether it will land heads or tails.
Likewise, when we measure the diameter of a roller bearing, we don’t know in
advance what the exact measurement will be. In this chapter, the emphasis will
be on discrete random variables and their probability distributions. In the next,
we’ll cover variables of the continuous type.
The Nature of a Discrete Probability Distribution
A discrete probability distribution is a listing of all possible outcomes of an
experiment, along with their respective probabilities of occurrence. We’ll begin
our discussion of such distributions with a small-scale example.
EXAMPLE
Discrete Probability Distribution
An experiment is conducted in which a fair coin is flipped twice. The result of
the experiment will be the random variable x 5 the number of times that heads
comes up.
SOLUTION
Since the two tosses are independent, the probability of heads remains constant
at P(heads) 5 0.5 from one toss to the next. Each of four possible sequences is
equally likely, and each has a probability of 0.25:
Two tosses of a coin: Possible sequences and their probabilities
Sequence x 5 Number of Heads Probability
HH 2 0.25 5 (0.5 3 0.5)
HT 1 0.25
TH 1 0.25
TT 0 0.25
1.00
Because the possible sequences are mutually exclusive (only one of them can
happen for a given series of two tosses), the sum of the joint probabilities is 1.00.
Two of the sequences (HT and TH) involve one head, and we can use the addition rule from Chapter 5 to find the probability that x 5 1. This can be expressed
as P(HT or TH) 5 P(HT) 1 P(TH) 5 0.25 1 0.25 5 0.50. We have just generated the following probability distribution for the discrete random variable x 5
number of heads in two coin tosses:
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXA
Two tosses of a coin: Discrete probability distribution
for the random variable x 5 number of heads
x P(x)
0 0.25
1 0.50
2 0.25
1.00170 Part 2: Probability
This distribution can also be described graphically, as in Figure 6.1. This discrete probability distribution reflects several important characteristics shared by
all such distributions:
FIGURE 6.1
Discrete probability distribution
for the random variable
x 5 number of heads in two
consecutive tosses of a coin.
0 1
x = Number of heads
2
0.00
0.10
0.20
0.30
0.40
0.50
P(x)
Characteristics of a discrete probability distribution:
1. For any value of x, 0 # P(x) # 1.
2. The values of x are exhaustive: The probability distribution includes all
possible values.
3. The values of x are mutually exclusive: Only one value can occur for a given
experiment.
4. The sum of their probabilities is one, or oP(xi) 5 1.0.
Although discrete probability distributions are often based on classical probabilities (as in Figure 6.1), they can be the result of relative frequencies observed
from past experience or even from subjective judgments made about future events
and their likelihood of occurrence. As an example of a discrete probability distribution based on relative frequencies, consider the following situation.
EXAMPLE
Discrete Probability Distribution
A financial counselor conducts investment seminars with each seminar limited
to 6 attendees. Because of the small size of the seminar group and the personal
attention each person receives, some of the attendees become clients following the
seminar. For the past 20 seminars she has conducted, x (for x 5 the number of
attendees who become clients) has had the relative frequency distribution shown
in Figure 6.2.
SOLUTION
The probability distribution for x 5 the number of attendees who become
clients can also be described in tabular form. The events and their respective
EXAMPLEEXAMPLEEXAMPLEEChapter 6: Discrete Probability Distributions 171
The Mean and Variance of a Discrete
Probability Distribution
For the observed data in Chapter 3, we used measures of central tendency and
dispersion to describe typical observations and scatter in the data values, respectively. Similarly, it can be useful to describe a discrete probability distribution
in terms of its central tendency and dispersion. For example, the investment
counselor of Figure 6.2 may be interested both in how many attendees from a
seminar typically become clients and in the extent to which the number who
become clients varies.
The mean of a discrete probability distribution for a discrete random variable
is called its expected value, and this is referred to as E(x), or . In the following
formula, it can be seen to be a weighted average of all the possible outcomes, with
each outcome weighted according to its probability of occurrence. The variance
probabilities are as follows:
Twenty investment seminars: Discrete probability
distribution for the random variable x 5 number
of attendees who become clients
x P(x)
0 0.05
1 0.10
2 0.20
3 0.25
4 0.15
5 0.15
6 0.10
1.00
As in the classical relative frequency distribution of Figure 6.1, the events in
Figure 6.2 are mutually exclusive and exhaustive. Interpreting their relative frequencies as probabilities, as shown in Figure 6.2, the sum of these probabilities
will be 1.0. In a given seminar, the probability that nobody will become a client is
P(0) 5 0.05, while the probability that at least 4 will become clients is P(x $ 4) 5
P(x 5 4) 1 P(x 5 5) 1 P(x 5 6), or 0.15 1 0.15 1 0.10 5 0.40.
FIGURE 6.2
Relative frequency distribution
for the number of attendees in
the seminar group who become
clients of the seminar leader.
0 1
x = Number of attendees who become clients
2 3 4 5 6
0.00
0.10
0.20
0.30
P(x)
AMPLEEXAMPLEEXAMPLEEXAMPLE172 Part 2: Probability
is the expected value of the squared difference between the random variable and
its mean, or E[(x 2 )2]:
General formulas for the mean and variance of a discrete probability distribution:
• Mean:
 5 E(x) or  5 oxi P(xi) for all possible values of x
• Variance:
 2 5 E[(x 2 )2] or  2 5 o(xi 2 )2 P(xi)
for all possible values of x, and the standard deviation is
 5 Ï
___
 2
An alternative formula for 2 is 2 5 [ox2iP(xi)] 2 2. It provides greater ease
of calculation when the data values are large or when there are a great many
possible values of x.
To demonstrate how the mean and variance are determined, we will compute
these descriptors for the discrete probability distribution in Figure 6.2.
Mean and Variance for x 5 the Number of Attendees Who Become
Clients (Represented by the Discrete Probability Distribution
of Figure 6.2)
• Mean:
 5 E(x) 5 oxiP(xi)
5 0(0.05) 1 1(0.10) 1 2(0.20) 1 3(0.25) 1 4(0.15) 1 5(0.15) 1 6(0.10)
5 3.2 attendees who become clients
• Variance:
2 5 E[(x 2 )2] 5 o(xi 2 )2P(xi)
5 (0 2 3.2)2(0.05) 1 (1 2 3.2)2(0.10) 1 (2 2 3.2)2(0.20)
1 (3 2 3.2)2(0.25) 1 (4 2 3.2)2(0.15) 1 (5 2 3.2)2(0.15)
1 (6 2 3.2)2(0.10)
5 2.66
and
 5 Ï
_____
2.66 5 1.63 attendees who become clients
In a given seminar, there will never be exactly x 5 3.2 attendees who become
clients. However, this is the expected value, or mean of x, and it will tend to be
the average result if the probability distribution remains unchanged and a great
many such seminars are conducted.
The formulas in this section can be used to determine the mean and variance for
any discrete probability distribution. For some distributions, however, the mean and
variance can be described and computed by using expressions that are both more
practical and more easily calculated. As we discuss the distributions to follow, such
expressions for the mean and variance will be presented as they become relevant.
Before we move on to examine three particular discrete probability distributions, the binomial, the hypergeometric, and the Poisson, consider the probability
distribution represented by a state’s three-digit daily lottery number. There are
NOTEChapter 6: Discrete Probability Distributions 173
1000 possible outcomes (000 to 999) and, because all of the outcomes are
equally likely, this is an example of what is called a discrete uniform distribution.
Attempts to tamper with this uniformity are not well received by state officials.
See Statistics in Action 6.1.
A feature of the Pennsylvania State Lottery is the “daily
number,” a three-digit number selected at 7 P.M. each
weekday evening. The number is obtained from the first
ball that pops from each of three machines containing balls
numbered from 0 to 9. The big event is televised, with a
senior-citizen witness and a lottery official present to verify
the number of each ball.
Since the daily number can range from 000 to 999, the
chance of any specific result is just 0.001. Lottery officials
rigorously weigh the balls and conduct extensive testing to
make sure some numbers don’t have a higher probability
than others. A number of years ago, an individual was convicted for injecting a latexlike substance that made some
of the balls heavier. (The author and many others watching
television noticed that some of the bouncing balls seemed
a little “tired.”) As a result of this one-day nonuniformity of
the probability distribution, security was heightened considerably, and officials are now more scrupulous than ever
in their efforts to ensure that the probability distribution is
indeed uniform.
Despite the uniformity of this discrete probability distribution, some players still insist on playing a “lucky” number based on their birth date, wedding date, or digits from
an auto license number. Some even avoid playing yesterday’s number because they think it’s been “used up” for
awhile. Many seem to prefer 333 or 777, although 923,
147, or any other choice would have the same probability
of being selected.
The Discrete Uniform Distribution That Took a Day Off
STATISTICS6.1IN ACTION
6.1 Why is a random variable said to be “random”?
6.2 Indicate whether each of the following random
variables is discrete or continuous.
a. The diameter of aluminum rods coming off a
production line
b. The number of years of schooling employees have
completed
c. The Dow Jones Industrial Average
d. The volume of milk purchased during a supermarket
visit
6.3 Indicate whether each of the following random
variables is discrete or continuous.
a. The number of drive-through customers at your local
McDonald’s today
b. The amount spent by all customers at the local
McDonald’s today
c. The time, as reported by your digital watch
d. The length of a pedestrian’s stride
6.4 The values of x in a discrete probability distribution
must be both exhaustive and mutually exclusive. What is
meant by these two terms?
6.5 A discrete random variable can have the values
x 5 3, x 5 8, or x 5 10, and the respective probabilities
are 0.2, 0.7, and 0.1. Determine the mean, variance, and
standard deviation of x.
6.6 Determine the mean, variance, and standard deviation of the following discrete probability distribution:
x 0 1 2
P(x) 0.60 0.30 0.10
6.7 Determine the mean, variance, and standard deviation of the following discrete probability distribution:
x 0 1 2 3 4
P(x) 0.10 0.30 0.30 0.20 0.10
6.8 Ed Tompkins, the assistant dean of a business school,
has applied for the position of dean of the school of
business at a much larger university. The salary at the
new university has been advertised as $200,000, and
Ed is very excited about this possibility for making a
big career move. He has been told by friends within the
administration of the larger university that his resume
is impressive and his chances of getting the position he
seeks are “about 60%.” If Ed stays at his current position, his salary next year will be $120,000. Assuming
that his friends have accurately assessed his chances of
success, what is Ed’s expected salary for next year?
6.9 Scam artists sometimes “stage” accidents by purposely walking in front of slow-moving luxury cars
whose drivers are presumed to carry large amounts of
liability insurance. The “victim” then offers to accept a
E X E R C I
S E S174 Part 2: Probability
small monetary cash settlement so the driver can avoid
an insurance claim, a blot on his driving record, and
increased insurance premiums. Marlin has just tried
to stage such a scheme by walking into a slow-moving
Mercedes and faking a minor injury. Assume that the
possible outcomes for Marlin are 2$500 (i.e., he is
arrested and fined for his scam), $0 (the driver recognizes
the scam and refuses to pay), and $100 (the driver simply pays up to get rid of Marlin). If the probabilities for
these outcomes are 0.1, 0.3, and 0.6, respectively, what is
Marlin’s expected monetary outcome for trying to scam
the Mercedes driver?
6.10 A regional manager visits local fast-food franchises
and evaluates the speed of the service. If the manager
receives her meal within 45 seconds, the server is given
a free movie-admission coupon. If it takes more than
45 seconds, the server receives nothing. Throughout the
company’s franchises, the probability is 0.60 that a meal
will be served within 45 seconds. What is the expected
number of coupons a counter employee will receive when
serving the regional manager?
6.11 A consultant has presented his client with three
alternative research studies that could be conducted. If
the client chooses alternative a, the consultant will have
to hire 1 additional employee. Two new employees will
have to be hired if the client chooses b, and 5 if the client
chooses c. The probability that the client will select alternative a is 0.1, with P(b) 5 0.5 and P(c) 5 0.4. Identify
the discrete random variable in this situation and determine its expected value.
6.12 A contractor must pay a $40,000 penalty if construction of an expensive home requires more than 16 weeks.
He will receive a bonus of $10,000 if the home is completed within 8 weeks. Based on experience with this type of
project, the contractor feels there is a 0.2 chance the home
will require more than 16 weeks for completion, and
there is a 0.3 chance it will be finished within 8 weeks. If
the price of the home is $550,000 before any penalty or
bonus adjustment, how much can the buyer expect to pay
for her new home when it is completed?
6.13 A music shop is promoting a sale in which the
purchaser of a compact disk can roll a die, then deduct
a dollar from the retail price for each dot that shows on
the rolled die. It is equally likely that the die will come
up any integer from 1 through 6. The owner of the music
shop pays $5.00 for each compact disk, then prices them
at $9.00. During this special promotion, what will be the
shop’s average profit per compact disk sold?
6.14 Laura McCarthy, the owner of Riverside Bakery,
has been approached by insurance underwriters trying
to convince her to purchase flood insurance. According
to local meteorologists, there is a 0.01 probability that
the river will flood next year. Riverside’s profits for the
coming year depend on whether Laura buys the flood
insurance and whether the river floods. The profits
(which take into consideration the $10,000 premium for
the flood insurance) for the four possible combinations
of Laura’s choice and river conditions are:
The River
Does Not
Flood Floods
Insurance
Decision
No Flood
Insurance
Get Flood
Insurance
$200,000 2$1,000,000
$190,000 $200,000
a. If Laura decides not to purchase flood insurance, use
the appropriate discrete probability distribution to
determine Riverside’s expected profit next year.
b. If Laura purchases the flood insurance, what will be
Riverside’s expected profit next year?
c. Given the results in parts (a) and (b), provide Laura
with a recommendation.
6.15 In 2007, nearly 24 million tons of steel mill products went to construction and contracting companies.
Transco Steel, a hypothetical manufacturer specializing
in the production of steel for this market, is faced with
aging production facilities and is considering the possible
renovation of its plant. The company has reduced its
alternatives to just three: (1) no renovation at all,
(2) minor renovation, and (3) major renovation.
Whichever strategy it chooses, the company’s profits will
depend on the size of the construction and contracting
industry in future years. Company executives have a subjective probability of 0.3 that the industry will not grow
at all, a probability of 0.5 that it will grow moderately,
and a probability of 0.2 that it will show a high rate of
growth. The company’s estimated profits (in millions of
dollars) for each combination of renovation strategy and
industry growth possibility are:
Source: Statistical Abstract of the United States 2009, p. 630.
Industry Growth
None Moderate High
Level of Plant
Renovation
None $35 $38 $42
Minor $28 $45 $55
Major $21 $40 $70
Given the executives’ subjective probabilities for the
three possible states of industry growth, what will be the
company’s expected profit if it chooses not to renovate
the plant? If it chooses minor renovations to the plant? If
it chooses major renovations to the plant? Based on these
expected values, provide the company’s management
with a recommendation as to which level of renovation
they should choose.Chapter 6: Discrete Probability Distributions 175
THE BINOMIAL DISTRIBUTION
One of the most widely used discrete distributions, the binomial distribution deals
with consecutive trials, each of which has two possible outcomes. This latter
characteristic is the reason for the “bi” part of the descriptor binomial. The outcomes can be identified in general terms, such as “success” versus “failure” or
“yes” versus “no,” or by more specific labels, such as “milk in the refrigerator”
versus “no milk in the refrigerator.” The binomial distribution relies on what is
known as the Bernoulli process:
Characteristics of a Bernoulli process:
1. There are two or more consecutive trials.
2. In each trial, there are just two possible outcomes—usually denoted as success
or failure.
3. The trials are statistically independent; that is, the outcome in any trial is not
affected by the outcomes of earlier trials, and it does not affect the outcomes
of later trials.
4. The probability of a success remains the same from one trial to the next.
Description and Applications
In the binomial distribution, the discrete random variable, x, is the number of
successes that occur in n consecutive trials of the Bernoulli process. The distribution, which has a wide variety of applications that will be demonstrated through
examples and chapter exercises, can be summarized as follows:
(6.2)
The binomial probability distribution:
The probability of exactly x successes in n trials is
P(x) 5 }} n!
x!(n 2 x)! x(1 2 )n2x where n 5 number of trials
x 5 number of successes
 5 the probability of
success in any given trial
(1 2  ) 5 the probability of failure in
any given trial
• Mean:
 5 E(x) 5 n
• Variance:
 2 5 E[(x 2 )2] 5 n(1 2 )
In the binomial distribution, we can view the n trials as comprising (1) the x
trials that are successes and (2) the (n 2 x) trials that are failures. The second
component in the formula, x, is the probability that x of x trials will all be successful. The third component, (1 2 )n2x, is the probability that (n 2 x) of (n 2 x)176 Part 2: Probability
1Source: Statistical Abstract of the United States 2006, pp. 367, 635.
trials will all be failures. The first component is the number of possible combinations for n items taken x at a time; in the context of the binomial distribution,
this is the number of possible sequences where there are exactly x successes in
n trials.
The binomial distribution is really a family of distributions, and the exact
member of the family is determined by the values of n and . The following observations may be made regarding the Bernoulli process and the requirement that
the probability of success () remain unchanged:
1. If sampling is done with replacement (the person or other item selected from
the population is observed, then put back into the population),  will be constant from one trial to the next.
2. If, when sampling without replacement, the number of trials (n) is very small
compared to the population (N) of such trials from which the sample is
taken, as a practical matter  can be considered to be constant from one trial
to the next. This would occur, for example, if n 5 5 interview participants
were randomly selected from a population of N 5 1000 persons. As a rule of
thumb, if the population is at least 20 times as large as the number of trials,
we will assume the “constant ” assumption has been satisfactorily met.
EXAMPLE
Binomial Distribution
Of the 18,000,000 part-time workers in the United States, 20% participate in retirement benefits.1 A group of 5 people is to be randomly selected from this group,
and the discrete random variable is x 5 the number of persons in the group who
participate in retirement benefits.
SOLUTION
This example can be considered a Bernoulli process, with n 5 5 trials and  5
0.2. The size of the population (N 5 18,000,000) is extremely large compared
to the number of trials, so  can be assumed to be constant from one trial to the
next. A worker either participates in retirement benefits or does not participate,
so there are only two possible outcomes for each of the 5 trials.
What Is the Expected Value of x?
The expected number of people in the group who participate can be calculated
as E(x) 5 n 5 5(0.2), or E(x) 5 1.0. The expectation would be that 1.0 persons
in the group would participate in retirement benefits. (In other words, if we were
to randomly select a great many such groups, we would expect the average value
of x to be 1.0.)
What Is the Probability That the Group Will Include Exactly Two
Participants?
Using the binomial formula, with n 5 5,  5 0.2, and (1 2 ) 5 (1.0 2 0.2) 5
0.8, the probability of x 5 exactly 2 participants is
P(x) 5}n!
x!(n 2 x)!
x (1 2 )n2x
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLChapter 6: Discrete Probability Distributions 177
and
P(2) 5}5!
2!(5 2 2)!
(0.2)2(0.8)522 5 0.205
i.e., }} 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
(2 ∙ 1)(3 ∙ 2 ∙ 1)
5
5 ∙ 4
}
2 ∙ 1
For Each of the Possible Values of x, What Is the Probability That Exactly
This Number of People Out of the Five Will Be Participants?
The values x 5 0 through x 5 5 include all of the possibilities for the number of
persons in the sample who are participants; their probabilities would be
P(0) 5}5!
0!(5 2 0)!
(0.2)0(0.8)5 5 0.328
(Note: Remember that 0! 5 1 and that any nonzero number raised to the zero
power is also equal to 1.)
P(1) 5}5!
1!(5 2 1)!
(0.2)1(0.8)4 5 0.410
P(2) 5}5!
2!(5 2 2)!
(0.2)2(0.8)3 5 0.205 (as calculated earlier)
P(3) 5}5!
3!(5 2 3)!
(0.2)3(0.8)2 5 0.051
P(4) 5}5!
4!(5 2 4)!
(0.2)4(0.8)1 5 0.006
P(5) 5}5!
5!(5 2 5)!
(0.2)5(0.8)0 5 0.000 (rounded)
We can verify that  5 E(x) 5 n, or 5(0.20) 5 1.0, by calculating E(x) 5
ox
iP(xi). Because we have rounded the individual probabilities, the result differs
slightly from the actual value,  5 1.000 persons:
E(x) 5 oxiP(xi)
5 0(0.328) 1 1(0.410) 1 2(0.205) 1 3(0.051) 1 4(0.006) 1 5(0.000)
5 0.997 persons (differs from  5 1.000 due to rounding of the
individual probabilities)
What Is the Probability That the Group Will Include at Least Three Persons
Who Are Participants?
Because x can take on only one of the possible values between 0 and 5, the
outcomes are mutually exclusive. (Notice that the sum of the six probabilities just
calculated will be 1.000.) Thus P(x $ 3) will be
P(x $ 3) 5 [P(x 5 3) 1 P(x 5 4) 1 P(x 5 5)]
5 0.051 1 0.006 1 0.000 5 0.057
The binomial distribution for the discrete random variable in this example is
shown in Figure 6.3. The distribution is positively skewed. Binomial distributions
can have different shapes, including being positively skewed, negatively skewed,
or even symmetrical.
XAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE178 Part 2: Probability
Using the Binomial Tables
As the preceding example indicates, calculating binomial probabilities can be
tedious. Fortunately, binomial distribution tables, such as those in Appendix A,
make the task a lot easier. We will continue with our example in demonstrating
their use.
A given binomial probability distribution depends on the values of  and n,
so a separate listing is necessary for various combinations of their values. For the
five randomly selected part-time workers, n 5 5 trials, and we will rely on the
individual binomial probabilities table (Table A.1), a portion of which is reproduced here as Table 6.1. The “” columns in the table range from 0.1 to 0.9 in
increments of 0.1. The value of  in our example was  5 0.2, so this will be the
TABLE 6.1
A portion of the table of
individual binomial probabilities
from Appendix A. For n 5 5
trials, each entry is P(x 5 k)
for the specified value of the
population proportion, .
Binomial Distribution, Individual Probabilities
for x 5 number of successes in n trials, prob(x 5 k)
n 5 5
 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0 0.5905 0.3277 0.1681 0.0778 0.0313 0.0102 0.0024 0.0003 0.0000
1 0.3281 0.4096 0.3601 0.2592 0.1562 0.0768 0.0284 0.0064 0.0005
k 2 0.0729 0.2048 0.3087 0.3456 0.3125 0.2304 0.1323 0.0512 0.0081
3 0.0081 0.0512 0.1323 0.2304 0.3125 0.3456 0.3087 0.2048 0.0729
4 0.0005 0.0064 0.0284 0.0768 0.1562 0.2592 0.3601 0.4096 0.3281
5 0.0000 0.0003 0.0024 0.0102 0.0313 0.0778 0.1681 0.3277 0.5905
For n 5 5 and  5 0.2, P(x 5 2) is 0.2048.
FIGURE 6.3
Binomial probability
distribution for text example
with n 5 5 and  5 0.20. Of
18 million part-time workers,
only 20% participate in
retirement benefits. Because
the population is very large
compared to the number in the
sample,  can be considered as
remaining constant from one
trial to the next.
0 1
x = Number of persons participating in retirement benefits
2 3 4 5
0.00
0.10
0.20
0.30
0.40
0.50
P(x)Chapter 6: Discrete Probability Distributions 179
appropriate column for determining the probabilities that we’ve just calculated
the hard way. Briefly summarizing our earlier treatment of this example, n 5 5,
 5 0.2, and the probabilities calculated for the possible values of x were
x 0 1 2 3 4 5
P(x) 0.328 0.410 0.205 0.051 0.006 0.000
Using the individual binomial probabilities in Table 6.1, we refer to the
“ 5 0.2” column within the “n 5 5” section. For example, Table 6.1 shows
P(x 5 0) to be 0.3277. Likewise, P(x 5 1) and P(x 5 2) are 0.4096 and 0.2048,
respectively.
Appendix A also includes a table for cumulative binomial probabilities
(Table A.2), a portion of which is reproduced as Table 6.2. Using this table, we
find that P(x # 3) 5 0.9933 for x 5 the number of participants among the five
randomly selected workers.
Using the table of cumulative binomial probabilities is the best way to
determine “greater than or equal to” probabilities, such as P(x $ 2). For example,
to find P(x $ 2),
P(x # 1) 1 P(x $ 2) 5 1.000
and
From the table of cumulative
probabilities, P(x # 1) 5 0.7373.
P(x $ 2) 5 1.0000 2 P(x # 1)
so
P(x $ 2) 5 1.0000 2 0.7373 5 0.2627
As an alternative, the individual binomial probabilities in Table 6.1 could
also be used in finding P(x $ 2). Using this table would be more cumbersome,
TABLE 6.2
A portion of the table
of cumulative binomial
probabilities from Appendix A.
For n 5 5 trials, each entry is
P(x # k) for the specified value
of the population proportion, .
Binomial Distribution, Cumulative Probabilities
for x 5 number of successes in n trials, prob (x  k)
n 5 5
 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0 0.5905 0.3277 0.1681 0.0778 0.0313 0.0102 0.0024 0.0003 0.0000
1 0.9185 0.7373 0.5282 0.3370 0.1875 0.0870 0.0308 0.0067 0.0005
k 2 0.9914 0.9421 0.8369 0.6826 0.5000 0.3174 0.1631 0.0579 0.0086
3 0.9995 0.9933 0.9692 0.9130 0.8125 0.6630 0.4718 0.2627 0.0815
4 1.0000 0.9997 0.9976 0.9898 0.9688 0.9222 0.8319 0.6723 0.4095
5 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
For n 5 5 and  5 0.2, P(x # 3) is 0.9933.180 Part 2: Probability
however, since we would have to add up each of the probabilities for x 5 2
through x 5 5, or
P(x $ 2) 5 P(x 5 2) 1 P(x 5 3) 1 P(x 5 4) 1 P(x 5 5)
5 0.2048 1 0.0512 1 0.0064 1 0.0003
5 0.2627
Excel and Minitab can provide either individual or cumulative binomial
probabilities. The procedures and the results are shown in Computer Solutions 6.1.
Binomial Probabilities
These procedures show how to get binomial probabilities associated with n 5 5 trials when the probability
of a success on any given trial is p 5 0.20.
For each software package:
Procedure I: Individual or cumulative probabilities for x 5 2 successes.
Procedure II: Complete set of individual or cumulative probabilities for x 5 0 successes through
x 5 5 successes.
EXCEL
COMPUTER 6.1 SOLUTIONS
I. Excel procedure for just one individual or cumulative binomial probability
1. Click on the cell into which Excel will place the probability. From the Formulas ribbon, click on the Insert Function
symbol at the extreme left. Click Statistical in the Category box. Click BINOMDIST in the Function box. Click OK.
2. Enter the number of successes (2) into the Number_s box. Enter the number of trials (5) into the Trials box. Enter
the probability of success on any given trial (0.20) into the Probability_s box. To obtain the individual probability,
P(x 5 2), enter false into the Cumulative box. To obtain the cumulative probability, P(x # 2), enter true into the
Cumulative box. Click OK.
II. Excel procedure for a complete set of individual or cumulative binomial probabilities
1. Enter the column labels into A1:C1 as shown in the printout above. Enter the lowest number of successes in the set
(in this case, 0) into A2. Enter 1 into A3. Click on cell A2 and highlight cells A2:A3. Click at the lower right corner
of A3 (the cursor will turn to a “1” sign) and drag downward so cells A2:A7 will contain the values 0 through 5.
2. To get the set of individual probabilities: Click on cell B2 (this is where the first individual probability will go). Follow the
remainder of the steps in Excel procedure I, except enter A2 into the Number_s (number of successes) box and enter
false into the Cumulative box. The first probability now appears in B2. Click to select cell B2. Click the lower right
corner of B2 (the cursor turns to a “1” sign) and drag down until cells B2:B7 contain the set of individual probabilities.
3. To get the set of cumulative probabilities: Click on cell C2 (this is where the first cumulative probability will go).
Follow the remainder of the steps in Excel procedure I, except enter A2 into the Number_s (number of successes)
box and enter true into the Cumulative box. The first probability now appears in C2. Click to select cell C2.
(continued)Chapter 6: Discrete Probability Distributions 181
Additional Comments on the Binomial Distribution
Under certain conditions, the binomial distribution can be approximated by other
probability distributions. This is especially useful since there is a practical limit to
the sheer volume of printed tables that can be included in a statistical appendix. A
later section of this chapter discusses how another discrete probability distribution
called the Poisson distribution can be used to approximate the binomial. Introduced
in the next chapter, a continuous distribution called the normal distribution is especially useful as a satisfactory substitute in many practical applications.
Click the lower right corner of C2 (the cursor turns to a “+” sign) and drag down until cells C2:C7 contain the set of
cumulative probabilities.
MINITAB
I. Minitab procedure for just one individual or cumulative binomial probability
1. Click Calc. Select Probability Distributions. Click Binomial.
2. In the Binomial Distribution menu, enter 5 into the Number of trials box. Enter 0.20 into the Probability of success
box. Select Input constant and enter the desired x value (e.g., 2) into the box. Select either Probability or Cumulative
probability, depending on whether you want an individual probability or a cumulative probability. Click OK.
II. Minitab procedure for a complete set of individual or cumulative binomial probabilities
Enter the desired x values (in this case 0 through 5) into column C1. Follow steps 1 and 2 above, except select
Input column and enter C1 into the box.
6.16 What is necessary for a process to be considered a
Bernoulli process?
6.17 When we are sampling without replacement, under
what conditions can we assume that the constant p
assumption has been satisfactorily met?
6.18 Seven trials are conducted in a Bernoulli process in
which the probability of success in a given trial is 0.7.
If x 5 the number of successes, determine the
following:
a. E(x) b. x c. P(x 5 3)
d. P(3 # x # 5) e. P(x . 4)
6.19 Twelve trials are conducted in a Bernoulli process
in which the probability of success in a given trial is 0.3.
If x 5 the number of successes, determine the following:
a. E(x) b. x c. P(x 5 3)
d. P(2 # x # 8) e. P(x . 3)
E X E R C I
S E S182 Part 2: Probability
6.20 A city law-enforcement official has stated that 20%
of the items sold by pawn shops within the city have
been stolen. Ralph has just purchased 4 items from one
of the city’s pawn shops. Assuming that the official is
correct, and for x 5 the number of Ralph’s purchases
that have been stolen, determine the following:
a. P(x 5 0) b. P(2 # x)
c. P(1 # x # 3) d. P(x # 2)
6.21 The regional manager in Exercise 6.10 plans to
evaluate the speed of service at five local franchises of the
restaurant. For the five servers, determine the probability
that
a. none will receive a movie coupon.
b. two of the five will receive a coupon.
c. three of the five will receive a coupon.
d. all five will receive a coupon.
6.22 According to the National Marine Manufacturers
Association, 50.0% of the population of Vermont were
boating participants during the most recent year. For a
randomly selected sample of 20 Vermont residents, with
the discrete random variable x 5 the number in the sample who were boating participants that year, determine
the following: Source: National Marine Manufacturers Association,
NMMA Boating Industry Report, p. 3.
a. E(x) b. P(x # 8) c. P(x 5 10)
d. P(x 5 12) e. P(7 # x # 13)
6.23 In the town of Hickoryville, an adult citizen has a
1% chance of being selected for jury duty next year. If
jury members are selected at random, and the Anderson
family includes three adults, determine the probability
that
a. none of the Andersons will be chosen.
b. exactly one of the Andersons will be chosen.
c. exactly two of the Andersons will be chosen.
d. all three Andersons will be chosen.
6.24 It has been reported that the overall graduation rate
for football players at Division I-A colleges and universities is 65%. Source: “How Football Rates,” USA Today, December
20, 2005, p. 12C.
a. If five players are randomly selected from the entire
population of Division I-A football players, what
is the probability that at least 2 of the 5 will
graduate?
b. In part (a), what is the probability that all 5 of the
players will graduate?
6.25 An association of coffee producers has estimated that 50% of adult workers drink coffee at least
occasionally while on the job. Given this estimate:
a. For a randomly selected group of 4 adult workers,
what is the probability that no more than 3 of them
drink coffee at least occasionally while on the job?
b. In part (a), what is the probability that at least 2 of
these individuals drink coffee at least occasionally
while on the job?
6.26 The U.S. Department of Labor has reported that
30% of the 2.1 million mathematical and computer
scientists in the United States are women. If 3 individuals
are randomly selected from this occupational group, and
x 5 the number of females, determine P(x 5 0), P(x 5 1),
P(x 5 2), and P(x 5 3). Source: Bureau of the Census, Statistical
Abstract of the United States 2002, p. 381.
6.27 Alicia’s schedule includes three Tuesday/Thursday
courses in which each professor uses a single coin flip to
decide whether to give a short quiz each class day. Alicia
is active on the university’s student advisory council,
works two part-time jobs, and is helping to build a float
for the upcoming homecoming parade. As a result of these
activities and slightly inadequate planning on her part,
she has had very little time to prepare for her Thursday
classes being held tomorrow. What is the probability that
Alicia will be “lucky” and have no surprise quizzes tomorrow? What is the probability that the worst will occur and
she will have surprise quizzes in all three classes? What is
the probability that she will escape with minimal damage
and have a quiz in only one of the three classes?
6.28OfficeQuip is a small office supply firm that is currently
bidding on furniture and office equipment contracts with
four different potential customers who are of comparable
size. For each contract, OfficeQuip would gain a profit of
$50,000 if that contract were accepted, so the company
could make as little as $0 or as much as $200,000. The four
potential customers are making independent decisions, and
in each case the probability that OfficeQuip will receive the
contract is 0.40. When all the decisions have been made,
what is the probability that OfficeQuip will receive none
of the contracts? Exactly one of the contracts? Exactly two
of the contracts? Exactly three of the contracts? All four
contracts? Overall, what is OfficeQuip’s expected profit in
this business-procurement venture?
6.29 Four wheel bearings are to be replaced on a company
vehicle, and the mechanic has selected the four replacement
parts from a large supply bin. Unknown to the mechanic,
10% of the bearings are defective and will fail within the
first 100 miles after installation. What is the probability
that the company vehicle will have a breakdown due to
defective wheel bearings before it completes tomorrow’s
200-mile delivery route?
6.30 It has been estimated that one in five Americans suffers from allergies. The president of Hargrove University
plans to randomly select 10 students from the undergraduate population of 1400 to attend a dinner at his home.
Assuming that Hargrove students are typical in terms of
susceptibility to allergies and that the college president’s
home happens to contain just about every common allergen to which afflicted persons react, what is the probability that at least 8 of the students will be able to stay for
the duration of the dinner event? Source: World Almanac and
Book of Facts 2009, p. 187.Chapter 6: Discrete Probability Distributions 183
6.31 Airlines book more seats than are actually available,
then “bump” would-be passengers whenever more people
show up than there are seats. Through the first quarter of
2009, the rate at which passengers were bumped was 1.40
per 1000 passengers. Assuming that, on average, the probability of any given passenger being bumped is 1.40/1000,
or 0.00140: Source: U.S. Bureau of Transportation Statistics, June 2009.
a. Emily is flying nonstop to visit Besco, Inc., for a job
interview. What is the probability that she will be
bumped?
b. Ten members of the Besco board of directors will be
flying nonstop to the firm’s annual meeting, and their
flights are independent of each other. What is the
probability that at least one of them will be bumped?
c. Besco sales personnel made a total of 220 nonstop
flights last year, and they experienced a total of four
bumpings. Could the company’s experience be considered as very unusual?
6.32 Every day, artists at Arnold’s House of Fine
Figurines produce 5 miniature statues that must be
sanded and painted. Past experience has shown that 10%
of the statues have a defect that does not show up until
after the statue has been sanded and painted. Whenever
a statue is found to have this defect, Arnold must pay a
specialist $20 to correct the defect and repaint the statue.
Describe the probability distribution for the daily amount
that Arnold must pay the specialist, and determine the
average daily amount the specialist will receive.
THE HYPERGEOMETRIC DISTRIBUTION
Description and Applications
The hypergeometric probability distribution is similar to the binomial in focusing
on the number of successes in a given number of consecutive trials, but there are
two major differences: First, the consecutive trials are not independent; that is,
the outcome in a trial can be affected by the outcomes in earlier trials, and can
in turn affect the outcomes in later trials. Second, the probability of success does
not remain constant from one trial to the next. Unlike the binomial distribution,
the hypergeometric distribution does not rely on a Bernoulli process.
The distinction between the binomial and hypergeometric distributions can
also be viewed in terms of sampling with replacement versus sampling without
replacement. In sampling without replacement, the object selected from the population is not returned to the population. When sampling is done without replacement and the population is not very large, the probability of a success can change
quite drastically from one trial to the next.
Situations like these can be analyzed by means of tree diagrams, along with
their marginal, conditional, and joint probabilities, but the hypergeometric distribution is a lot easier. The distribution can be described as follows:
The hypergeometric distribution:
The probability of exactly x successes in n trials that are not independent is
P(x) 5
( sx) ∙ ( N n 22 xs)
}}
( Nn )
where N 5 size of the population
n 5 size of the sample
s 5 number of successes in
the population
x 5 number of successes in the sample
• Mean:  5 E(x) 5 }ns
N
• Variance: 2 5 E[(x 2 )2] 5}ns(N 2 s)
N2
.
N 2 n
}N2
1
(6.3)184 Part 2: Probability
EXAMPLE
Hypergeometric Distribution
Twenty businesses located in a small community have filed tax returns for the
preceding year. Six of the returns were filled out incorrectly. An Internal Revenue
Service (IRS) auditor randomly selects four returns for further examination, and
the random variable is x 5 the number of examined returns that are incorrectly
filled out.
What Is the Expected Value of x?
For a sample size of n 5 4, with s 5 6 incorrect returns in the population, and
a population size of N 5 20, the expected number of incorrect returns in the
sample is
E(x) 5 }ns
N
5
4(6)
}
20
5 1.20 incorrect returns
What Is the Probability That There Will Be Exactly Two Incorrect Returns
Among the Four That Are Chosen for Further Examination?
The quantity to be determined is P(x 5 2), with N 5 20 businesses in the population, s 5 6 incorrect returns in the population, and n 5 4 returns randomly
selected from the population. Using the hypergeometric formula for P(x):
P(x) 5
1s x2 • 1Nn 22xs 2
}}
1N n 2
and P(2) 5
16 22 • 120 4 22 26 2
}}
120 4 2
5
16 22 • 114 2 2
}1
20
4 2
so
P(2) 5
6!
}
2!(6 2 2)!
•
14!
}}
2!(14 2 2)!
}}}
20!
}}
4!(20 2 4)!
5
6 • 5
}
2 • 1
•
14 • 13
}
2 • 1
}}
20 • 19 • 18 • 17
}}
4 • 3 • 2 • 1
5
1365
}
4845 5 0.2817
The probability P(x 5 2) is 0.2817 that exactly two of the four returns will
be incorrect. If we were to repeat this procedure for the other possible values of
x, we would find the following probabilities for x 5 0 through 4 incorrect returns
in the sample:
x 5 Number of
incorrect returns: 0 1 2 3 4
P(x): 0.2066 0.4508 0.2817 0.0578 0.0031
As the preceding example illustrates, the calculation of hypergeometric
probabilities is rather cumbersome. Unfortunately, with all the possible combinations of the determining characteristics, N, n, and s, tables of hypergeometric
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEChapter 6: Discrete Probability Distributions 185
Hypergeometric Probabilities
These procedures show how to get hypergeometric probabilities associated with x 5 the number of successes in n 5 4 trials from a population of N 5 20 when there are s 5 6 successes within the population.
The trials are without replacement. For each software package:
Procedure I: Individual or cumulative probabilities for x 5 2 successes.
Procedure II: Complete set of individual or cumulative probabilities for x 5 0 successes through
x 5 4 successes.
EXCEL
I. Excel procedure for obtaining an individual hypergeometric probability
1. Click on the cell into which Excel is to place the individual probability. From the Formulas ribbon, click on the
Insert Function symbol at the extreme left. Click Statistical in the Category box. Click HYPGEOMDIST in the
Function box. Click OK.
2. Enter the number of successes in the sample (2) into the Sample_s box. Enter the number of trials (4) into the
Number_sample box. Enter the number of successes in the population (6) into the Population_s box. Enter the
size of the population (20) into the Number_pop box. Click OK. The probability, P(x 5 2) 5 0.2817, will appear in
the cell selected in step 1.
[Alternatively, you can use Excel worksheet template tmhyper.]
II. Excel procedure for a complete set of individual or cumulative hypergeometric probabilities
1. Enter the column labels into A1:C1 as shown in the printout above. Enter the lowest possible number of successes
in the sample (in this case, 0) into A2. Enter 1 into A3. Click on cell A2 and highlight cells A2:A3. Click at the
lower right corner of A3 (the cursor will turn to a “1” sign) and drag downward so cells A2:A6 will contain the
values 0 through 4. (Note that the highest possible number of successes in the sample is 4.)
2. To get the set of individual probabilities: Click on cell B2 (this is where the first individual probability will go). From
the Formulas ribbon, click on the Insert Function symbol at the extreme left. Click Statistical in the Category
box. Click HYPGEOMDIST in the Function box. Click OK. Now follow step 2 in Excel procedure I, except enter A2
into the Sample_s (number of successes in the sample) box. The first individual probability now appears in B2.
3. Click to select cell B2. Click the lower right corner of B2 (the cursor turns to a “1” sign) and drag down until cells
B2:B6 contain the set of individual probabilities from P (x 5 0) through P (x 5 4).
4. To get the set of cumulative probabilities: Copy the value in cell B2 to cell C2 (this will be the first cumulative probability). Click on cell C3 and enter the equation 5C21B3. Click the lower right corner of cell C3 (the cursor turns to
a “1” sign) and drag down until cells C2:C6 contain the set of cumulative probabilities.
COMPUTER 6.2 SOLUTIONS
probabilities are not a realistic alternative. It is much easier to use the computer,
and Computer Solutions 6.2 describes how we can use Excel or Minitab to determine individual or cumulative hypergeometric probabilities.
(continued)186 Part 2: Probability
MINITAB
I. Minitab procedure for just one individual or cumulative hypergeometric probability
1. Click Calc. Select Probability Distributions. Click Hypergeometric.
2. In the Hypergeometric Distribution menu, enter 20 into the Population size (N) box. Enter 6 into the Event
count in population (M) box. Enter 4 into the Sample size (n) box. Select Input constant and enter the desired
x value (e.g., 2) into the box. Select either Probability or Cumulative probability, depending on whether you
want an individual probability or a cumulative probability. Click OK.
II. Minitab procedure for a complete set of individual or cumulative hypergeometric probabilities
Enter the desired x values (in this case, 0 through 4) into column C1. Follow steps 1 and 2 above, except select Input
column and enter C1 into the box.
6.33 Under what circumstances should the hypergeometric distribution be used instead of the binomial
distribution?
6.34 Using the hypergeometric distribution, with N 5 4,
n 5 2, and s 5 3, determine the following:
a. P(x 5 0) b. P(x 5 1) c. P(x 5 2)
6.35 Using the hypergeometric distribution, with N 5 5,
n 5 2, and s 5 3, determine the following:
a. P(x 5 0) b. P(x 5 1) c. P(x 5 2)
6.36 In a criminal trial, there are 25 persons who have
been approved by both parties for possible inclusion
in the eventual jury of 12. Of those who have been
approved, there are 14 women and 11 men. If the judge
forms the final jury of 12 by randomly selecting individuals from the approved listing, what is the probability
that at least half of the eventual jurors will be males?
6.37 A computer firm must send a group of three
specialists to give a technical presentation to a potential
customer. Five such individuals are anxious to attend,
but only two are considered to have strong interpersonal
skills. Since the specialists have equal levels of seniority,
they have agreed to place their names in a hat and allow
management to randomly select the three to attend the
presentation. What is the probability that there will be at
least one person with strong interpersonal skills among
the group that is sent?
6.38 Unknown to a rental car office, 3 of the 12
subcompact models now available for rental are subject to a safety recall soon to be announced by the
National Highway Traffic Safety Administration.
Five subcompacts will be rented today, and the cars
will be randomly selected from those available in the
pool. What is the probability that exactly one of the
recall-affected cars will be rented today? What is the
probability that all three of the recall-affected cars will
be rented today?
E X E R C I
S E SChapter 6: Discrete Probability Distributions 187
THE POISSON DISTRIBUTION
Description and Applications
The Poisson distribution is a discrete probability distribution that is applied to
events for which the probability of occurrence over a given span of time, space,
or distance is extremely small. The discrete random variable, x, is the number of
times the event occurs over the given span, and x can be 0, 1, 2, 3, and so on,
with (theoretically) no upper limit. Besides its ability to approximate the binomial distribution when n is large and  is small, the Poisson distribution tends to
describe phenomena like these:
• Customer arrivals at a service point during a given period of time, such as the
number of motorists approaching a tollbooth, the number of hungry persons
entering a McDonald’s restaurant, or the number of calls received by a company switchboard. In this context it is also useful in a management science
technique called queuing (waiting-line) theory.
• Defects in manufactured materials, such as the number of flaws in wire or
pipe products over a given number of feet, or the number of knots in wooden
panels for a given area
• The number of work-related deaths, accidents, or injuries over a given number of production hours
• The number of births, deaths, marriages, divorces, suicides, and homicides
over a given period of time
Although it is closely related to the binomial distribution, the Poisson distribution has a number of characteristics that make it unique. Like the binomial
distribution, it is a family of distributions, but its shape is determined by just one
descriptor, its mean. In the Poisson distribution, the mean is called  (the Greek
lowercase letter lambda) instead of . Also, the mean of the distribution is numerically equal to its variance. The distribution can be summarized as follows:
(6.4)
The Poisson distribution:
The probability that an event will occur exactly x times over a given span of
time, space, or distance is
P(x) 5 }x · e2
x!
where  5 the mean, or E(x); the expected number
of occurrences over the given span
e 5 the mathematical constant, 2.71828
(e is the base of the natural logarithm
system)
(Note: In the Poisson distribution, the mean and the variance are equal.)
1. When using a pocket calculator to make the computation, remember that
2.718282l is the same as 1 divided by 2.71828l.
2. Appendix A also contains tables for individual and cumulative Poisson probabilities. The application of these tables will be discussed after we use an
example in demonstrating how the Poisson distribution works.
NOTES188 Part 2: Probability
EXAMPLE
Poisson Distribution
In an urban county, health care officials anticipate that the number of births this
year will be the same as last year, when 438 children were born—an average of
438y365, or 1.2 births per day. Daily births have been distributed according to
the Poisson distribution.
SOLUTION
What Is the Mean of the Distribution?
E(x) 5  can be expressed in a number of ways, because it reflects here the
number of occurrences over a span of time. Accordingly, the “span of time”
can be months ( 5 438y12 5 36.5 births per month), weeks ( 5 438y52 5
8.42 births per week), or days ( 5 438y365 5 1.2 births per day). For purposes of our example, we will be using  5 1.2 births per day in describing
the distribution.
For Any Given Day, What Is the Probability That No Children Will Be Born?
Using the Poisson formula, with  5 1.2 births per day, to find P(x 5 0):
P(x) 5 }x ? e2l
x!
and P(0) 5 1.2 }0 ? e21.2
0!
5
1 ? 0.30119
}}
1
5 0.3012
Calculate Each of the Following Probabilities: P(x 5 1), P(x 5 2), P(x 5 3),
P(x 5 4), and P(x 5 5)
Using the same approach as in the preceding calculations, with  5 1.2 mean
births per day and values of x from 1 to 5 births per day:
P(x 5 1 birth) 5 }} 1.21 • e21.2
1!
5
(1.2000)(0.30119)
}}
1
5 0.3614
P(x 5 2 births) 5 }} 1.22 • e21.2
2!
5
(1.4400)(0.30119)
}}
2 • 1
5 0.2169
P(x 5 3 births) 5 }} 1.23 • e21.2
3!
5
(1.7280)(0.30119)
}}
3 • 2 • 1
5 0.0867
P(x 5 4 births) 5 }} 1.24 • e21.2
4!
5
(2.0736)(0.30119)
}}
4 • 3 • 2 • 1
5 0.0260
P(x 5 5 births) 5 }} 1.25 • e21.2
5!
5
(2.4883)(0.30119)
}}
5 • 4 • 3 • 2 • 1
5 0.0062
If we were to continue, we’d find that P(x 5 6) 5 0.0012 and P(x 5 7) 5
0.0002, with both rounded to four decimal places. Including these probabilities
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAChapter 6: Discrete Probability Distributions 189
with those just calculated, the result would be the following Poisson probability
distribution. As Figure 6.4 shows, the distribution is positively skewed.
x 0 1 2 3 4 5 6 7
P(x) 0.3012 0.3614 0.2169 0.0867 0.0260 0.0062 0.0012 0.0002
What Is the Probability That No More Than One Birth Will Occur
on a Given Day?
Since the events are mutually exclusive, this can be calculated as P(x 5 0) 1
P(x 5 1), or 0.3012 1 0.3614, and P(x # 1) 5 0.6626.
Using the Poisson Tables
The Poisson calculations are much easier than those for the binomial. For example,
you need compute e2l only one time for a given value of l, since the same term
is used in finding the probability for each value of x. Using either the individual
or the cumulative Poisson distribution table in Appendix A (Tables A.3 and A.4,
respectively) makes the job even easier. Table 6.3 shows a portion of Table A.3,
while Table 6.4 shows its cumulative counterpart. If the problem on which you
are working happens to include one of the  values listed, you need only refer to
the P(x 5 k) or the P(x # k) values in the appropriate table.
In the previous example,  was 1.2 births per day. Referring to the  5 1.2
column of Table 6.3, we see the same probabilities just calculated for P(x 5 1)
through P(x 5 7). For example, in the fifth row of the  5 1.2 column, the entry
for P(x 5 4) is 0.0260. To quickly find the probability that there would be no
more than 2 births on a given day, we can refer to the cumulative Poisson probabilities in Table 6.4 and see that P(x # 2) is 0.8795.
In some cases, the sum of the individual probabilities in a particular Poisson
table will not add up to 1.0000. This is due to (1) rounding and (2) the fact that
there is no upper limit to the possible values of x. Some values not listed in the
table have P(x) . 0, but would appear as 0.0000 because of their extremely small
chance of occurrence.
0 1 2
x = Number of births during a given day
3 4 5 6 7
0.00
0.10
0.20
0.30
0.40
P(x)
FIGURE 6.4
For the example discussed
in the text, the Poisson
probability distribution for the
number of births per day is
skewed to the right. The mean
of the distribution is 1.2 births
per day, and this descriptor is
all that is needed to determine
the Poisson probability for each
value of x.
PLEEXAMPLEEXAMPLE190 Part 2: Probability
Poisson Distribution, Individual Probabilities
for x 5 number of occurrences, prob (x 5 k)
 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
0 0.3329 0.3012 0.2725 0.2466 0.2231 0.2019 0.1827 0.1653 0.1496 0.1353
1 0.3662 0.3614 0.3543 0.3452 0.3347 0.3230 0.3106 0.2975 0.2842 0.2707
2 0.2014 0.2169 0.2303 0.2417 0.2510 0.2584 0.2640 0.2678 0.2700 0.2707
3 0.0738 0.0867 0.0998 0.1128 0.1255 0.1378 0.1496 0.1607 0.1710 0.1804
4 0.0203 0.0260 0.0324 0.0395 0.0471 0.0551 0.0636 0.0723 0.0812 0.0902
k 5 0.0045 0.0062 0.0084 0.0111 0.0141 0.0176 0.0216 0.0260 0.0309 0.0361
6 0.0008 0.0012 0.0018 0.0026 0.0035 0.0047 0.0061 0.0078 0.0098 0.0120
7 0.0001 0.0002 0.0003 0.0005 0.0008 0.0011 0.0015 0.0020 0.0027 0.0034
8 0.0000 0.0000 0.0001 0.0001 0.0001 0.0002 0.0003 0.0005 0.0006 0.0009
9 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0002
10 0.0000 0.0000 0.0000 0.0000
TABLE 6.3
A portion of the table of
individual Poisson probabilities
from Appendix A.
For  5 1.2 births per day, the probability that
there will be exactly 3 births on a given day is
P(x 5 3), or 0.0867.
Poisson Distribution, Cumulative Probabilities
for x 5 number of occurrences, prob (x  k)
 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
0 0.3329 0.3012 0.2725 0.2466 0.2231 0.2019 0.1827 0.1653 0.1496 0.1353
1 0.6990 0.6626 0.6268 0.5918 0.5578 0.5249 0.4932 0.4628 0.4337 0.4060
2 0.9004 0.8795 0.8571 0.8335 0.8088 0.7834 0.7572 0.7306 0.7037 0.6767
3 0.9743 0.9662 0.9569 0.9463 0.9344 0.9212 0.9068 0.8913 0.8747 0.8571
4 0.9946 0.9923 0.9893 0.9857 0.9814 0.9763 0.9704 0.9636 0.9559 0.9473
k 5 0.9990 0.9985 0.9978 0.9968 0.9955 0.9940 0.9920 0.9896 0.9868 0.9834
6 0.9999 0.9997 0.9996 0.9994 0.9991 0.9987 0.9981 0.9974 0.9966 0.9955
7 1.0000 1.0000 0.9999 0.9999 0.9998 0.9997 0.9996 0.9994 0.9992 0.9989
8 1.0000 1.0000 1.0000 1.0000 0.9999 0.9999 0.9998 0.9998
9 1.0000 1.0000 1.0000 1.0000
TABLE 6.4
A portion of the table
of cumulative Poisson
probabilities from
Appendix A.
For  5 1.2 births per day, the probability that
there will be no more than 3 births on a given
day is P(x # 3), or 0.9662.Chapter 6: Discrete Probability Distributions 191
As with the binomial distribution, Excel and Minitab can provide either
individual or cumulative Poisson probabilities. The procedures and the results
are shown in Computer Solutions 6.3.
The Poisson Approximation to the Binomial
Distribution
When n is relatively large and  (the probability of a success in a given trial) is
small, the binomial distribution can be closely approximated by the Poisson
distribution. As a rule of thumb, the binomial distribution can be satisfactorily
approximated by the Poisson whenever n $ 20 and  # 0.05. Under these conditions, we can just use  5 n and find the probability of each value of x using
the Poisson distribution.
Poisson Probabilities
These procedures show how to get probabilities associated with a Poisson distribution with mean 5 1.2.
For each software package:
Procedure I: Individual or cumulative probabilities for x 5 2 occurrences.
Procedure II: Complete set of individual or cumulative probabilities for x 5 0 occurrences
through x 5 8 occurrences.
EXCEL
I. Excel procedure for just one individual or cumulative Poisson probability
1. Click on the cell into which Excel will place the probability. From the Formulas ribbon, click on the Insert Function
symbol at the extreme left. Click Statistical in the Category box. Click POISSON in the Function box. Click OK.
2. Enter the number of occurrences (2) into the X box. Enter the mean of the distribution (1.2) into the Mean box. To
obtain the individual probability, P(x 5 2), enter false into the Cumulative box. To obtain the cumulative probability, P(x # 2), enter true into the Cumulative box. Click OK.
COMPUTER 6.3 SOLUTIONS
(continued)192 Part 2: Probability
II. Excel procedure for a complete set of individual or cumulative Poisson probabilities
1. Enter the column labels into A1:C1 as shown in the printout. Enter the lowest number of occurrences in the set (in
this case, 0) into A2. Enter 1 into A3. Click on cell A2 and highlight cells A2:A3. Click at the lower right corner of
A3 (the cursor will turn to a “1” sign) and drag downward so cells A2:A10 will contain the values 0 through 8.
2. To get the set of individual probabilities: Click on cell B2 (this is where the first individual probability will go). Follow
the remainder of the steps in Excel procedure I, except enter A2 into the X (number of occurrences) box and enter
false into the Cumulative box. The first probability now appears in B2. Click to select cell B2. Click the lower
right corner of B2 (the cursor turns to a “1” sign) and drag down until cells B2:B10 contain the set of individual
probabilities.
3. To get the set of cumulative probabilities: Click on cell C2 (this is where the first cumulative probability will go).
Follow the remainder of the steps in Excel procedure I, except enter A2 into the X (number of occurrences) box and
enter true into the Cumulative box. The first probability now appears in C2. Click to select cell C2. Click the lower
right corner of C2 (the cursor turns to a “1” sign) and drag down until cells C2:C10 contain the set of cumulative
probabilities.
MINITAB
I. Minitab procedure for just one individual or cumulative Poisson probability
1. Click Calc. Select Probability Distributions. Click Poisson.
2. In the Poisson Distribution menu, enter 1.2 into the Mean box. Select Input constant and enter the desired
x value (e.g., 2) into the box. Select either Probability or Cumulative probability, depending on whether you
want an individual probability or a cumulative probability. Click OK.
II. Minitab procedure for a complete set of individual or cumulative binomial probabilities
Enter the desired x values (in this case, 0 through 8) into column C1. Follow steps 1 and 2 above, except select Input
column and enter C1 into the box.
EXAMPLE
Poisson Approximation to Binomial
Past experience has shown that 1.0% of the microchips produced by a certain
firm are defective. A sample of 30 microchips is randomly selected from the firm’s
production. If x 5 the number of defective microchips in the sample, determine
P(x 5 0), P(x 5 1), P(x 5 2), P(x 5 3), and P(x 5 4).
EXAMPLEEXAMPLEEChapter 6: Discrete Probability Distributions 193
SOLUTION
The Binomial Approach
The descriptors for the binomial distribution are  5 0.01, the chance that any
given microchip will be defective, and n 5 30, the number of microchips in the
sample. Although we will skip the actual binomial calculations, we would use
 5 0.01 and n 5 30 in arriving at the probabilities shown below.
The Poisson Approximation Approach
The only descriptor for the Poisson distribution is  5 n 5 30(0.01), or  5 0.3
microchips expected to be defective in the sample. Referring to the Poisson
table for  5 0.3, we fnd the following probabilities for x 5 0 through x 5 4.
Note the very close similarity between each binomial probability and its Poisson
approximation:
P(x), Using the
x 5 Number of
Defective Microchips Binomial Poisson
in the Sample Distribution Approximation
0 0.7397 0.7408
1 0.2242 0.2222
2 0.0328 0.0333
3 0.0031 0.0033
4 0.0002 0.0003
6.39 For a discrete random variable that is Poisson
distributed with l 5 2.0, determine the following:
a. P(x 5 0) b. P(x 5 1) c. P(x # 3)
d. P(x $ 2)
6.40 For a discrete random variable that is Poisson distributed with l 5 9.6, determine the following:
a. P(x 5 7) b. P(x 5 9) c. P(x # 12)
d. P(x $ 10)
6.41 In 2006, there were about 490 motor vehicle thefts
for every 100,000 registrations. Assuming (1) a Poisson
distribution, (2) a community with a comparable theft
rate and 1000 registered motor vehicles, and (3) x 5 the
number of vehicles stolen during the year in that community, determine the following: Source: Statistical Abstract of the
United States 2009, pp. 188, 672.
a. E(x) b. P(x 5 3) c. P(x 5 5)
d. P(x # 8) e. P(3 # x # 10)
6.42 Arrivals at a walk-in optometry department in
a shopping mall have been found to be Poisson
distributed with a mean of 2.5 potential customers
arriving per hour. If x 5 number of arrivals during a
given hour, determine the following:
a. E(x) b. P(x 5 1) c. P(x 5 3)
d. P(x # 5) e. P(2 # x # 6)
6.43 The U.S. divorce rate has been reported as 3.6
divorces per 1000 population. Assuming that this rate
applies to a small community of just 500 people and
is Poisson distributed, and that x 5 the number of
divorces in this community during the coming year,
determine the following: Source: New York Times Almanac
2009, p. 294.
a. E(x) b. P(x 5 1) c. P(x 5 4)
d. P(x # 6) e. P(2 # x # 5)
6.44 During the 12 p.m.–1 p.m. noon hour, arrivals at a
curbside banking machine have been found to be Poisson
distributed with a mean of 1.3 persons per minute. If
x 5 number of arrivals during a given minute, determine
the following:
a. E(x) b. P(x 5 0) c. P(x 5 1)
d. P(x # 2) e. P(1 # x # 3)
6.45 During the winter heating season in a northern state,
Howard’s Heating and Cooling Service receives an
average of 3.1 emergency service calls per day from heating customers in their area. Howard has increased personnel and equipment resources to better handle such calls,
and his company is now able to satisfy a maximum of 8
emergency service calls per day. During a typical day during the heating season, and assuming the daily number of
E X E R C I
S E S
MPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE194 Part 2: Probability
emergency service calls to be Poisson distributed, what is
the probability that Howard will receive more emergency
service calls tomorrow than his company is able to handle? Considering this probability, would it seem advisable
for Howard to increase or to decrease the level of personnel and equipment he devotes to handling such calls?
6.46 Over the past year, a university’s computer system
has been struck by a virus at an average rate of 0.4
viruses per week. The university’s information technology managers estimate that each time a virus occurs, it
costs the university $1000 to remove the virus and repair
the damages it has caused. Assuming a Poisson distribution, what is the probability that the university will have
the good fortune of being virus-free during the upcoming week? During this same week, what is the expected
amount of money that the university will have to spend
for virus removal and repair?
6.47 According to the Mortgage Bankers Association
of America, the foreclosure rate on home mortgages in
2007 was 2%. Assuming that this rate is applicable to a
community where 500 homes have mortgages, use
the Poisson approximation to the binomial distribution
to determine the following for x 5 the number of
foreclosures in this community during the coming year:
Source: Statistical Abstract of the United States 2009, p. 727.
a. E(x) b. P(x 5 7) c. P(x 5 10)
d. P(x # 12) e. P(8 # x # 15)
6.48 The Federal Aviation Administration reports that
American Airlines got 6.06 complaints of mishandled
baggage per 1000 passengers in 2008. Assuming this
rate applies to the next 500 passengers who depart from
Portland, Oregon, on American Airlines flights, what
is the probability that at least 5 people will have their
baggage mishandled? That no more than 3 people will
have their baggage mishandled? Source: The New York Times
Almanac 2009, p. 426.
6.49 Taxicab drivers are more likely to be murdered
on the job than members of any other occupation,
even police officers and security guards. The annual
murder rate for cab drivers is 30.0 homicides per
100,000 workers, compared to just 0.6 per 100,000 for
all occupations. Assume that taxicab companies in a
large Northeastern city employ a total of 3000 drivers
and that the family of any driver murdered on the job
receives $50,000 from the city. What is the probability
that at least two of the city’s taxicab drivers will be
murdered on the job during the next year? What is the
approximate expected value for the amount of money
the city will be providing to families of murdered
taxicab drivers during the next year? Source: “Occupational
Homicide Rate Is Highest Among Taxi Drivers,” USA Today, May 2, 2000,
p. 3A.
6.50 A manufacturing plant’s main production
line breaks down an average of 2.4 times per day.
Whenever the line goes down, it costs the company
$500 in maintenance, repairs, and lost production.
What is the probability that the production line will
break down at least 3 times tomorrow? What is the
approximate expected value for the amount of money
that production line breakdowns will cost the company
each day?
6.51 The company in Exercise 6.50 has disciplined a
worker who was suspected of pilfering tools and supplies
from the plant. The very next day, the production line
broke down 9 times. Management has confronted the
union with accusations of sabotage on the line, but the
union president says it’s just a coincidence that the production line happened to break down so many times the
day after the worker was disciplined. Using probabilities
appropriate to your discussion, comment on how much
of a coincidence this high number of breakdowns would
appear to be.
SIMULATING OBSERVATIONS FROM
A DISCRETE PROBABILITY DISTRIBUTION
Besides describing the exact probability distribution for a discrete random variable,
statistical software can also randomly select observations from the distribution itself. For example, we can use Excel or Minitab to randomly select observations
from a given Poisson distribution, then display them in the form of a frequency
distribution. Computer Solutions 6.4 describes the necessary procedures. Because
we are using two different software packages, there will be two different sets of
randomly selected data. We will display and discuss the results for Minitab.
Table 6.5 shows the individual P(x) probabilities as computed by Minitab for
three different Poisson distributions (l 5 0.1, l 5 1.2, and l 5 5.0). Next to each
(6.5)Chapter 6: Discrete Probability Distributions 195
theoretical probability is the relative frequency with which the x value occurred
when Minitab randomly selected 10,000 observations from that distribution.
Note that the relative frequencies for the randomly selected x values are very
close, but not necessarily the same as the theoretical probabilities listed in the
probability distribution. As we discussed in the previous chapter, P(heads) 5 0.5
for a coin flip, but flipping a coin 10 times doesn’t mean that you’re certain to
get heads exactly 5 times. However, due to the law of large numbers, observed
relative frequencies will more closely approach the theoretical probability distribution as the number of observations grows larger.
Parts A through C of Table 6.6 show Minitab frequency distributions for
10,000 observations randomly selected from the corresponding probability
distribution of Table 6.5. In comparing the three frequency distributions of
Table 6.6, notice how they become more symmetrical as  is increased when
we proceed from A to C. Although the Poisson distribution is always positively
skewed, it becomes more symmetrical as its mean, , becomes larger.
The ability to select random observations from a known or assumed probability distribution also has applications in a number of management science
areas. Such applications typically involve using one or more such distributions
to mathematically represent real-world phenomena of interest. In general, their
purpose is to help the management scientist gain insight into how the processes
Simulating Observations from a Discrete Probability Distribution
These procedures simulate 10,000 observations from a Poisson distribution with mean 5 1.2. The procedures
for simulating observations from a binomial distribution will differ only slightly from those described here.
EXCEL
1. Start with a blank worksheet. From the Data ribbon, click Data Analysis in the rightmost menu section. Click
Random Number Generation. Click OK.
2. Enter 1 into the Number of Variables box. Enter 10000 into the Number of Random Numbers box. Within the
Distribution box, select Poisson. Enter the mean (1.2) into the Lambda box. Select Output Range and enter A1
into the box. Click OK. The 10,000 simulated observations will be located in A1:A10000.
MINITAB
1. Click Calc. Select Random Data. Click Poisson. Enter 10000 into the Number of rows of data to generate box.
Enter C1 (the destination column) into the Store in columns box. Enter 1.2 into the Mean box. Click OK. The
10,000 simulated observations will be located in C1.
2. To obtain the frequencies for the different values of x, click Stat, select Tables, and click Tally Individual
Variables. Enter C1 into the Variables box and select Counts. Click OK. The relative frequencies shown in part B
of the right side of Table 6.5 are obtained by dividing each of the frequencies by 10,000.
The frequency distributions in Table 6.6 can be generated using the Minitab procedure described previously in Computer
Solutions 2.1. As shown in this earlier procedure, we have the option of displaying the frequencies on the chart istelf.
COMPUTER 6.4 SOLUTIONS196 Part 2: Probability
TABLE 6.5
Besides describing a
probability distribution, the
computer can also randomly
select observations from
the distribution itself.
Minitab output. Poisson probability
distribution for the discrete random
variable, x:
When Minitab randomly
selected 10,000
observations from the
distribution, the relative
frequency with which each x
Probability Density Function value appeared:
Poisson with mean 5 0.1
x P( X 5 x )
0 0.904837
1 0.090484
2 0.004524
3 0.000151
4 0.000004
Probability Density Function
Poisson with mean 5 1.2
x P( X 5 x )
0 0.301194
1 0.361433
2 0.216860
3 0.086744
4 0.026023
5 0.006246
6 0.001249
7 0.000214
8 0.000032
0.9056
0.0892
0.0050
0.0002
0.0000
0.3060
0.3588
0.2085
0.0918
0.0273
0.0063
0.0011
0.0002
0.0000
0.0067
0.0343
0.0848
0.1424
0.1824
0.1693
0.1433
0.1059
0.0637
0.0345
0.0195
0.0086
0.0029
0.0010
0.0005
0.0001
0.0001
Probability Density Function
Poisson with mean 5 5
x P( X 5 x )
0 0.006738
1 0.033690
2 0.084224
3 0.140374
4 0.175467
5 0.175467
6 0.146223
7 0.104445
8 0.065278
9 0.036266
10 0.018133
11 0.008242
12 0.003434
13 0.001321
14 0.000472
15 0.000157
16 0.000049
A.
B.
C.Chapter 6: Discrete Probability Distributions 197
TABLE 6.6
Minitab frequency distributions
for 10,000 observations
randomly taken from each of
the three Poisson probability
distributions in Table 6.5.
A. 10,000 random observations from Poisson distribution with mean  5 0.1
B. 10,000 random observations from Poisson distribution with mean  5 1.2
(continued)198 Part 2: Probability
C. 10,000 random observations from Poisson distribution with mean  5 5.0
interact, as well as to determine how these interactions might be influenced by
managerial decisions. Most introductory management science texts include extensive treatment of the nature, the construction, and the applications of such
mathematical models of the real world.
Note: These exercises require a computer and statistical
software capable of selecting random observations from
discrete probability distributions.
6.52 Using your statistical software package:
a. Simulate 100 observations from a Poisson distribution having  5 2.35.
b. Generate the individual Poisson probabilities for a
Poisson distribution having  5 2.35.
c. Repeat part (a).
d. Repeat part (b).
e. When part (a) is repeated, the results are now
different. However, when part (b) is repeated, the
results are the same. What does this say about frequency distributions versus probability distributions?
6.53 Using your statistical software package:
a. Simulate 50 observations from a binomial distribution for which n 5 5 and  5 0.2.
b. Generate a frequency distribution for these observations. Is the distribution symmetrical? If not, is it
positively skewed or negatively skewed?
c. Repeat parts (a) and (b), but simulate 500 observations. Is the frequency distribution more symmetrical
or less symmetrical than the one previously obtained?
d. Comment on how the shape of the distribution
would change if you were to continue increasing the
number of simulated observations.
E X E R C I
S E SChapter 6: Discrete Probability Distributions 199
SUMMARY
• Discrete random variable and its probability distribution
A discrete random variable can take on only certain values along a range or continuum, and there are gaps between these possible values. On the other hand, a
continuous random variable can have any value along a range or continuum. This
chapter has focused on discrete random variables, while the next concentrates on
continuous random variables.
The probability distribution for a discrete random variable lists all possible
values the variable can assume, along with their respective probabilities of occurrence. There are several types of discrete distributions, each with its own characteristics and applications.
• Binomial distribution and the Bernoulli process
The binomial distribution deals with consecutive trials, each of which has only
two possible outcomes, generally referred to as success or failure. This distribution assumes a Bernoulli process, in which the trials are independent of each
other, and the probability of success remains constant from one trial to the next.
• Hypergeometric distribution
Like the binomial distribution, the hypergeometric distribution deals with consecutive trials, and there are only two possible outcomes for each trial. The trials
are not statistically independent, however, and the probability of a success is not
constant from one trial to the next.
• Poisson distribution
The Poisson distribution applies to events for which the probability of occurrence
over a given span of time, space, or distance is extremely small. In this distribution, the random variable is the number of occurrences in the specified span of
time, space, or distance. For a Bernoulli process in which the number of trials is
large and the probability of success on any given trial is very small, the Poisson
distribution can be used as an approximation to the binomial distribution.
• Determining probabilities for a discrete random variable
Appendix A contains tables for the binomial and Poisson distributions. However, computer statistical packages enable the user to determine the probabilities
associated with a given distribution without the need to refer to such tables.
• Simulation of observations from a discrete probability distribution
Some computer packages can randomly select a sample of observations from
a probability distribution that has already been specified. In addition to being
useful in more advanced management science applications such as simulation
models intended to represent real-world phenomena, this capability enables the
user to easily view the approximate shape of a given probability distribution.
Based on the law of large numbers, the observed relative frequencies will more
closely approach the theoretical probability distribution as the number of observations grows larger.
(6.6)200 Part 2: Probability
Characteristics of a Discrete Probability Distribution
• For any value of x, 0 # P(x) # 1.
• The values of x are exhaustive: The probability distribution includes all possible
values.
• The values of x are mutually exclusive: Only one value can occur for a given
experiment.
• The sum of their probabilities is 1, or oP(xi) 5 1.0.
General Formulas: Mean and Variance of a Discrete Probability Distribution
• Mean:
 5 E(x) or  5 oxiP(xi) for all possible values of x
• Variance:
 2 5 E[(x 2 )2] or 2 5 o(xi 2 )2P(xi) for all possible values of x,
and the standard deviation
is  5 Ï
___
2
The Binomial Probability Distribution
The probability of exactly x successes in n trials is
P(x) 5
n!
}x
!(n 2 x)! x(1 2 )n2x where n 5 number of trials
x 5 number of successes
 5 the probability of success in
any given trial
(1 2 ) 5 the probability of failure in
any given trial
• Mean:
 5 E(x) 5 n
• Variance:
2 5 E[(x 2 )2] 5 n(1 2 )
The Hypergeometric Distribution
The probability of exactly x successes in n trials that are not independent is
P(x) 5
1 s x2 ? 1nN22xs2
}}
1 N n 2 where Nn 5 5 size of the population size of the sample
s 5 number of successes in the population
x 5 number of successes in the sample
• Mean:
 5 E(x) 5 }ns
N
E Q U A T I O N SChapter 6: Discrete Probability Distributions 201
• Variance:
2 5 E[(x 2 )2] 5 ns }(N 2 s)
N2
∙
N 2 n
}N2
1
The Poisson Probability Distribution
The probability that an event will occur exactly x times over a given span of time, space,
or distance is
P(x) 5 l}x • e2l
x!
where l 5 the mean, or E(x); the expected number of
occurrences over the given span
e 5 the mathematical constant, 2.71828 (e is the
base of the natural logarithm system)
(Note: In the Poisson distribution, the mean and the variance are equal.)
6.54 The owner of a charter fishing boat has found that
12% of his passengers become seasick during a halfday fishing trip. He has only two beds below deck to
accommodate those who become ill. About to embark
on a typical half-day trip, he has 6 passengers on
board. What is the probability that there will be
enough room below deck to accommodate those who
become ill?
6.55 During 2007, the crash rate for commuter air carriers was 1.0 per 100,000 flying hours. Assuming this
rate continues, what is the probability that there will be
no more than one crash in the next 50,000 flying hours?
Source: Statistical Abstract of the United States 2009, p. 662.
6.56 Bill has to sell three more cars this month in order
to meet his quota. Tonight he has after-dinner appointments with five prospective customers, each of whom
happens to be interested in a different car. If he has a
30% chance of success with each customer, what is the
probability that he will meet his quota by tomorrow
morning?
6.57 In Exercise 6.56, suppose that three of the customers are interested in the same car, and that they will go
elsewhere if it has already been sold. Would it be appropriate to use the binomial distribution under these conditions? Why or why not?
6.58 During the 2009 tax filing season, 15.8% of all
individual U.S. tax returns were prepared by H&R
Block. For a random selection of 3 tax returns, describe
the probability distribution for x 5 the number in the
sample whose returns were prepared by H&R Block.
Source: H&R Block, Inc., Fast Facts 2009.
6.59 Twenty percent of the population of Maryland
describe themselves as binge drinkers. For a randomly
selected group of 5 Maryland residents, use the appropriate statistical table to describe the probability distribution for x 5 the number in the sample who describe
themselves as binge drinkers. Source: Statistical Abstract of the
United States 2009, p. 130.
6.60 A company has installed five intrusion-detection
devices in an office building. The devices are very
sensitive and, on any given night, each one has a 0.1
probability of mistakenly sounding an alarm when no
intruder is present. If two or more of the devices are
activated on the same night, the intrusion system
automatically sends a signal to the local police. If no
intruder is present on a given night, what is the
probability that the system will make the mistake of
calling the police?
6.61 A mining company finds that daily lost-work injuries average 1.2. If the local union contract has a clause
requiring that the mine be shut down as soon as three
workers incur lost-work injuries, on what percentage
of the days will the mine be operational throughout the
day?
6.62 Unknown to a quality-control inspector, 20% of
a very large shipment of electric switches are defective.
The inspector has been told to reject the shipment if, in
a sample of 15 switches, 2 or more are defective. What is
the probability that the shipment will be rejected?
6.63 Repeat Exercise 6.29, but under the assumption
that there are only 20 bearings in the supply bin, 2 of
which are defective.
CHAPTER EXERCISES202 Part 2: Probability
6.64 A town council includes 10 Democrats and 15
Republicans. In response to a developer’s request for a
variance that would allow her to build a convenience
store in an area of town currently zoned as residential,
the mayor is going to randomly select 5 council members for a committee to examine the issue. What is the
probability that there will be at least 3 Republicans on
the committee?
6.65 Fred and Emily are among the 50 finalists in a lottery in which 5 winners will be randomly selected to
receive a free vacation in Hawaii. Using the hypergeometric distribution, determine the probability that exactly
one of these two individuals will win a free vacation.
6.66 Seven of the 15 campus police officers available for
assignment to the auditorium in which a local politician
is to speak have received advanced training in crowd
control. If 5 officers are randomly selected for service
during the speech, what is the probability that exactly
2 of them will have had advanced training in crowd
control? What is the probability that at least 3 of them
will have had advanced training?
6.67 A producer of copper tubing has found that, on the
average, five flaws occur per 2000 feet of tubing. The
tubing is sold in continuous rolls of 200 feet each. What
is the probability that a randomly selected 200-foot roll
will have no flaws?
6.68 A large plumbing contractor buys eight 200-foot
rolls from the producer described in Exercise 6.67.
Determine the probability that no more than three of the
eight rolls will be flawless.
6.69 In 2004, 50% of all U.S. families had a net worth
less than $93,000. What is the probability that exactly
three of a randomly selected sample of five families would
have a net worth less than $93,000? Source: Statistical
Abstract of the United States 2009, p. 459.
6.70 A trucking company has found that its trucks average 0.2 breakdowns during round trips from New York
to Los Angeles.
a. What is the probability that a single truck will make
the complete trip without experiencing a breakdown?
b. If 3 trucks are assigned to a NY/LA round trip, what
is the probability that at least 2 of them will make
the complete trip without experiencing a breakdown?
6.71 The Bureau of Labor Statistics says that baggage
handlers, trash haulers, and coal miners are among the
occupations having the highest risk of a debilitating
injury or illness. For coal miners working over the course
of a year, the injury rate has been reported as 3.6 injuries
per 100 coal miners. For a company that employs 200
coal miners, what is the probability that at least 10 of
its workers will experience a debilitating injury or illness over the next year? Source: “Workers at Risk,” USA Today,
March 15, 2006, p. 6D.
6.72 According to the U.S. Centers for Disease Control
and Prevention, Americans 85 years of age and older
experience flu-associated respiratory and circulatory
illness at an annual rate of approximately 1200 cases
per 100,000 persons. If a rural county has 500 seniors
in this age group, what is the probability that no more
than 5 of them will experience such an illness during
the coming year? If, in such a county, 15 seniors in this
age group were to come down with such an illness
during the coming year, would this be considered rather
unusual? Explain your conclusion. Source: Anita Manning,
“Tens of Thousands More People Are Hospitalized for Flu Than
Expected,” USA Today, September 15, 2004, p. 5D.
6.73 Although they have no commercials, public
television stations periodically conduct campaigns during which programming is occasionally interrupted by
appeals for subscriber donations. A local public
television station has found that, during these breaks,
the number of calls averages 3 per minute. It has
recently upgraded the switchboard so it can handle as
many as 5 calls per minute. The station is now in the
middle of a program break. What is the probability
that the switchboard will be overloaded during the next
minute?
6.74 A gorilla once made headlines in 400 newspapers
and appeared on three national television shows after
correctly “selecting” the winner of 13 out of 20 football
games. He had made his “choice” for each game by
grabbing one of two pieces of paper on which the names
of the teams were written. If a great many gorillas had
the same opportunity, what percentage of them would
be at least this lucky? Source: “Gone Ape,” Pittsburgh Press,
June 8, 1986.
6.75 It has been estimated that over 90% of the world’s
large corporations are actively involved in data warehousing. Using the 90% figure, and considering a random sample of 10 large corporations, what is the probability that at least 7 of them are actively involved in
data warehousing?
6.76 A country club has found that the number of complete rounds played by a member during a given outing
averages 1.3 rounds. What is the probability that a golfer
selected at random will complete at least 2 rounds of golf
on his or her next outing?
6.77 A computer manufacturer that sells notebook
computers directly to customers has been having
problems with one of its models—it seems that about
10% of the units are inoperable upon arrival to the
customer. They are functional when tested at the factory,
but there is some problem associated with the shipping/
handling process that the company’s engineers have not
yet been able to identify and correct. A highly valued
customer places an emergency order for 8 of these
notebooks, and wants them delivered overnight andChapter 6: Discrete Probability Distributions 203
ready for use in the field by tomorrow. Wanting to be
honest with this customer, and wishing to help ensure
that the customer receives at least 8 working computers, the manufacturer offers to send 9 units instead of
the 8 that the customer needs. The customer accepts this
gesture of goodwill and 9 units are sent. What is the
probability that the customer will end up with at least 8
working computers?
6.78 It has been estimated that 5% of people who are
hospitalized pick up infections while in the hospital, and
that 2.2 million people a year get sick and 88,000 die
because of these infections. Experts advise patients to be
sure that health care personnel have washed their hands,
to select surgeons with higher levels of experience, and
to minimize the amount of time that they are treated
with catheterization. Grove Tree Mountain Hospital
anticipates admitting 100 patients this month for surgical and other treatments. Suppose that 14 of the patients
in the sample actually end up getting a hospital-borne
infection. Treating the 100 patients as a random sample
of all hospital patients, and applying the infection rate
cited above, what is the probability that a typical hospital would have 14 or more infections in a sample of this
size? Considering this probability, comment on whether
Grove Tree Mountain might not be a typical hospital in
terms of its patients’ likelihood of picking up an infection. Source: Anita Manning, “Hospital Infections Jump as Outpatient
Care Rises,” USA Today, March 6, 2000, p. 8D.
6.79 A tire manufacturer has advertised that only 2%
of its tires have surface blemishes. Further, the manufacturer argues that tires with such blemishes are perfectly
safe, and with only the physical appearance suffering
very slightly. A consumer magazine randomly purchases
40 of the tires for an upcoming tire performance comparison report. Upon examining the tires, the magazine’s
test engineers find that 5 of them have surface blemishes.
If it were true that just 2% of the tires produced have
surface blemishes, what would be the probability of getting 5 or more with such blemishes in a random sample
of 40 tires? Considering this probability, comment on the
believability of the tire manufacturer’s advertising claim.
6.80 J.D. Power and Associates’ Initial Quality Study
reports that the industry-average problem rate for
vehicles is 108 problems per 100 vehicles. The highestrated brand was Lexus, with a rate of just 84 problems
per 100 vehicles. On the other hand, BMW’s Mini was
reported as having 165 problems per 100 vehicles. Larry
is on his way to pick up his brand-new Lexus. At the
same time, Mark is on his way to pick up his brand-new
Mini. What is the probability that Larry’s Lexus will
have exactly 2 problems? What is the probability that
Mark’s Mini will have no more than 1 problem? What is
the probability that neither Larry nor Mark will have a
problem with his new car? Source: Sharon Silke Carty, “U.S.
Autos Power Forward with Gains in Quality Survey,” USA Today, June 23,
2009, p. 3B.
Arriving at the office a bit early, Ted finds Luke Thorndike
in a fit of rage. Mr. Thorndike is sipping coffee and looking over Ted’s notes on the defective racquets that were
constructed with aluminum purchased from each of the
company’s three suppliers.
In rummaging through the basement crate, Ted has
found a total of 30 defective racquets. Of these, 5 were
made of aluminum supplied by the Snowmet Corporation
and 5 were made of aluminum purchased from Barstow
Aluminum, Inc. These were the two suppliers from whom
Thorndike Sports Equipment had been buying aluminum
for many years, and with good results.
The cause of Mr. Thorndike’s anger is the total of
20 defective racquets made from aluminum purchased
from the company’s newest and lowest-priced supplier,
Darwood Discount Metals, Inc. Luke recently placed a big
order with Darwood, and he is counting the minutes until
10 a.m. so that he can call the West Coast firm, cancel
the order, and give Mr. Darwood a big piece of his mind.
The arrival of the morning mail only serves to heighten
Luke’s anger, as three more complaints are received from
disgruntled customers demanding immediate replacements
for their broken racquets.
Ten o’clock finally rolls around, and Luke places a
person-to-person collect call to Mr. Darwood. After berating Mr. Darwood and his products, Luke demands that
the order be canceled immediately. Mr. Darwood does not
appreciate the elder Thorndike’s tirade, but he is quite cognizant of the fact that 10% of his company’s profits come
from Thorndike Sports Equipment. Though irritated, he
patiently tries to reason with Mr. Thorndike.
According to Mr. Darwood, independent metallurgists
have conducted a lot of tests in which they found Darwood
aluminum to be every bit as good as the product supplied
Thorndike Sports Equipment
INTEGRATED C
ASE204 Part 2: Probability
by Snowmet and Barstow. He suggests that the unusually
high number of defective racquets found by Luke is merely
a fluke. Already operating on a short fuse, Mr. Thorndike
responds that he doesn’t much care for silly rhymes in the
middle of the morning, and he warns Darwood not to load
any more aluminum onto Thorndike-bound trucks until
he and Ted have a chance to further examine the information Ted collected over the weekend. He promises to call
Mr. Darwood with a final decision by 3 p.m., Pacific Time.
Ted spends a very busy morning and skips lunch, but
by 2 p.m. he comes up with some data that might prove
useful. Most important, he has uncovered a research study
in which it was found that 0.8% of all aluminum racquets
end up being returned as defective. The number of Snowmet
and Barstow racquets in the “defectives” crate is about 1%
of those produced. However, of the 1200 racquets made
from Darwood aluminum, 20 (about 1.7%) are defective.
Ted decides to consider the situation as a binomial probability distribution in which the probability of a defect on
a given trial is 0.008, corresponding to the research finding
that 0.8% of all the aluminum racquets produced are defective. Using the estimate for the entire industry, the expected
number of defectives among the 1200 racquets made of
Darwood aluminum would be n 5 1200(0.008), or 9.6,
but 20 defectives were observed.
To determine how unusual this result really is, Ted
uses Excel and n 5 1200,  5 0.008 to find the cumulative probabilities for the number of defects. The results are
shown in Table 6.7.
Receiving a telephone call from home and finding out
that the water pipes have just burst and his house is flooding, Ted has to flee the office. On his way out, he asks you
to use the information he has found, along with Table 6.7,
in giving advice to Mr. Thorndike before he makes his call
to Darwood.
Table 6.7 Ted Thorndike has used Excel in generating these
cumulative probabilities for a binomial distribution in which the
number of trials is 1200 and the probability of a defect in any
given trial is 0.8%, or 0.008.AP Photo/IIHS
205
ATM ARRIVAL TIMES:
A CONTINUOUS DISTRIBUTION
In the opening vignette to Chapter 6, we described how the number of automatic
teller machine (ATM) customers over a 5-minute period represents a discrete probability distribution, with x 5 the number of arrivals (0, 1, 2, etc.) during the period.
Without getting into trouble with campus security, stake out your local ATM
as you did for the opener to Chapter 6—except now use the chronograph feature
of your (or a friend’s) watch to record the times between the arrivals of successive
customers. For example, x1 is the time between the arrival of customer A and customer B, x2 is the time between customers B and C, and so on. What you’re doing
now is creating a continuous probability distribution, the topic of this chapter.
As a matter of fact, our discussion here is the flip side of the vignette that
opened Chapter 6. The inverse of arrivals-per-minute is minutes-between-arrivals.
Where the discrete Poisson distribution of Chapter 6 focused on the former,
the continuous exponential distribution of this chapter is concerned with the
latter. This unique relationship between the Poisson and
exponential distributions will be
examined later in this chapter.
Measuring and predicting time
between arrivals can be important
to planners and service providers
in many real-world situations,
especially those dealing with
emergency medical services.
C H A P T E R 7
ContiDniusotruibs uPtrioobnasbility
Some things look a little like a normal curve.• Understand the nature and the applications of the normal distribution.
• Use the standard normal distribution and z-scores to determine
probabilities associated with the normal distribution.
• Use the normal distribution to approximate the binomial distribution.
• Understand the nature and the applications of the exponential
distribution, including its relationship to the Poisson distribution of
Chapter 6.
• Use the computer in determining probabilities associated with the
normal and exponential distributions.
LEARNING
OBJECTIVES
After reading this chapter,
you should be able to:
206
INTRODUCTION
Chapter 6 dealt with probability distributions for discrete random variables,
which can take on only certain values along an interval, with the possible values
having gaps between them. This chapter presents several continuous probability
distributions; these describe probabilities associated with random variables that
are able to assume any of an infinite number of values along an interval.
Discrete probability distributions can be expressed as histograms, where the
probabilities for the various x values are expressed by the heights of a series of
vertical bars. In contrast, continuous probability distributions are smooth curves,
where probabilities are expressed as areas under the curves. The curve is a function of x, and f(x) is referred to as a probability density function. Since the continuous random variable x can be in an infinitely small interval along a range or
continuum, the probability that x will take on any exact value may be regarded
as zero. Therefore, we can speak of probabilities only in terms of the probability
that x will be within a specified interval of values. For a continuous random variable, the probability distribution will have the following characteristics:
(7.1)
The probability distribution for a continuous random variable:
1. The vertical coordinate is a function of x, described as f(x) and referred to as the
probability density function.
2. The range of possible x values is along the horizontal axis.
3. The probability that x will take on a value between a and b will be the area
under the curve between points a and b, as shown in Figure 7.1.
FIGURE 7.1
For a continuous random
variable, the probability
distribution is described by a
curve called the probability
density function, f(x). The total
area beneath the curve is 1.0,
and the probability that x will
take on some value between a
and b is the area beneath the
curve between points a and b.
a b
x
f(x)
Area = P(a ≤ x ≤ b)The first continuous probability distribution we will examine is the normal
distribution. As suggested by the commentary in Statistics in Action 7.1, this is
a very important continuous distribution in both the study of statistics and its
application to business. The normal distribution is a bell-shaped, symmetrical
curve, the use of which is facilitated by a standard table listing the cumulative
areas beneath.
As demonstrated in Chapter 6, binomial probabilities can involve either
tedious computations or extensive tables. We will describe a method by which
the normal distribution can be used as a convenient approximation for determining the probabilities associated with discrete random variables that are of the
binomial distribution.
The next topic will be the exponential distribution. This is related to the
(discrete) Poisson distribution of the previous chapter but is a continuous probability distribution describing probabilities associated with the (continuous) time
or distance intervals between occurrences of the so-called rare events of a Poisson
process.
Finally, we will use Excel and Minitab to randomly select observations from
a specified continuous probability distribution. This is similar to the simulated
samples obtained from selected discrete probability distributions in Chapter 6.
I was flattered but kind of surprised when Dr. Weiers asked
Thorndike Sports Equipment to do a Statistics in Action
commentary for his textbook. Doesn’t he realize we’re
just imaginary? After all, he made us up in the first place.
Anyway, I usually don’t spend a lot of time thinking about
the normal curve, but preparing this commentary helped
open my eyes to how often it comes up.
In many facets of our business, we find a normal
curve closely approximates the data we’ve collected. For
example, in the “bounce test” we use in monitoring the
liveliness of our golf balls, we find the bounce distances
tend to be normally distributed. Although we don’t like
to talk about it much, we try to make sure our sponsored
professionals get balls that are from the most lively 0.5% of
the output. My grandson Ted tells me that corresponds to
z 5 12.58 standard deviation units from the mean.
According to our human resources manager, a lot of
the aptitude and performance tests we give potential
and current employees also tend to result in scores that
are distributed according to that classic “bell-shaped”
curve. She’s been with us for many years, and she finds
it very interesting to compare the means and standard
deviations today with the ones she got ten or twenty
years ago.
Ted says we also rely on the normal distribution in
constructing and interpreting the statistical process control charts that have helped Thorndike racquets gain their
reputation for consistently high quality. Here at Thorndike
Sports Equipment, we greatly appreciate the applications
and value of the normal distribution, and we are happy to
pass our experience on to you.
Luke Thorndike
President and Founder
Thorndike Sports
Equipment
7.1
Navigating Normal Curves with Thorndike Sports Equipment
STATISTICS I
N ACTION
The probability density function, f(x), for a given continuous distribution is
expressed in algebraic terms, and the areas beneath are obtained through
the mathematics of calculus. However, tables are provided in the text for
readily identifying or calculating such areas.
4. The total area under the curve will be equal to 1.0.208 Part 2: Probability
THE NORMAL DISTRIBUTION
Description and Applications
The normal distribution is the most important continuous distribution in statistics.
It occupies this position for three key reasons: (1) many natural and economic
phenomena tend to be approximately normally distributed; (2) it can be used
in approximating other distributions, including the binomial; and (3) as we will
see in Chapter 8, sample means and proportions tend to be normally distributed
whenever repeated samples are taken from a given population of any shape.
This latter characteristic is related to the central limit theorem, a concept also
discussed in the next chapter.
The normal distribution is really a family of distributions, each member of
which is bell shaped and symmetrical (the left side is a mirror image of the right).
As Figure 7.2 shows, (1) the mean, median, and mode are all at the same position
on the horizontal axis; (2) the curve is asymptotic, approaching the horizontal axis
at both ends, but never intersecting with it; and (3) the total area beneath the curve
is equal to 1.0. The specific member of the family depends on just two descriptors:
(7.2)
7.1 What is the difference between a continuous probability
distribution and a discrete probability distribution?
7.2 What is a probability density function and how is it
relevant to the determination of probabilities associated
with a continuous random variable?
7.3 Why is the total area beneath a probability density
function equal to 1.0?
7.4 What is the probability that a continuous random
variable will take on any specific value? Explain your
answer.
E X E R C I
S E S
FIGURE 7.2
The normal distribution is
actually a family of bell-shaped
distributions, each of which
has these characteristics.
The specific member of the
family depends on just two
descriptors, the mean () and
the standard deviation ().
Symmetric: Each side is the
mirror image of the other
Mean, median,
and mode
Asymptotic:
Tail approaches
the x axis but
never reaches it
Asymptotic:
Tail approaches
the x axis but
never reaches it
Area = 0.5 Area = 0.5Chapter 7: Continuous Probability Distributions 209
FIGURE 7.3
The shape of a normal curve
and its position on the x axis
will depend on its mean ()
and standard deviation ().
Although curve B has a greater
mean, its standard deviation is
smaller.
A B
mA mB
x
In this equation,  is the geometric constant for the ratio of the circumference
of a circle to its diameter, and is not related to the symbol used for a population
proportion.
NOTE
the mean () and the standard deviation (). The normal distribution can be
described as follows:
The normal distribution for the continuous random variable, x (with
2` # x # 1`):
f(x) 5 _______ 1
Ï
___
2
e2(12)[(x2)]2 where  5 mean
 5 standard deviation
e 5 the mathematical constant, 2.71828
 5 the mathematical constant, 3.14159
Not all normal curves will look exactly like the one in Figure 7.2. The shape
of the curve and its location along the x axis will depend on the values of the
standard deviation and mean, respectively. You can demonstrate this by using
Seeing Statistics Applet 4 at the end of the chapter. Figure 7.3 shows two normal
curves representing output diameters for two tubing extrusion machines:
• The average diameter of tubing from machine B exceeds that from machine A.
Looking along the x axis, it can be seen that B . A.
• The variability of the output from machine A is greater than that from
machine B. In symbolic terms, this can be expressed as A . B.
Areas Beneath the Normal Curve
Regardless of the shape of a particular normal curve, the areas beneath it can be
described for any interval of our choice. This makes it unnecessary to get involved
with calculus and the complexities of the probability density function presented
earlier. For any normal curve, the areas beneath the curve will be as follows (see
also part (a) in Figure 7.4):
• About 68.3% of the area is in the interval  2  to  1 .
• About 95.5% of the area is in the interval  2 2 to  1 2.
• Nearly all of the area (about 99.7%) is in the interval  2 3 to  1 3.210 Part 2: Probability
FIGURE 7.4
For any normal distribution,
the approximate areas shown
in part (a) will lie beneath the
curve for the intervals shown.
In part (b), the areas show
how these intervals could
apply to the amount of time
logged by general-aviation
aircraft during the year.
Assuming  5 120 hours and
 5 30 hours, about 95.5% of
the planes logged between 60
and 180 hours of flying time.
(a) Approximate areas beneath the normal curve
(b) Approximate areas applied to annual flying hours
68.3%
95.5%
99.7%
68.3%
95.5%
99.7%
m – 3s m – 2s m – 1s m m + 1s m + 2s m + 3s
Assuming:
m = 120 hours
s = 30 hours
30 60 90 120 150 180 210 x (hours)
EXAMPLE
Normal Distribution
The General Aviation Manufacturers Association has reported the average annual
number of hours flown by general-aviation aircraft to be approximately 120 hours.1
For purposes of our example, we will assume a normal distribution and a standard
deviation of 30 hours for the flying hours of aircraft within this population.
EXAMPLEEXAMPLEEXAMPL
1Source: General Aviation Manufacturers Association, General Aviation Statistical Databook, 2008
Edition, pp. 28–29.Chapter 7: Continuous Probability Distributions 211
SOLUTION
Based on the reported mean,  5 120 hours, and the assumed standard deviation,
 5 30 hours, we can use the known properties of the normal distribution
to determine that about 95.5% of these aircraft logged between (120 2 2[30])
and (120 1 2[30]) hours, or between 60 and 180 hours of flying time during
that year. Likewise, we could determine that about 68.3% of the aircraft logged
between 90 and 150 ( 6 ) hours and that about 99.7% logged between 30
and 210 ( 6 3) hours. These intervals and approximate areas are shown in
part (b) of Figure 7.4.
XAMPLEEXAMPLEEXAMPLE
7.5 It has been stated that the normal distribution is
really a “family” of distributions. Explain.
7.6 In the normal distribution, the probability that x
will exceed ( 1 2) is the same as the probability that
x will be less than ( 2 2). What characteristic of the
normal distribution does this reflect?
7.7 Sketch two different normal distributions along a
single x axis so that both of the following conditions are
satisfied: (a) a 5 b and (b) a . b.
7.8 If x is normally distributed with  5 20.0 and
 5 4.0, determine the following:
a. P(x $ 20.0) b. P(16.0 # x # 24.0)
c. P(x # 12) d. P(x 5 22.0)
e. P(12.0 # x # 28.0) f. P(x $ 16)
7.9 If x is normally distributed with  5 25.0 and
 5 5.0, determine the following:
a. P(x $ 25.0) b. P(20.0 # x # 30.0)
c. P(x # 30) d. P(x 5 26.2)
e. P(15.0 # x # 25.0) f. P(x $ 15)
7.10 The Canada Urban Transit Association has reported
that the average revenue per passenger trip during a given
year was $1.55. If we assume a normal distribution and
a standard deviation of  5 $0.20, what proportion of
passenger trips produced a revenue of Source: American
Public Transit Association, APTA 2009 Transit Fact Book, p. 35.
a. less than $1.55?
b. between $1.15 and $1.95?
c. between $1.35 and $1.75?
d. between $0.95 and $1.55?
7.11 In 2007, the average conventional first mortgage for
new single-family homes was $360,000. Assuming a normal
distribution and a standard deviation of  5 $30,000,
what proportion of the mortgages were Source: Bureau of the
Census, Statistical Abstract of the United States 2009, p. 726.
a. more than $360,000?
b. between $300,000 and $420,000?
c. between $330,000 and $390,000?
d. more than $270,000?
7.12 In 2009, the average charge for tax preparation
by H&R Block, Inc. was $187. Assuming a normal
distribution and a standard deviation of  5 $20, what
proportion of H&R Block’s tax preparation fees were
Source: hrblock.com, July 14, 2009.
a. more than $187?
b. between $147 and $227?
c. between $167 and $207?
d. more than $227?
7.13 It has been reported that the average hotel check-in
time, from curbside to delivery of bags into the room,
is 12.0 minutes. An Li has just left the cab that brought
her to her hotel. Assuming a normal distribution with a
standard deviation of 2.0 minutes, what is the probability
that the time required for An Li and her bags to get to
the room will be:
a. greater than 14.0 minutes?
b. between 10.0 and 14.0 minutes?
c. less than 8.0 minutes?
d. between 10.0 and 16.0 minutes?
7.14 The average American family of four spends $5000
per year on food prepared at home. Assuming a normal
distribution with a standard deviation of $1000 and
a randomly selected American family of four, what is
the probability that the family’s annual spending for
food prepared at home will be: Source: “Business Briefing,”
Pittsburgh Tribune-Review, July 5, 2009, p. E1.
a. more than $8000?
b. between $5000 and $7000?
c. less than $6000?
d. between $3000 and $6000?
7.15 On average, commuters in the Los Angeles,
California, area require 30.0 minutes to get to work.
Assume a normal distribution with a standard deviation
of 5.0 minutes and a randomly selected Los Angeles-area
E X E R C I
S E S212 Part 2: Probability
THE STANDARD NORMAL DISTRIBUTION
Description and Applications
Because there is a different normal curve for every possible pair of  and , the
number of statistical tables would be limitless if we wished to determine the areas
corresponding to possible intervals within all of them. Fortunately, we can solve
this dilemma by “standardizing” the normal curve and expressing the original x
values in terms of their number of standard deviations away from the mean. The
result is referred to as a standard (or standardized) normal distribution, and it
allows us to use a single table to describe areas beneath the curve. The key to the
process is the z-score:
(7.3)
commuter named Jamal. Source: The World Almanac and Book
of Facts 2006, p. 474.
a. What is the probability that Jamal will require more
than 45.0 minutes to get to work on any given day?
b. Jamal has just left home and must attend an important
meeting with the CEO in just 25.0 minutes. If the CEO
routinely fires employees who are tardy, what is the
probability that Jamal will be going to work tomorrow?
The z-score for a standard normal distribution:
z 5
x 2 
______

where z 5 the distance from the mean, measured in
standard deviation units
x 5 the value of x in which we are interested
 5 the mean of the distribution
 5 the standard deviation of the distribution
In the aircraft flying hours example discussed earlier, the mean and (assumed)
standard deviation were  5 120 hours and  5 30 hours, respectively. Using the
z-score equation, we can convert this distribution into a standard normal distribution in which we have z values instead of x values. For example:
x Value in the Original Corresponding z Value in the
Normal Distribution Standard Normal Distribution
x 5 120 hours z 5 x______ 2 

5
120 2 120
__________
30
5 0.00
x 5 160 hours z 5 x______ 2 

5
160 2 120
__________
30
5 1.33
x 5 90 hours z 5 x______ 2 

5
90 2 120
_________
30
5 21.00
In Figure 7.5, there are two scales beneath the curve: an x (hours of flying time)
scale and a z (standard deviation units) scale. Note that the mean of the z scale is
0.00. Regardless of the mean and standard deviation of any normal distribution,
we can use the z-score concept to express the original values in terms of standard
deviation multiples from the mean. It is this transformation that allows us to use
the standard normal table in determining probabilities associated with any normal
distribution.You can use Seeing Statistics Applet 5, at the end of the chapter, to see
how the normal distribution areas respond to changes in the beginning and ending
z values associated with Figure 7.5.Chapter 7: Continuous Probability Distributions 213
Using the Standard Normal Distribution Table
Regardless of the units involved (e.g., pounds, hours, dollars, and so on) in the
original normal distribution, the standard normal distribution converts them into
standard deviation distances from the mean. As we proceed through the next
example, our original units will be grams.
In demonstrating the use of the standard normal distribution table, we will
rely on a situation involving the weight added to balance a generator shaft. For
each of the related questions, we will go through these steps:
1. Convert the information provided into one or more z-scores.
2. Use the standard normal table to identify the area(s) corresponding to the
z-score(s). A portion of the standard normal distribution table is shown in
Table 7.1. The table provides cumulative areas to the z value of interest.
3. Interpret the result in such a way as to answer the original question.
The cumulative standard normal distribution table is on the rear endsheet
of the text, and an abbreviated portion of it is shown in Table 7.1. The row
labels (e.g., 1.4) refer to the integer and first decimal place of z, and the column
labels refer to the second decimal place (e.g., 0.05); thus, we can look in row 1.4
and column 0.05 and see that the cumulative probability (or area) to the left of
z 5 1.45 is 0.9265. Here are a few more examples of how to find cumulative
areas for given z values:
• Cumulative area associated with z 5 21.37: Referring to the 21.3 row and
the 0.07 column of Table 7.1, we find this probability is 0.0853.
• Cumulative area associated with z 5 20.08: Referring to the 20.0 row and
the 0.08 column of Table 7.1, we find this probability is 0.4681.
• Cumulative area associated with z 5 0.09: Referring to the 0.0 row and the
0.09 column of Table 7.1, we see this probability is 0.5359.
• Cumulative area associated with z 5 1.73: Referring to the 1.7 row and the
0.03 column of Table 7.1, we see this probability is 0.9582.
FIGURE 7.5
For the distribution of annual
flying hours in part (b) of
Figure 7.4, the upper of
the two horizontal scales
shown here represents
the standardized normal
distribution, in which x values
(hours) have been converted
to z-scores (standard deviation
units from the mean). The use
of z-scores makes it possible
to use a single table for all
references to the normal
distribution, regardless of the
original values of  and .
Assuming:
m = 120 hours
s = 30 hours
z: z = –3 z = –2 z = –1 z = 0 z = +1 z = +2 z = +3 (s units)
x: 60 90 150 180 210 (hours) 30 120
For x = 90 hours, z = x–––– – m
s
90 – 120
––––––––
30
= = –1.00214 Part 2: Probability
TABLE 7.1
A portion of the cumulative
standard normal distribution
table from the back endsheet
0
For z = 1.34, refer to the
1.3 row and the 0.04 column
to find the cumulative area, 0.9099.
z
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
22.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183
21.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233
21.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294
21.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367
21.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455
21.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559
21.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681
21.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823
21.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985
21.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170
21.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379
20.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611
20.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867
20.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148
20.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451
20.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776
20.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121
20.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483
20.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859
20.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247
20.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319
1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817Chapter 7: Continuous Probability Distributions 215
In applications involving the standard normal table, actual areas of interest
may lie to the left of the mean or to the right of the mean, and may even require
that one cumulative probability be subtracted from another. Because of (1) the
variety of possibilities, and (2) the importance of knowing how to use the standard normal table, we will examine a number of applications that relate to the
following example.
EXAMPLE
Normal Probabilities
Following their production, industrial generator shafts are tested for static and
dynamic balance, and the necessary weight is added to predrilled holes in order to
bring each shaft within balance specifications. From past experience, the amount
of weight added to a shaft has been normally distributed, with an average of
35 grams and a standard deviation of 9 grams.
SOLUTIONS
What Is the Probability That a Randomly Selected Shaft Will Require
Between 35 and 40 Grams of Weight for Proper Balance? [P(35  x  40)]
1. Conversion of x 5 40 grams to z-score:
z 5
x 2 
______

5
40 2 35
________
9
5 0.56
Conversion of x 5 35 grams to z-score:
z 5
x 2 
______

5
35 2 35
________
9
5 0.00
2. Reference to the standard normal table: Referring to the 0.5 row and the 0.06
column of the table, we find the cumulative area to z 5 0.56 is 0.7123. Referring to the 0.0 row and the 0.00 column, we can verify that the cumulative
area to z 5 0.00 (the mean) is 0.5000.
3. Solution: The probability that the randomly selected shaft will require between
35 and 40 grams is 0.7123 2 0.5000 5 0.2123, as illustrated in Figure 7.6.
What Is the Probability That a Randomly Selected Shaft Will Require at
Least 50 Grams of Weight for Proper Balance? [P(x  50)]
1. Conversion of x 5 50 grams to z-score:
z 5
x 2 
______

5
50 2 35
________
9
5 1.67
2. Reference to the standard normal table: Referring to the 1.6 row and the 0.07
column, we find the cumulative area to z 5 1.67 is 0.9525. Also, we must
consider that the total cumulative area (i.e., to z approaching positive infinity)
is, by definition, 1.0000.
3. Solution: The probability that the randomly selected shaft will require at least
50 grams is 1.0000 2 0.9525 5 0.0475, as illustrated in Figure 7.7.
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPL216 Part 2: Probability
FIGURE 7.6
The probability is 0.2123 that
a randomly selected generator
shaft will require between
35 and 40 grams of weight
in order to achieve proper
balance.
z: 0.00 0.56 (s units)
x: 35 40 (grams)
m = 35 grams
s = 9 grams
For x = 40 grams, z = x–––– – m
s
40 – 35
––––––––
9
= = 0.56
P(35 ≤ x ≤ 40) = P(0.00 ≤ z ≤ 0.56)
or 0.7123 (Cumulative area to z = 0.56)
– 0.5000 (Cumulative area to z = 0.00)
= 0.2123
FIGURE 7.7
The probability is 0.0475 that
a randomly selected generator
shaft will require at least
50 grams for proper balance.
z: 0.00 1.67 (s units)
x: 35 50 (grams)
m = 35 grams
s = 9 grams
P(x ≥ 50) = P(z ≥ 1.67)
or 1.0000 (Cumulative area to z = +∞)
–0.9525 (Cumulative area to z = 1.67)
= 0.0475
What Is the Probability That a Randomly Selected Shaft Will Require
Between 41 and 49 Grams of Weight for Proper Balance? [P(41  x  49)]
1. Conversion of x 5 49 grams to z-score:
z 5
x 2 
______

5
49 2 35
________
9
5 1.56
Conversion of x 5 41 grams to z-score:
z 5
x 2 
______

5
41 2 35
________
9
5 0.67
2. Reference to the standard normal table: Referring to the 1.5 row and the
0.06 column of the table, we find the cumulative area to z 5 1.56 is 0.9406.
Referring to the 0.6 row and the 0.07 column of the table, the cumulative
area to z 5 0.67 is 0.7486.
3. Solution: The probability that the randomly selected shaft will require between
41 and 49 grams is 0.9406 2 0.7486 5 0.1920, as illustrated in Figure 7.8.
EXAMPLEEXAMPLEEXAMPLEEXChapter 7: Continuous Probability Distributions 217
What Is the Probability That a Randomly Selected Shaft Will Require No
More Than 26 Grams of Weight for Proper Balance? [P(x  26)]
1. Conversion of x 5 26 grams to z-score:
z 5
x 2 
______

5
26 2 35
________
9
5 21.00
2. Reference to the standard normal table: Referring to the 21.0 row and the
0.00 column of the table, we find the cumulative area to z 5 21.00 is 0.1587.
3. Solution: As Figure 7.9 illustrates, the probability that a randomly selected
shaft will require no more than 26 grams is 0.1587.
In the preceding cases, we proceeded from x values to the determination of
a corresponding area beneath the standard normal curve. However, it is also
possible to specify an area, then identify a range of x values that corresponds to
this area.
FIGURE 7.8
The probability is 0.1920 that
a randomly selected generator
shaft will require between
41 and 49 grams for proper
balance.
z: 0.00 1.56 0.67 (s units)
x: 35 49 41 (grams)
m = 35 grams
s = 9 grams
P( 41 ≤ x ≤ 49) = P(0.67 ≤ z ≤ 1.56)
or 0.9406 (Cumulative area to z = 1.56)
–0.7486 (Cumulative area to z = 0.67)
= 0.1920
FIGURE 7.9
The probability is 0.1587 that
a randomly selected generator
shaft will require no more than
26 grams for proper balance.
z: –1.00 ( 0.00 s units)
x: 26 (grams) 35
m = 35 grams
s = 9 grams
P(x ≤ 26) = P(z ≤ –1.00)
= 0.1587 (Cumulative area to z = –1.00)
MPLEEXAMPLEEXAMPLE218 Part 2: Probability
EXAMPLE
x Value for a Cumulative Probability
In the setting just described, the average amount of weight added to the generator
shafts was 35 grams, with a standard deviation of 9 grams. Lower weights are
more desirable, and management has just directed that the best 5% of the output
be reserved for shipment to aerospace customers. Translating “the best 5%” into
an amount of balancing weight, what weight cutoff should be used in deciding
which generator shafts to reserve for aerospace customers?
SOLUTION
The quantity to be determined is the amount of weight (w) such that only 5%
of the shafts require no more than w grams. Expressed in probability terms, the
value of w must be such that P(x # w) 5 0.05.
The solution is to find out what value of z corresponds to a cumulative area
of 0.05, then convert this z value into w grams. Reversing the steps of the previous examples:
1. Find the value of z that corresponds to a cumulative area of 0.05, or 0.0500:
Of the values listed in the standard normal table, we see that the cumulative area to z 5 21.64 is 0.0505, while the cumulative area to z 5 21.65
is 0.0495. Interpolating between the two, the number of grams in the cutoff
corresponds to z 5 21.645.
2. Continuing to work backward, and knowing that  5 35 grams and
 5 9 grams, the next step is to substitute w into the equation for the previously determined z score, z 5 21.645:
z 5
w 2 
______

or 21.645 5 w_______ 2 35
9
and w 5 35 2 9(1.645) 5 20.195 grams
As Figure 7.10 shows, the manufacturer should set the cutoff for aerospace generator shafts at 20.195 grams, since 95% of the shafts produced will require this
amount or more in order to achieve proper balance.
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMP
FIGURE 7.10
If the best 5% of the output
is to be reserved for aerospace
customers, shafts requiring no
more than 20.195 grams for
balancing would be allocated
to this industry.
z: –1.645 ( 0.00 s units)
x: 20.195 = w 35 (grams)
m = 35 grams
Area = 0.05 s = 9 grams
For a cumulative area of 0.05, z must be –1.645.
If w = the maximum number of grams for generator
shafts being sold to the aerospace industry,
z =
w – m
–––––
s
w – 35
–––––
9
or –1.645 = and w = 20.195Chapter 7: Continuous Probability Distributions 219
Excel and Minitab can be used to obtain cumulative normal probabilities
associated with specified values for x—the procedures and results are shown
in Computer Solutions 7.1. These software packages can also be used in doing
the reverse—that is, specifying a cumulative probability, then finding the x
value associated with it. These procedures and results are shown in Computer
Solutions 7.2.
EEXAMPLEEXAMPLE
Normal Probabilities
These procedures show how to get cumulative probabilities.
For each software package:
Procedure I: Getting the cumulative probability associated with one x value.
Procedure II: Getting cumulative probabilities associated with a set of x values.
EXCEL
I. Excel procedure for the cumulative probability associated with one x value
1. Click on the cell into which Excel will place the probability. From the Formulas ribbon, click on the Insert Function
symbol at the extreme left. Click Statistical in the Category box. Click NORMDIST in the Function box. Click OK.
2. Enter the x value of interest (e.g., 45) into the X box. Enter the mean of the distribution (e.g., 35) into the Mean
box. Enter the standard deviation (e.g., 9) into the Standard_dev box. Enter True into the Cumulative box.
Click OK.
II. Excel procedure for cumulative probabilities associated with a set of x values
1. Enter the labels into A4:B4 as shown in the printout above. Enter the x values of interest (e.g., 45, 35, 25) starting
with cell A5. Click on cell B5.
2. From the Formulas ribbon, click on the Insert Function symbol at the extreme left. Click Statistical in the
Category box. Click NORMDIST in the Function box. Click OK.
3. Enter A5 into the X box. Enter the mean of the distribution (e.g., 35) into the Mean box. Enter the standard deviation (e.g., 9) into the Standard_dev box. Enter True into the Cumulative box. Click OK. The first cumulative probability will appear in cell B5.
4. Click at the lower right corner of B5 (the cursor will turn to a “1” sign) and drag downward so cells B5:B7 will
contain the set of cumulative probabilities.
COMPUTER 7.1 SOLUTIONS
(continued)220 Part 2: Probability
MINITAB
I. Minitab procedure for cumulative probability associated with one x value
1. Click Calc. Select Probability Distributions. Click Normal.
2. In the Normal Distribution menu, enter the mean of the distribution (e.g., 35) into the Mean box. Enter the standard deviation (e.g., 9) into the Standard deviation box. Select Input constant and enter the x value of interest
(e.g., 45) into the box. Select Cumulative. Click OK.
II. Minitab procedure for cumulative probabilities associated with a set of x values
Enter the desired x values (e.g., 45, 35, 25) into column C1. Follow steps 1 and 2 above, except select Input column
and enter C1 into the box.
Inverse Normal Probabilities
These procedures show how to get x values associated with cumulative normal probabilities.
For each software package:
Procedure I: Getting the x value associated with one cumulative probability.
Procedure II: Getting the x values associated with a set of cumulative probabilities.
EXCEL
I. Excel procedure for the x value associated with one cumulative probability
1. Click on the cell into which Excel will place the x value. From the Formulas ribbon, click on the Insert Function
symbol at the extreme left. Click Statistical in the Category box. Click NORMINV in the Function box. Click OK.
2. Enter the cumulative probability of interest (e.g., 0.9750) into the Probability box. Enter the mean of the distribution
(e.g., 35) into the Mean box. Enter the standard deviation (e.g., 9) into the Standard_dev box. Click OK.
COMPUTER 7.2 SOLUTIONSChapter 7: Continuous Probability Distributions 221
II. Excel procedure for x values associated with a set of cumulative probabilities
1. Enter the labels into A4:B4 as shown in the printout above. Enter the cumulative probabilities of interest (e.g.,
0.9750, 0.5000, 0.2500) starting with cell A5. Click on cell B5.
2. From the Formulas ribbon, click on the Insert Function symbol at the extreme left. Click Statistical in the
Category box. Click NORMINV in the Function box. Click OK.
3. Enter A5 into the Probability box. Enter the mean of the distribution (e.g., 35) into the Mean box. Enter the
standard deviation (e.g., 9) into the Standard_dev box. Click OK. The first cumulative probability will appear in
cell B5.
4. Click at the lower right corner of B5 (the cursor will turn to a “1” sign) and drag downward so cells B5:B7 will
contain the set of cumulative probabilities.
MINITAB
I. Minitab procedure for the x value associated with one cumulative probability
1. Click Calc. Select Probability Distributions. Click Normal.
2. In the Normal Distribution menu, enter the mean of the distribution (e.g., 35) into the Mean box. Enter the standard deviation (e.g., 9) into the Standard deviation box. Select Input constant and enter the cumulative probability of interest (e.g., 0.9750) into the box. Select Inverse cumulative probability. Click OK.
II. Minitab procedure for the x values associated with a set of cumulative probabilities
Enter the cumulative probabilities of interest (e.g., 0.9750, 0.5000, 0.2500) into column C1. Follow steps 1 and 2
above, except select Input column and enter C1 into the box.
7.16 The normal distribution is really a family of
distributions. Is the standard normal distribution also a
family of distributions? Explain.
7.17 In the standard normal distribution, approximately
what values of z correspond to the
a. first and third quartiles?
b. first and ninth deciles?
c. 23rd and 77th percentiles?
7.18 A continuous random variable, x, is normally distributed with a mean of $1000 and a standard deviation
of $100. Convert each of the following x values into its
corresponding z-score:
a. x 5 $1000 b. x 5 $750
c. x 5 $1100 d. x 5 $950
e. x 5 $1225
7.19 A continuous random variable, x, is normally
distributed with a mean of 200 grams and a standard
deviation of 25 grams. Convert each of the following
x values into its corresponding z-score:
a. x 5 150 b. x 5 180
c. x 5 200 d. x 5 285
e. x 5 315
E X E R C I
S E S222 Part 2: Probability
7.20 Using the standard normal table, find the following
probabilities associated with z:
a. P(0.00 # z # 1.25)
b. P(21.25 # z # 0.00)
c. P(21.25 # z # 1.25)
7.21 Using the standard normal table, find the following
probabilities associated with z:
a. P(0.00 # z # 1.10)
b. P(z $ 1.10)
c. P(z # 1.35)
7.22 Using the standard normal table, find the following
probabilities associated with z:
a. P(20.36 # z # 0.00)
b. P(z # 20.36) c. P(z $ 20.43)
7.23 Using the standard normal table, find the following
probabilities associated with z:
a. P(21.96 # z # 1.27)
b. P(0.29 # z # 1.00)
c. P(22.87 # z # 21.22)
7.24 Using the standard normal table, determine a z
value (to two decimal places) such that the
a. cumulative area to z is 0.7486.
b. cumulative area to z is 0.0735.
c. area between z and positive infinity is 0.3508.
d. area between z and positive infinity is 0.0212.
7.25 Using the standard normal table, determine a z
value (to two decimal places) such that the
a. cumulative area to z is 0.70.
b. cumulative area to z is 0.10.
c. area between z and negative infinity is 0.54.
d. area between z and positive infinity is 0.30.
7.26 For the normal distribution described in Exercise
7.10, what is the probability that a randomly selected
passenger trip would have produced a revenue
a. between $1.55 and $1.80?
b. over $1.90? c. under $1.25?
7.27 For the normal distribution described in Exercise
7.11, what is the probability that a randomly selected
first mortgage would have been for an amount
a. between $305,000 and $325,000?
b. over $325,000? c. under $390,000?
7.28 For the normal distribution described in Exercise
7.12, what is the probability that a randomly selected
tax preparation customer would have paid a fee
a. between $150 and $180?
b. under $140? c. over $220?
7.29 For the normal distribution described in Exercise
7.15, what is the probability that Jamal’s commuting
time to work today will be
a. less than 29 minutes?
b. between 29 and 33 minutes?
c. more than 32 minutes?
7.30 For the normal distribution described in Exercise
7.15, what commuting time will be exceeded on only 10%
of Jamal’s commuting days?
7.31 For the normal distribution described in Exercise
7.11, what first-mortgage amount would have been
exceeded by only 5% of the mortgage customers?
7.32 For the normal distribution described in Exercise
7.12, what tax preparation fee would have been
exceeded by 90% of the tax preparation customers?
7.33 It has been reported that households in the West
spend an annual average of $6050 for groceries. Assume
a normal distribution with a standard deviation of
$1500. Source: David Stuckey and Sam Ward, “Home Cooked,” USA
Today, July 25, 2006, p. 1A.
a. What is the probability that a randomly selected
Western household spends more than $6350 for
groceries?
b. How much money would a Western household have
to spend on groceries per year in order to be at the
99th percentile (i.e., only 1% of Western households
would spend more on groceries)?
7.34 Andre is a fearless circus performer who gets shot
from a special cannon during the grand finale of the
show and is supposed to land on a safety net at the
other side of the arena. The distance he travels varies,
but is normally distributed with a mean of 150 feet
and a standard deviation of 10 feet. The landing net is
30 feet long.
a. To maximize Andre’s probability of landing on the
net, how far away from the cannon should he
position the nearest edge of the net?
b. Given the net position in part (a), what is the probability that Andre will be able to return for tomorrow
night’s show?
7.35 Drying times for newly painted microwave oven
cabinets are normally distributed with a mean of
2.5 minutes and a standard deviation of 0.25 minutes.
After painting, each cabinet is mated with its electronic
modules and mechanical components. The production
manager must decide how much time to allow after
painting before these other components are installed.
If the time is too short, the paint will smudge and the
unit will have to be refinished. If the time is too long,
production efficiency will suffer. A consultant has
concluded that the time delay should be just enough
to allow 99.8% of the cabinets to dry completely,
with just 0.2% ending up being smudged and sent
back for refinishing. Given this information, for what
time setting should the production manager set the
automatic timer that pauses the production line while
each cabinet dries?
7.36 KleerCo supplies an under-hood, emissionscontrol air pump to the automotive industry. The
pump is vacuum powered and works while the engineChapter 7: Continuous Probability Distributions 223
is operating, cleaning the exhaust by pumping extra
oxygen into the exhaust system. If a pump fails before
the vehicle in which it is installed has covered 50,000
miles, federal emissions regulations require that it be
replaced at no cost to the vehicle owner. The company’s
current air pump lasts an average of 63,000 miles, with
a standard deviation of 10,000 miles. The number of
miles a pump operates before becoming ineffective has
been found to be normally distributed.
a. For the current pump design, what percentage of
the company’s pumps will have to be replaced at no
charge to the consumer?
b. What percentage of the company’s pumps will fail at
exactly 50,000 miles?
c. What percentage of the company’s pumps will fail
between 40,000 and 55,000 miles?
d. For what number of miles does the probability
become 80% that a randomly selected pump will no
longer be effective?
7.37 The KleerCo company in Exercise 7.36 would like
to design a more durable pump so that no more than
2% of original-equipment pumps are returned under free
warranty, and the standard deviation continues to be
10,000 miles. What will the new average service life have
to be in order for the new pump design to comply with
this requirement?
7.38 A battery manufacturer has just begun distribution
of its SuperVolt, a 6-volt lantern battery for outdoor
enthusiasts. Prototype tests of the new battery found the
average lifetime in continuous use to be 8 hours, with a
standard deviation of 2 hours. Battery lifetimes in such
applications have been approximately normally distributed. Edgar Evans, the consumer reporter for a large metropolitan newspaper, has purchased one of the batteries
to take along on a camping trip. On the first day of his
trip, Edgar goes fishing in his rowboat, becomes disoriented, and gets lost among the many small islands in the
large lake near his camp. Fortunately, Edgar has brought
along his trusty flashlight and its new SuperVolt battery.
At 9 p.m., Edgar turns on his flashlight and shines it into
the air, hoping that someone will see his signal and rescue him. Unfortunately, it is not until 3 a.m. that searchers begin their flight over the lake. If Edgar’s light is still
shining as they become airborne, they will easily spot it
from the air. When Edgar finally gets back to the city,
what is the probability that he will have an exciting story
to tell about how his flashlight and its SuperVolt battery
were two of the heroes in his rescue?
THE NORMAL APPROXIMATION TO THE
BINOMIAL DISTRIBUTION
The binomial distribution, discussed in Chapter 6, is symmetrical whenever the
population proportion, , is 0.5 and approaches symmetry for values that are
close to 0.5. In addition, whenever the number of trials, n, becomes larger, the
binomial distribution takes on a bell-shaped appearance that resembles very
closely the normal distribution discussed in this chapter.
Although the binomial distribution is discrete and the normal distribution is continuous, the normal distribution is a very good approximation to the binomial whenever both n and n(1 2 ) are 5. This can be quite useful when n and  are not
among the values listed in the binomial probability tables. As we have seen, such tables
are already quite extensive for even relatively small upper limits for n. In addition,
calculating binomial probabilities in the absence of such tables is a tedious proposition.
The Binomial Distribution: A Brief Review
The Bernoulli Process
The binomial distribution assumes a Bernoulli process, which has the following
characteristics:
1. There are two or more consecutive trials.
2. On each trial, there are just two possible outcomes—denoted as “success” or
“failure.”
3. The trials are statistically independent; that is, the outcome in any trial is not
affected by the outcomes of earlier trials, and it does not affect the outcomes
of later trials.
(7.4)224 Part 2: Probability
4. The probability of a success remains the same from one trial to the next. This is
satisfied if either we sample with replacement or the sample is small compared
to the size of the population. As a rule of thumb, if we are sampling without
replacement and the population is at least 20 times as large as the number of
trials, the constant  assumption can be considered to have been met.
The Mean and Standard Deviation
The mean and standard deviation of the binomial distribution are the basis for
the normal distribution that will be used in making the approximation. As
described in Chapter 6, these are
• Mean:
 5 n
• Standard deviation:
 5 Ï
__________
n(1  ) where n 5 the number of trials
 5 the probability of success
on any given trial
x 5 the number of successes
in n trials
Correction for Continuity
Because the binomial distribution has gaps between possible values of x, while
the normal distribution is continuous, the normal approximation to the binomial
involves a correction for continuity. The correction consists of expanding each
possible value of the discrete variable, x, by 0.5 in each direction. In continuous
distributions like the normal, the probability of x taking on any exact value (e.g.,
x 5 12.000) is zero. This is why we must consider each of the discrete values of
x as being an interval.
The continuity correction process where the discrete random variable, x, can take
on integer values from 0 to 15:
The discrete value of x is converted to the interval:
x 5 0 20.5  x  0.5
x 5 1 0.5  x  1.5
1.5  x  2.5
...
x 5 2
x 5 14 13.5  x  14.5
x 5 15 14.5  x  15.5
...Chapter 7: Continuous Probability Distributions 225
The Approximation Procedure
In using the normal distribution to approximate the binomial, we first determine
the mean and standard deviation for the binomial distribution, then use these
values as  and  in finding the area(s) of interest beneath the standard normal
curve. In particular, we can find
1. the probability for an individual value for x, now expressed in terms of the
interval from [x 2 0.5] to [x 1 0.5], or
2. the probability that x will lie within or beyond a given range of values. (Note:
In this application, the correction for continuity becomes less important
whenever x can take on a large number of possible values.)
EXAMPLE
Normal Approximation
This example demonstrates the use of the normal distribution to approximate
the binomial. It represents a situation in which both n and n(1  ) are $5,
which is requisite for using the approximation. For purposes of our example,
we will assume that 60% of the Americans who exercise each week by walking
are women. A randomly selected group comprises 15 persons who exercise each
week by walking, with x 5 the number of women in the group.
SOLUTION
Determine the Mean and Standard Deviation of x.
The expected number of females in the group is 9.0.
 5 n or 15(0.60) 5 9
 5 Ï
__________
n(1  ) 5 Ï
_________________
15(0.60)(1  0.60) 5 1.897
What Is the Probability That There Will Be Exactly 11 Females in the Group?
Using the normal distribution to approximate the binomial, with  5 9.0 and
 5 1.897, this probability is expressed as P(10.5 # x # 11.5). The z-score
associated with x 5 11.5 is
z 5
x 2 
______

5
11.5 2 9.0
__________
1.897
5 1.32
The z-score associated with x 5 10.5 is
z 5
x 2 
______

5
10.5 2 9.0
__________
1.897
5 0.79
Referring to the standard normal table, we find the cumulative area to
z 5 1.32 is 0.9066 and the cumulative area to z 5 0.79 is 0.7852. As Figure 7.11
shows, P(10.5 # x # 11.5) is the difference between these areas, or 0.9066 2
0.7852 5 0.1214.
Note that this approximation is quite close to the actual binomial probability of 0.1268, obtained from the binomial probability tables in Appendix
A. Keep in mind, however, that this example was one in which both  and n
happened to be available in these tables. This will not always be the case, hence
the usefulness of being able to use the normal distribution as an approximation
to the binomial.
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEX226 Part 2: Probability
What Is the Probability That There Will Be at Least 10 Females in the
Group of 15?
Again using the normal distribution to approximate the binomial, the probability of interest is now expressed as P(x $ 9.5). (The continuity correction has
replaced the discrete value of 10 with the interval from 9.5 to 10.5.) The z-score
associated with x 5 9.5 is
z 5
x 2 
______

5
9.5 2 9.0
_________
1.897
5 0.26
Referring to the standard normal table, we see that the cumulative area to
z 5 0.26 is 0.6026. Because the total area beneath the normal curve is 1.0000,
the area to the right of z 5 0.26 is (1.0000 2 0.6026) 5 P(x $ 9.5) 5 0.3974.
The normal approximation is again very close to the actual probability (0.4032)
obtained from the binomial tables.
The normal approximation to the binomial distribution is quite close whenever
both n and n(1  ) are $5. However, the approximation becomes even better
as the value of n increases and whenever  is closer to 0.5. Also, when we are
determining probabilities for x being within or beyond a given interval, the correction for continuity becomes less important when x can take on a great many
different values. For example, if n 5 1000 and  5 0.3, adding and subtracting
0.5 from each x value will make virtually no difference in the results. Refer to
Seeing Statistics Applet 6, at the end of the chapter, for a visual demonstration of
the normal approximation to the binomial distribution.
FIGURE 7.11
Normal approximation to a
binomial distribution with
n 5 15 and  5 0.60. The
continuity correction represents
x 5 11 as the interval from
10.5 to 11.5, then the
standard normal table is used
in obtaining the approximate
value of 0.1214 for P(x 5 11).
0
0.00
0.05
0.10
0.15
0.20
0.25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
x = Number of females in the group of 15
f(x)
Using the normal approximation with m = 9.0 and s = 1.897,
P(x = 11) is estimated as P(10.5 ≤ x ≤ 11.5), or P(0.79 ≤ z ≤ 1.32),
Shaded area =
0.9066 (Cumulative area to z = 1.32)
–0.7852 (Cumulative area to z = 0.79)
= 0.1214
MPLEEXAMPLEEXAMPLEChapter 7: Continuous Probability Distributions 227
7.39 What is the correction for continuity and why is
it employed when the normal distribution is used in
approximating the binomial distribution?
7.40 Under what circumstances is it permissible to use
the normal distribution in approximating the binomial
distribution?
7.41 In a certain binomial distribution,  5 0.25 and
n 5 40. In using the normal approximation,
a. What are the mean and standard deviation of the
corresponding normal distribution?
b. If x 5 the number of “successes” among the 40
observations, determine the following: P(x 5 8),
P(12 # x # 16), P(10 # x # 12), P(x $ 14).
7.42 In a certain binomial distribution,  5 0.30 and
n 5 20. In using the normal approximation,
a. What are the mean and standard deviation of the
corresponding normal distribution?
b. If x 5 the number of “successes” among the 20
observations, determine the following: P(x 5 5),
P(4 # x # 7), P(1 # x # 5), P(x $ 7).
7.43 Approximately 80% of American families have
some form of life insurance coverage. For a randomly
selected sample of n 5 15 families, and the discrete random variable x 5 the number of families in the sample
who are covered by life insurance: Source: American Council
of Life Insurers, 2008 Life Insurers Fact Book, p. 78.
a. Calculate the mean and the standard deviation of the
binomial distribution.
b. Using the binomial distribution, calculate the probability that x will be exactly 12 families.
c. Repeat part (b), using the normal approximation to
the binomial. (Hint: The area of interest under the
normal curve will be between x 5 11.5 and
x 5 12.5.)
d. Using the normal approximation to the binomial,
find the probability that at least 10 families in
the sample will be covered by some form of life
insurance.
7.44 About 40% of new single-family homes completed
in the United States during 2007 were heated with electricity. For a randomly selected sample of 20 new singlefamily homes completed during that year and the discrete
random variable, x 5 the number of homes in this group
heated with electricity: Source: Statistical Abstract of the United
States 2009, p. 593.
a. Calculate the mean and standard deviation of the
binomial distribution.
b. Using the binomial distribution tables, determine the
probability that x will be exactly 8 homes heated
with electricity.
c. Using the binomial distribution tables, determine the
probability that x will be at least 6 but no more than
9 homes.
d. Repeat part (b), using the normal approximation to
the binomial distribution.
e. Repeat part (c), using the normal approximation to
the binomial distribution.
7.45 Of all individual tax returns filed in the United
States during the 2009 tax filing season, 15.8% were
prepared by H&R Block. For a randomly selected sample
of 900 tax returns filed during this period, use the normal
approximation to the binomial distribution in determining the probability that between 110 and 140 of these
returns were prepared by H&R Block. Source: hrblock.com,
July 14, 2009.
7.46 The Electronic Industries Association reports that
about 50% of U.S. households have a camcorder. For a
randomly selected sample of 800 U.S. households, use
the normal approximation to the binomial distribution
in determining the probability that at least 410 of these
households have a camcorder. Source: New York Times
Almanac 2009, p. 409.
7.47 Repeat Exercise 7.45, but without using the
correction for continuity (i.e., the discrete random
variable, x, is not replaced by the interval x  0.5 to
x 1 0.5). Does your answer differ appreciably from that
obtained in Exercise 7.45? If not, comment on the need
for using the correction for continuity when x can take
on a large number of possible values.
7.48 Repeat Exercise 7.46, but without using the correction for continuity (i.e., the discrete random variable,
x, is not replaced by the interval x  0.5 to x 1 0.5).
Does your answer differ appreciably from that obtained
in Exercise 7.46? If not, comment on the need for using
the correction for continuity when x can take on a large
number of possible values.
E X E R C I
S E S228 Part 2: Probability
THE EXPONENTIAL DISTRIBUTION
Description and Applications
Chapter 6 discussed the Poisson distribution, in which the discrete random
variable was the number of rare events occurring during a given interval of time,
space, or distance. For a Poisson process, a distribution called the exponential
distribution describes the continuous random variable, x 5 the amount of time,
space, or distance between occurrences of these rare events. For example, in the
context of arrivals at a service counter, x will be a continuous random variable
describing the time between successive arrivals.
As with the Poisson distribution, the exponential distribution has application to the management science topic of queuing, or waiting-line theory. Like the
Poisson distribution, it also assumes that the arrivals are independent from one
another. The distribution can be summarized as follows:
(7.5)
The exponential distribution:
• For x 5 the length of the interval between occurrences:
f(x) 5 le2lx for x  0 and l  0 where l 5 the mean and variance
of a Poisson distribution
1l 5 the mean and standard
deviation of the corresponding
exponential distribution
e 5 the mathematical constant,
2.71828, the base of the natural
logarithm system
• Areas beneath the curve:
P(x  k) 5 e2lk where k 5 the time, space, or distance until the next
occurrence
Like the Poisson distribution, the exponential distribution is really a family
of distributions, and the particular member of the family is determined by just
one descriptor: its mean. Actually, the mean of the exponential distribution is the
inverse of that of the Poisson. For example, if the average number of arrivals at a
tollbooth is l 5 5 persons per minute, the mean of the corresponding exponential
distribution will be 1yl 5 1y5, or 0.20 minutes between persons. Note that the
Poisson random variable is discrete (number of persons), whereas the random
variable for the exponential distribution is continuous, in this case, time.
It is important that we be consistent in our use of physical-measurement units
when we invert  to arrive at the mean of the corresponding exponential distribution. For example, if arrivals tend to occur at the rate of 15 vehicles per hour, we
can express this as 15 arrivals divided by 60 minutes, or 0.25 arrivals per minute.
If we then consider  5 0.25 arrivals per minute as the Poisson mean, the mean
of the corresponding exponential distribution must also be based on minutes, and
1yl will be 1y0.25, or 4 minutes between arrivals.
NOTEChapter 7: Continuous Probability Distributions 229
FIGURE 7.12
An exponential distribution
in which the mean time
between occurrences is
6 minutes. The probability is
0.4347 that at least 5 minutes
will elapse before the next call
to the “911” switchboard.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1/6
f (x)
x = Minutes between calls to "911"
= 0.4347
P(x ≥ 5) = e–(1/6)(5) = 2.71828–0.833
EXAMPLE
Exponential Distribution
Calls to the “911” emergency number of a large community have been found to
be Poisson distributed with an average of 10 calls per hour.
SOLUTIONS
Determine the Mean and Standard Deviation of the Corresponding
Exponential Distribution
The mean of the Poisson distribution is  5 10 calls per hour, and the mean of the
corresponding exponential distribution is the inverse of this, 1y 5 1y10, or 0.10
hours between calls. The standard deviation is numerically equal to the mean.
For greater clarity in answering the questions that follow, we will express
the means of the Poisson and exponential distributions using minutes instead of
hours. Accordingly, the mean of the Poisson distribution is 10 calls per 60 minutes, or l 5 1y6 calls per minute. The mean of the corresponding exponential
distribution is 1yl 5 6 minutes between calls.
What Is the Probability That the Next Call Will Occur at Least 5 Minutes
from Now?
Referring to the continuous probability distribution of Figure 7.12, this probability corresponds to the area to the right of k 5 5 minutes on the x axis. Applying
the formula for determining areas beneath the curve:
P(x $ k) 5 e2lk where k 5 5 minutes
or
P(x $ 5) 5 e2(1y6)(5) 5 2.7182820.833 or 0.4347
There is a 0.4347 chance that at least 5 minutes will elapse before the next
call is received. Alternatively, the probability that the next call will occur within
the next 5 minutes is (1.000 2 0.4347), or 0.5653. This is because (1) the total
area beneath the curve is 1.0, and (2) the exponential distribution is continuous
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE230 Part 2: Probability
and the probability is zero that a continuous random variable will take on any
exact value [i.e., P(x 5 5) is 0.0].
What Is the Probability That the Next Call Will Occur Between 3 and
8 Minutes from Now?
The area beneath the curve in the interval 3 # x # 8 can be determined by subtracting the area representing x $ 8 from the area representing x $ 3. This is
shown in Figure 7.13 and is calculated as
P(3 # x # 8) 5 P(x $ 3) 2 P(x $ 8) 5 e2(1y6)(3) 2 e2(1y6)(8)
5 2.7182820.500 2 2.7182821.333 or 0.6065 2 0.2637 5 0.3428
Excel and Minitab can be used to obtain cumulative exponential probabilities
associated with specified values for x—the procedures and results are shown in
Computer Solutions 7.3. These software packages can also be used in doing the
reverse—that is, specifying a cumulative probability, then finding the x value associated with it. These procedures and results are shown in Computer Solutions 7.4.
XAMPLEEXAMPLEEXAMPLE
Exponential Probabilities
These procedures show how to get cumulative exponential probabilities.
For each software package:
Procedure I: Getting the cumulative probability associated with one x value.
Procedure II: Getting cumulative probabilities associated with a set of x values.
COMPUTER 7.3 SOLUTIONS
FIGURE 7.13
For the exponential distribution
in which the mean time
between calls is 6 minutes,
the probability is 0.3428 that
between 3 and 8 minutes will
elapse before the next call to
the “911” switchboard.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1/6
f(x)
x = Minutes between calls to "911"
= 0.3428
P(3 ≤ x ≤ 8) = P(x ≥ 3) – P(x ≥ 8) and shaded area =
0.6065 (Area to the right of x = 3)
–0.2637 (Area to the right of x = 8)Chapter 7: Continuous Probability Distributions 231
EXCEL
I. Excel procedure for the cumulative probability associated with one x value
1. Click on the cell into which Excel will place the probability. From the Formulas ribbon, click on the Insert
Function symbol at the extreme left. Click Statistical in the Category box. Click EXPONDIST in the Function
box. Click OK.
2. Enter the x value of interest (e.g., 5) into the X box. Enter the inverse of the exponential mean (e.g., if the mean
is 6 minutes between calls, enter 1y6, or 0.16667) into the Lambda box. Enter True into the Cumulative box.
Click OK.
II. Excel procedure for cumulative probabilities associated with a set of x values
1. Enter the labels into A3:B3 as shown in the printout above. Enter the x values of interest (e.g., 7, 6, 5) starting with
cell A4. Click on cell B4.
2. From the Formulas ribbon, click on the Insert Function symbol at the extreme left. Click Statistical in the
Category box. Click EXPONDIST in the Function box. Click OK.
3. Enter A4 into the X box. Enter the inverse of the exponential mean (e.g., 1/6, or 0.16667) into the Lambda box.
Enter True into the Cumulative box. Click OK. The first cumulative probability will appear in cell B4.
4. Click at the lower right corner of B4 (the cursor will turn to a “1” sign) and drag downward so cells B4:B6 will
contain the set of cumulative probabilities.
MINITAB
I. Minitab procedure for cumulative probability associated with one x value
1. Click Calc. Select Probability Distributions. Click Exponential.
2. In the Exponential Distribution menu, enter the mean of the distribution (e.g., 6) into the Mean or Scale box.
Select Input constant and enter the x value of interest (e.g., 5) into the box. Select Cumulative probability.
Click OK.
II. Minitab procedure for cumulative probabilities associated with a set of x values
Enter the desired x values (e.g., 7, 6, 5) into column C1. Follow steps 1 and 2 above, except select Input column
and enter C1 into the box.232 Part 2: Probability
Inverse Exponential Probabilities
For Minitab only. Excel does not currently offer inverse exponential probabilities.
MINITAB
I. Minitab procedure for the x value associated with one cumulative probability
1. Click Calc. Select Probability Distributions. Click Exponential.
2. In the Exponential Distribution menu, enter the mean of the distribution (e.g., 6) into the Mean or Scale box.
Select Input constant and enter the cumulative probability of interest (e.g., 0.8000) into the box. Select Inverse
cumulative probability. Click OK.
II. Minitab procedure for the x values associated with a set of cumulative probabilities
Enter the cumulative probabilities of interest (e.g., 0.9750, 0.5000, 0.2500) into column C1. Follow steps 1 and 2
above, except select Input column and enter C1 into the box.
COMPUTER 7.4 SOLUTIONS
7.49 What is the relationship between the Poisson distribution and the exponential distribution?
7.50 Every day, drivers arrive at a tollbooth. If the
Poisson distribution were applied to this process, what
would be an appropriate random variable? What would
be the exponential-distribution counterpart to this
random variable?
7.51 The main switchboard at the Home Shopping
Network receives calls from customers. If the Poisson
distribution were applied to this process, what would
be an appropriate random variable? What would be the
exponential-distribution counterpart to this random
variable?
7.52 A random variable is Poisson distributed with
l 5 1.5 occurrences per hour. For the corresponding
exponential distribution, and x 5 hours until the next
occurrence, identify the mean of x and determine the
following:
a. P(x $ 0.5) b. P(x $ 1.0)
c. P(x $ 1.5) d. P(x $ 2.0)
7.53 A random variable is Poisson distributed with
l 5 0.02 occurrences per minute. For the corresponding
exponential distribution, and x 5 minutes until the next
occurrence, identify the mean of x and determine the
following:
a. P(x $ 30.0) b. P(x $ 40.0)
c. P(x $ 50.0) d. P(x $ 60.0)
E X E R C I
S E SChapter 7: Continuous Probability Distributions 233
7.54 A random variable is Poisson distributed with
 5 0.50 arrivals per minute. For the corresponding
exponential distribution, and x 5 minutes until the next
arrival, identify the mean of x and determine the
following:
a. P(x # 0.5) b. P(x # 1.5)
c. P(x $ 2.5) d. P(x $ 3.0)
7.55 The owner of a self-service carwash has found that
customers take an average of 8 minutes to wash and dry
their cars. Assuming that the self-service times tend to be
exponentially distributed, what is the probability that a
customer will require more than 10 minutes to complete
the job?
7.56 A taxi dispatcher has found that successive calls for
taxi service are exponentially distributed, with a mean time
between calls of 5.30 minutes. The dispatcher must disconnect the telephone system for 3 minutes in order to have
the push-button mechanism repaired. What is the probability that a call will be received while the system is out of
service?
7.57 During 2008, U.S. general aviation pilots had 1.20
fatal crashes per 100,000 flying hours. Harriet Arnold
is president of Arnold’s Flying Service, a company that
operates a total of 50 sightseeing planes based in 20
regions of the United States. Altogether, the planes in
this fleet are in the air about 40,000 hours per year. Ever
since a fatal crash last year involving one of the company’s planes that took off under questionable weather
conditions, the company has been getting close scrutiny
from both government safety officials and the national
media. Harriet believes her planes are just as safe as
anyone else’s, but is concerned that even one more fatal
accident within the coming year might be too much for
the now-struggling company to overcome. Assuming
an exponential distribution for x 5 thousands of flying
hours between fatal crashes, what is the probability that
Arnold’s Flying Service will not experience a fatal crash
until at least a year from today? Until at least 2 years
from today? What number of flying hours would be associated with a 90% probability of experiencing no crashes
within that number of flying hours from today? Source:
General Aviation Manufacturers Association, General Aviation Statistical
Databook, 2008 Edition, p. 58.
7.58 The Bureau of Labor Statistics reports the fatality
rate for workers in the rail transportation industry is
0.006 fatalities per 200,000 worker-hours. During 2007,
rail transportation workers put in a total of 545 million
worker-hours. Assuming that (1) an exponential distribution applies for the number of worker-hours between
fatalities in this industry, and (2) the annual 545 million
worker-hours of activity continues in the industry, what
is the expected number of worker-hours until the next
fatality occurs in this industry? If a given rail transportation company requires 1,000,000 worker-hours each
year, what is the probability that the next fatal injury in
this company will not occur until at least 30 years from
now? Source: Bureau of Labor Statistics, Census of Fatal Occupational
Injuries, 2007.
SIMULATING OBSERVATIONS FROM A CONTINUOUS
PROBABILITY DISTRIBUTION
Although practically all statistical packages can provide probabilities like those
in the Computer Solutions earlier in this chapter, some can also randomly
select observations from the continuous distribution itself. For example, Computer Solutions 7.5 lists the procedures and shows the results when Excel and
Minitab are used in randomly selecting 10,000 observations from a normal
distribution. Both sets of simulated observations are from a normal distribution examined earlier in the chapter—in this situation, the continuous random
variable is x 5 the amount of weight added so that a generator shaft would
have proper static and dynamic balance. In the probability distribution, the
average amount of weight added is  5 35 grams and the standard deviation
is  5 9 grams.
In each of the printouts in Computer Solutions 7.5, the mean and standard
deviation of the simulated observations are very close to the actual mean and
standard deviation of the theoretical distribution from which the 10,000 observations were drawn. In addition to differing slightly from the theoretical values,
the Excel and Minitab sample means and standard deviations also differ slightly
(7.6)234 Part 2: Probability
Simulating Observations from a Continuous Probability Distribution
These procedures simulate 10,000 observations from a normal distribution with mean 5 35 and standard
deviation 5 9. The procedures for simulating observations from an exponential distribution will differ only
slightly from those described here.
EXCEL
1. Start with a blank worksheet. From the Data ribbon, click Data Analysis in the rightmost menu section. Click
Random Number Generation. Click OK.
2. Enter 1 into the Number of Variables box. Enter 10000 into the Number of Random Numbers box. Select Normal
within the Distribution box. Enter 35 into the Mean box. Enter 9 into the Standard Deviation box. Select Output
Range and enter A1 into the box. Click OK. The 10,000 simulated observations will be located in A1:A10000.
3. The Excel descriptive statistics and histogram above can be generated using the procedures in Computer
Solutions 2.1 and 3.1.
MINITAB
COMPUTER 7.5 SOLUTIONSChapter 7: Continuous Probability Distributions 235
from each other. This is because each package generated its own sample of
10,000 observations, and each sample had a different set of observations from
the other.
According to the law of large numbers, histograms like those in Computer
Solutions 7.5 will more closely approach the theoretical distribution from
which they were drawn whenever the number of observations becomes larger.
The topic of how samples can differ from each other, and from the population
from which they were drawn, will be an important topic in the next two chapters.
1. Click Calc. Select Random Data. Click Normal. Enter 10000 into the Number of rows of data to generate box.
Enter C1 (the destination column) into the Store in columns box. Enter 35 into the Mean box. Enter 9 into the
Standard deviation box. Click OK. The 10,000 simulated observations will be located in column C1.
2. The Minitab descriptive statistics and histogram shown above can be generated using the procedures in Computer
Solutions 2.1 and 3.1.
7.59 A computer statistical package has simulated 1000
random observations from a normal distribution with
 5 50 and  5 10. Sketch the approximate box-andwhisker display for the resulting data.
7.60 A computer statistical package has simulated 2000
random observations from a normal distribution with
 5 80 and  5 20. Sketch the approximate box-andwhisker display for the resulting data.
7.61 If a computer statistical package were to simulate
2000 random observations from a normal distribution
with  5 150 and  5 25, what percentage of these
observations would you expect to have a value of 140 or
less? Do you think the actual number in the “#140” range
would equal the expected number in this range? If so,
why? If not, why not?
7.62 If a computer statistical package were to simulate
1000 random observations from a normal distribution
with  5 200 and  5 50, what percentage of these
observations would you expect to have a value of 300 or
more? Do you think the actual number in the “$300”
range would equal the expected number in this range? If
so, why? If not, why not?
E X E R C I
S E S
SUMMARY
• Nature and applications of the normal distribution
Continuous probability distributions describe probabilities associated with random variables that can take on any value along a given range or continuum and
for which there are no gaps between these possible values. The curve f(x) is a
function of the continuous random variable, and f(x) is called the probability
density function. For any specified interval of values, the probability that x will
(7.7)236 Part 2: Probability
assume a value within the interval is the area beneath the curve between the two
points describing the interval. The total area beneath the curve is 1.0, and the
probability that x will take on any specific value is 0.0.
The most important continuous distribution is the normal distribution. This
is because many natural and economic phenomena tend to be approximately
normally distributed; it can be used to approximate other distributions, such as
the binomial; and the distribution of sample means and proportions tends to be
normally distributed whenever repeated samples are taken from a given population of any shape.
The normal distribution is a symmetrical, bell-shaped curve, and the mean,
median, and mode are located at the same position. The curve is asymptotic in
that each end approaches the horizontal (x) axis, but never reaches it. The normal
distribution is actually a family of such distributions, and the specific member of
the family is determined by the mean () and standard deviation (). Regardless
of the shape of a particular normal curve, about 68.3% of the area is within the
interval  6 , 95.5% within  6 2, and 99.7% within  6 3.
• Standard normal distribution and z-scores
When the original x values are expressed in terms of the number of standard
deviations from the mean, the result is the standard (or standardized) normal distribution in which the z value (or z-score) replaces x. The standard normal distribution has a mean of 0 and a standard deviation of 1, and cumulative areas beneath
its curve are readily available from the standard normal table.
• Normal distribution to approximate the binomial distribution
Although the binomial distribution is discrete and the normal distribution is
continuous, the normal distribution is a good approximation to the binomial
whenever both n and n(1 2 ) are $ 5. The normal approximation is useful
whenever suitable binomial tables are not available, and makes it unnecessary to
engage in the tedious computations that the binomial distribution would otherwise require.
Because the binomial distribution has gaps between possible values of x,
the normal approximation to the binomial involves a correction for continuity,
which consists of “expanding” each possible value of x by 0.5 in each direction.
This correction becomes less important whenever the number of trials is large
and we are finding the probability that x lies within or beyond a given range of
values.
• Exponential distribution
For a Poisson process, the exponential distribution describes the continuous random variable, x 5 the amount of time, space, or distance between occurrences of
the rare events of interest. The exponential distribution is a family of distributions, all of which are positively skewed, and the specific member is determined
by the value of the mean.
• Computer-generated probabilities and simulations
As with discrete probability distributions, computer statistical packages can
determine the probabilities associated with possible values of x. In the case of
continuous distributions, this includes finding the exact area beneath a specified
portion of the curve, or probability density function, and eliminates the need
for calculations or table references. Some packages, such as Excel and Minitab,
can also simulate the random selection of observations from the distribution
itself.Chapter 7: Continuous Probability Distributions 237
The Normal Distribution
f(x) 5 ______ 1
Ï
___
2
e2(1y2)[(x2)y]2 where  5 mean
 5 standard deviation
e 5 the mathematical constant, 2.71828
 5 the mathematical constant, 3.14159
The z-Score for a Standard Normal Distribution
z 5
x 2 
_____

where z 5 the distance from the mean, measured in standard
deviation units
x 5 the value of x in which we are interested
 5 the mean of the distribution
 5 the standard deviation of the distribution
The Normal Approximation to the Binomial Distribution
• Mean:
 5 n
• Standard deviation:
 5 Ï
_________
n(1 2 ) where n 5 the number of trials
 5 the probability of success
on any given trial
x 5 the number of successes in n trials
(Note: This approximation is used when n and n(1 2 ) are both  5.)
The Exponential Distribution
• For x 5 the length of the interval between occurrences:
f(x) 5 le2lx for x . 0 and l . 0 where l 5 the mean and variance
of a Poisson distribution
1yl 5 the mean and standard
deviation of the corresponding
exponential distribution
e 5 the mathematical constant,
2.71828, the base of the
natural logarithm system
• Areas beneath the curve:
P(x  k) 5 e2lk where k 5 the time, space, or distance until the
next occurrence
E Q U A T I O N S
7.63 According to a recent survey, 35% of the adult
citizens of Mayville have a charge account at Marcy’s
Department Store. Mr. Marcy’s nephew is going to trial
for embezzlement, and nine jurors are to be randomly
selected from the adult citizenry. What is the probability
that at least four of the jurors will be charge-account
holders?
7.64 A food expert claims that there is no difference
between the taste of two leading soft drinks. In a taste
CHAPTER EXERCISES238 Part 2: Probability
test involving 200 persons, however, 55% of the subjects
say they like soft drink A better than B. If the food
expert is correct, we would expect that 50% of the
persons would have preferred soft drink A. Under the
assumption that the food expert is correct, what would
be the probability of a “taste-off” in which 110 or more
of the subjects preferred soft drink A?
7.65 In Exercise 7.64, the maker of soft drink A claims its
product to be “superior” to soft drink B. Given the results
obtained in Exercise 7.64, do they appear to have a solid
basis for their claim?
7.66 During an annual heating season, the average gas
bill for customers in a New England community heating
their homes with gas was $457. Assuming a normal distribution and a standard deviation of $80:
a. What proportion of homes heating with gas had a
gas bill over $382?
b. What proportion of homes heating with gas had a
gas bill between $497 and $537?
c. What amount was exceeded by only 2.5% of the
homes heating with gas?
d. What amount was exceeded by 95% of the homes
heating with gas?
7.67 It has been reported that the average monthly cell
phone bill is $50. Assuming a normal distribution and a
standard deviation of $10, what is the probability that a
randomly selected cell phone subscriber’s bill last month
was less than $35? More than $70? Source: Statistical
Abstract of the United States 2009, p. 706.
7.68 During 2004, the average number of flying hours for
aircraft operated by regional airlines in the United States
was 2389. Assume a normal distribution and a standard
deviation of 300 hours. Source: Statistical Abstract of the United
States 2006, p. 701.
a. What proportion of aircraft flew more than 2200
hours during the year?
b. What proportion of aircraft flew between 2000 and
2400 hours during the year?
c. What number of annual flying hours was exceeded
by only 15% of the aircraft?
d. What number of annual flying hours was exceeded
by 75% of the aircraft?
7.69 During fiscal 2008, the average daily volume for
FedEx Corporation was 7,000,000 packages per day.
Assuming a normal distribution and a standard deviation of 800,000 packages per day, on what proportion
of the days was the volume between 6,000,000 and
6,500,000 packages? Source: FedEx Corporation, 2008 Annual
Report, p. 28.
7.70 For itemized tax returns in the $60,000–$75,000
income group for the most recent year reported, the average charitable contribution was $1935. Assume a normal
distribution and a standard deviation of $400.
Source: Statistical Abstract of the United States 2002, p. 360.
a. For a randomly selected household from this income
group, what is the probability that the household
made charitable contributions of at least $1600?
b. For a randomly selected household from this income
group, what is the probability that the household
made charitable contributions of between $2200
and $2400?
c. What level of contributions was exceeded by only
20% of the itemized returns in this income group?
d. What level of contributions was exceeded by 40% of
the itemized returns in this income group?
7.71 Of the 1.0 million persons of voting age in Maine,
72% voted in the 2008 presidential election. For a randomly selected group of 30 Maine residents who were
of voting age at that time, what is the probability that
at least 20 of these persons voted in the presidential
election? Source: nonprofitvote.org, July 15, 2009.
7.72 A public relations agency tells its client that 80%
of the residents in a 50-mile radius believe the company
is “an industry leader.” Skeptical, the company commissions a survey in which just 320 of the 500 persons
interviewed felt the company was an industry leader. If
the public relations agency’s claim is correct, what is the
probability that interviews with 500 people would yield
320 or fewer who felt the company was an industry
leader?
7.73 Given the results of Exercise 7.72, evaluate the
claim made by the public relations agency.
7.74 A researcher is studying a particular process and
has identified the following times, in minutes, between
consecutive occurrences: 1.8, 4.3, 2.3, 2.1, 5.6, 1.5,
4.5, 1.8, 2.4, 2.6, 3.2, 3.5, 1.6, 2.1, and 1.9. Given
this information, what would be the researcher’s best
estimate for the mean of the exponential distribution?
What would be her best estimate for the mean of the
corresponding Poisson distribution?
7.75 A safety researcher has found that fourth to sixth
graders engaging in unorganized activities on school
grounds experienced injuries at the rate of 2.4 injuries
per 100 thousand student-days. On the average, how
many student-days elapse between injuries to fourth
to sixth graders in unorganized activities at school,
and what is the probability that the next such injury
will occur before 45 thousand student-days have
passed?
7.76 The number of defects in rolls of aluminum sheet
average 3.5 defects per 100 feet. The company president is accompanying overseas visitors on a tour of the
plant, and they have just arrived at the aluminum-sheet
machine. If they remain at the machine long enough to
observe 32 feet of aluminum being produced, what is the
probability that they will have left the area before the
next defect occurs?Chapter 7: Continuous Probability Distributions 239
7.77 The mileage death rate for motorcycle riders has
been estimated to be about 39 deaths per 100 million
miles of motorcycle travel. A national motorcycling
association has 1200 members who travel a total of
2 million miles each year on their motorcycles. What
is the probability that the next highway fatality experienced by the association will not occur until more than
a year from now? Until more than 2 years from now?
Source: USAToday.com, July 30, 2006.
7.78 The “20-ounce” package of mozzarella cheese distributed by a food company actually has a weight that is normally distributed, with  5 21 ounces and  5 0.2 ounces.
a. What is the probability that a randomly selected
package of this cheese will actually contain at least
20.5 ounces?
b. What is the probability that a randomly selected
package of this cheese will actually contain between
20.5 and 21.3 ounces?
c. A shopper randomly selects 8 of these packages.
What is the probability that 3 or more of them will
contain at least 21.2 ounces of cheese?
7.79 State Police Academy graduates from a large graduating class have scored an average of 81.0 on a proficiency test, and the scores are normally distributed with
 5 8.5. Graduates are randomly paired and assigned
to barracks throughout the state. Madeleine has scored
83.0 on the test. What is the probability that she will be
paired with someone whose score was within 5.0 points
of her own?
7.80 The trainer for a professional football team has
found that, on the average, his team experiences 2.3
sprained ankles per 1000 game minutes. If a game
lasts 60 minutes, and the number of game minutes
between ankle sprains is exponentially distributed,
determine the probability that the next ankle sprain
will occur
a. during the team’s next game.
b. sometime during the team’s next ten games.
c. at the moment the 2-minute warning is sounded in
the next game.
7.81 The precooked weight of hamburgers at a gourmet
hamburger restaurant has been normally distributed
with a mean of 5.5 ounces and a standard deviation of
0.15 ounces. A dining-column journalist from the local
newspaper enters the restaurant and orders a hamburger.
What is the probability that he will receive a hamburger
that had a precooked weight less than 5.3 ounces? If, in
a separate visit to the restaurant, the journalist is accompanied by three colleagues who also order a hamburger,
what is the probability that at least 2 of these 4 customers
will receive a hamburger that had a precooked weight
greater than 5.7 ounces?
7.82 Boxes are filled with sugar by a machine that is
set to deliver an average of 20.3 ounces. The weights
are normally distributed with a standard deviation of
0.3 ounces. If a consumer advocate buys a package of the
product and then proceeds to weigh the contents, what is
the probability that the package she has purchased will
have a content weight between 20 and 21 ounces? If the
consumer advocate plans to purchase 100 boxes of the
product, then file suit against the company if more than
5 of the boxes have content weights below the 20-ounce
stated weight on the package, what is the probability that
she will end up filing suit against the company?
7.83 In Exercise 7.82, the company would like to put
in a greater weight “cushion” to help protect itself from
consumer advocates. If the company wants to have just
2% of the packages contain less than 20 ounces, to what
average weight must the filling machine be set?
7.84 Discount Micros, a computer mail-order company,
has received 2000 desktop computers from a major
manufacturer. Unknown to the mail-order company,
250 of the computers have a hard disk that was improperly installed at the factory. A consulting firm purchases
40 computers from Discount Micros. If the order is
filled by random selection from the 2000 computers
just received, what is the probability that the shipment
to the consulting firm will include no more than 2 defective computers? At least 1, but no more than 3 defective
computers?
7.85 The U-Drive car rental corporation has found that
the cars in the company’s rental fleet experience punctures at the rate of 1.25 punctures per 10,000 miles, and
the number of punctures per distance traveled is Poisson
distributed. Ed and Harriet are taking advantage of a
U-Drive promotion in which there is no mileage charge.
During their vacation, they plan to drive from Boston to
Los Angeles and back, a total distance of 6164 miles.
a. What is the probability that Ed and Harriet will not
have to change any tires during their vacation?
b. While traveling west, what is the probability that Ed
and Harriet will experience a punctured tire before
they make it to Denver, 2016 miles away?
c. For what mileage is the probability 0.80 that Ed and
Harriet will experience a puncture before they reach
this distance?
7.86 Bonnie Rogers, the customer relations manager for
a large computer manufacturer, has recently studied the
duration of telephone calls received by the company’s
technical support department. The call lengths are exponentially distributed, with a mean duration of 8 minutes.
Experience has taught Bonnie that lengthy calls usually
can be more efficiently handled by a technical support
manager, rather than by the less experienced personnel
who receive the initial technical inquiry. To avoid having the higher-paid managers virtually take over the
task of handling technical support, Bonnie would like
to have only the longest 10% of the calls redirected to a240 Part 2: Probability
technical support manager. How long must a technical
support call last before it qualifies for redirection to a
technical support manager?
7.87 A storage warehouse in a relatively remote part of
a county is protected by a homemade burglary detection
system. Once actuated by an intruder, the system sounds
a horn and flashes a light for 15 minutes, then saves battery power by shutting down and resetting itself until the
next intrusion is detected. Police in three patrol cars have
this part of the county as one of their responsibilities,
and they routinely drive by the warehouse facility during the hours from dark to dawn. On average, a patrol
car passes the warehouse every 20 minutes, and the time
between patrol car arrivals is exponentially distributed.
It’s the middle of the night and a burglar has just broken
into the storage warehouse and set off the alarm. What
is the probability that the alarm will shut off before the
next police patrol car drives by?
Since Thorndike “Graph-Pro” racquetball racquets were
introduced several years ago, no attempt has ever been
made to offer a lightweight version. Ted has done some
research and found that some of the other manufacturers
offer a lighter version of their standard racquet. In addition, some of the racquetball players he has talked to have
said a lighter racquet would probably improve their game.
Talking with production engineers, Ted learns that the
racquets coming off the line don’t all weigh the same anyway. According to the chief engineer, the racquets weigh
an average of 240 grams, but the weights “vary all over
the lot,” and he takes advantage of the situation to press
for a high-tech $700,000 machine that would produce
racquets of equal quality while holding this variation
within narrower limits.
Old Luke isn’t too keen on buying a $700,000 machine, especially since Ted might have a better solution.
Ted has dusted off the literature that accompanied the
racquet-production machine and found that the machine
generates output that is approximately normally distributed, with a standard deviation of 10 grams. This is a little
sloppy these days, but maybe Thorndike Sports Equipment can turn this weakness into a strength.
Conversing further with Luke, talking to three sportinggoods retailers, and interviewing players at five different
racquetball facilities, Ted comes back to the office with a
terrific idea. Asking Luke to sit down, and bringing him
some coffee, Ted begins his pitch: “Say, grandfather, why
don’t we take the racquets at the lighter end of the range
and label them ‘Graph-Pro Light,’ with the ones more in
the middle being ‘Graph-Pro Regular,’ and the ones at the
heavier end can be ‘Graph-Pro Stout’?”
Luke responds that just because this works for the beer
companies, this doesn’t mean it will work for Thorndike
Sports Equipment. However, he’s willing to give it a try.
After accompanying Ted on another round of conversations with retailers and players, the elder Thorndike tells
Ted that for now, he’d like to see 15% of the production
consist of “Graph-Pro Light,” with 80% “Graph-Pro
Regular,” and 5% “Graph-Pro Stout.”
Ted’s next task is to tell the production people how
to go about selecting which racquets are to be put into
these three different categories. He’s also trying to follow
his grandfather’s directive to come up with a better name
for the “Stout” model, but he figures he can do that after
lunch.
Thorndike Sports Equipment
(Corresponds to Thorndike Video Unit Three)
INTEGRATED CAS
ES
Thorndike Golf Products Division
During an 8-hour shift of golf-ball production, one golf
ball is randomly selected from each 2 minutes’ worth of
output. The ball is then tested for “liveliness” by rolling it
down a grooved, stainless-steel surface. At the bottom, it
strikes a smooth iron block and bounces backward, up a
gradually sloped, grooved surface inscribed with distance
markings. The higher the ball gets on the rebound surface,
the more lively it is.
The production machine is set so that the average rebound distance will be  5 30.00 inches,
and the standard deviation is  5 2.00 inches. The
scores are normally distributed.1. A driving range has just purchased 100 golf balls. Use
the computer and the binomial distribution in determining the individual and cumulative probabilities
for x 5 the number of balls that scored at least 31.00
inches on the “bounce” test.
2. Repeat part (1), but use the normal approximation to
the binomial distribution. Do the respective probabilities appear to be very similar?
3. Given the distribution of bounce test scores described
above, what score value should be exceeded only 5%
of the time? For what score value should only 5% of
the balls do more poorly?
4. What is the probability that, for three consecutive
balls, all three will happen to score below the lower of
the two values determined in part (3)?
Two hundred forty golf balls have been subjected
to the bounce test during the most recent 8-hour
shift. From the 1st through the 240th, their
scores are provided in computer data file CDB07
as representing the continuous random variable,
BOUNCE. The variable BALL is a sequence from
1 to 240.
5. Using the computer, generate a line graph in which
BOUNCE is on the vertical axis and BALL is on the
horizontal axis.
6. On the graph obtained in part (5), draw two horizontal lines — one at each of the BOUNCE scores determined in part (3).
7. Examining the graph and the horizontal lines drawn
in part (6), does it appear that about 90% of the balls
have BOUNCE scores between the two horizontal
lines you’ve drawn, as the normal distribution would
suggest when  5 30.00 and  5 2.00?
8. Comparing the BOUNCE scores when BALL 5 1
through 200 to those when BALL 5 201 through 240,
does it appear that the process may have changed in
some way toward the end of the work shift? In what
way? Does it appear that the machine might be in need
of repair or adjustment? If so, in what way should the
adjustment alter the process as it appeared to exist at
the end of the work shift?
Note: Items (5) through (8) contain elements of the topic
known as statistical process control. It is an important
technique that will be examined more completely in
Chapter 20.
This applet has two normal curves, a blue one and a red
one. The blue one stays fixed with a mean of 0 and a standard deviation of 1. Using the sliders at the bottom, we can
change the mean and standard deviation of the red curve
and observe how these changes affect its position and
its shape. Moving the top slider to the left or right will
decrease or increase the mean of the red normal curve, and
moving the bottom slider to the left or right will decrease
or increase its standard deviation.
Applet Exercises
4.1 Position the top slider at the far left, then gradually
move it to the far right. What effect does this have on the
mean of the distribution?
4.2 Using the bottom slider, change the standard deviation
so that it is greater than 1. How does the shape of the red
curve compare to that of the blue curve?
4.3 Using the two sliders, position the red curve so that its
mean is as small as possible and its shape is as narrow as
possible. Compare the numerical values of the mean and
standard deviation to those of the blue curve (mean 5 0,
standard deviation 5 1).
SEEING STATISTICS: APPLET 4
Size and Shape of Normal Distribution
Try out the Applets: http://www.cengage.com/bstatistics/weiers242 Part 1: Part Title
This applet allows us to conveniently find the area corresponding to a selected interval
beneath a normal curve that
has been specified. We can
specify the mean by typing it
into the box at the lower left
(be sure to press the Return or
Enter key while the cursor is
still in the text box). Likewise,
we can specify the standard
deviation of the normal curve
by using the text box at the
lower right.
With the distribution
described by its mean and
standard deviation, we can
click on the shaded portion
of the curve and change the
left and right endpoints for
the interval. The resulting z
values and the probability
corresponding to the shaded
area are shown at the top of
the applet.
Applet Exercises
5.1 Using the text boxes, ensure that the mean is 120 and
the standard deviation is 30, as in the distribution of flying hours in Figure 7.5. (Don’t forget to hit the Enter or
Return key while the cursor is still in the text box.) Next,
ensure that the left and right boundaries of the shaded
area are 60 and 180, respectively; if necessary, click at the
left and right edges of the shaded area and drag them.
Observing the display at the top, what is the probability
that a randomly selected plane will have flown between 60
and 180 hours during the year? What values of z correspond
to 60 hours and 180 hours, respectively?
5.2 With the mean and standard deviation set at 120 and
30, respectively, drag the edges of the shaded area so that
its left boundary is at 120 and its right boundary is at 180.
What is the probability that a randomly selected plane will
have flown between 120 and 180 hours during the year?
What values of z correspond to 120 hours and 180 hours,
respectively?
5.3 By dragging the left and right edges of the shaded
area, determine the probability that a randomly selected
plane will have flown between 130 and 160 hours during
the year. What values of z correspond to 130 hours and
160 hours, respectively?
5.4 Drag the left and right edges of the shaded area so
that the left boundary corresponds to z 5 0 and the right
boundary corresponds to z 5 11.0. What probability is
now associated with the shaded area?
Normal Distribution Areas
SEEING STATISTICS: APPLET 5
Try out the Applets: http://www.cengage.com/bstatistics/weiersThis applet demonstrates the normal approximation to the
binomial distribution. We specify the desired binomial distribution by entering the sample size (n) and the population
proportion (, shown in the applet as pi). We enter the k
value of interest into the text box at the lower left. Once
again, when using the text boxes, remember to press Enter
or Return while the cursor is still within the box. Once we
have specified n, , and k, we see two results: (1) the binomial probability P(x # k), shown in the Prob text box, and
(2) the corresponding approximation using the normal distribution. Please note that probabilities shown in the applet
may sometimes differ very slightly from those we might
calculate using the pocket calculator and printed tables.
This is because the applet is using more exact values than
we are able to show within printed tables like the binomial
and normal tables in the text.
Applet Exercises
6.1 With n 5 15 and  5 0.6, as in the example in
Section 7.4, what is the probability that there are no more
than k 5 9 females in the sample of 15 walkers? How does
this compare with the corresponding probability using the
normal approximation to the binomial distribution?
6.2 With n 5 5 and  5 0.6, what is the actual binomial
probability that there are no more than k 5 3 females in
the sample of 5? How does this compare with the corresponding probability using the normal approximation?
What effect has the smaller sample size had on the closeness of the approximation?
6.3 With n 5 100 and  5 0.6, what is the actual binomial
probability that there are no more than k 5 60 females in
the sample of 100? How does this compare with the corresponding probability using the normal approximation?
What effect has the larger sample size had on the closeness of the approximation?
6.4 Repeat Applet Exercise 6.1 for k values of 6 through
12, in each case identifying the actual binomial probability
that there will be no more than k females in the sample.
For each value of k, also note the normal approximation
result and its closeness to the actual binomial probability.
SEEING STATISTICS: APPLET 6
Normal Approximation to Binomial Distribution
Try out the Applets: http://www.cengage.com/bstatistics/weiers244
Sampling Distributions
PROCESS CONTROL AND THE SAMPLING
DISTRIBUTION OF THE MEAN
C H A P T E R 8
When an industrial process is operating properly, it is said to be “in control.” However,
random variations from one sample to the next can lead to sample means that differ
both from each other and from the assumed population mean, . Applying the concepts
introduced in this chapter, we can state that if the process is in control, the means of
samples of n $ 30 should tend to be normally distributed, with x_ 5  and x_ 5 Ï__ n.
As long as the process is in control, there is a 95.5% chance that a sample mean
will be within z 5 62 standard errors of the population mean, a 99.73% chance that
it will be within z 5 63 standard errors of the population mean, and so on. Using this
approach, “control limits” can be established to monitor the production process.
For example, if a filling machine is known to have a process standard deviation of
 5 0.1 ounces and is set to fill boxes with 12.0 ounces of product, the 95.5% (“twosigma”) control limits for the means of samples of n 5 30 would be
12.0 6 2.0______ Ï0.1 ww 30C 5 12.0 6 0.037 ounces
or from 11.963 to 12.037 ounces
If sample means consistently tend
to fall outside these limits, this suggests
that the population mean may have
“drifted” from the desired 12.0 ounces
and that a filling-quantity adjustment is
in order. Such use of what is called an x}
control chart will be examined further in
Chapter 20, Total Quality Management.
Samples tend to vary in their proximity to the
population target.
Stockbyte/Getty ImagesChapter 8: Sampling Distributions 245
• Understand that the sample mean or the sample proportion can be
considered a random variable.
• Understand and determine the sampling distribution of means for
samples from a given population.
• Understand and determine the sampling distribution of proportions
for samples from a given population.
• Explain the central limit theorem and its relevance to the shape of
the sampling distribution of a mean or proportion.
• Determine the effect on the sampling distribution when the samples
are relatively large compared to the population from which they are
drawn.
LEARNING
OBJECTIVES
After reading this chapter,
you should be able to:
INTRODUCTION
In the two preceding chapters, we discussed populations in terms of (1) whether
they represented discrete versus continuous random variables and (2) their
specific probability distributions and the shape of these distributions. In both
chapters, we emphasized entire populations. More often, however, we must use
sample data to gain insight into the characteristics of a population. Using information from a sample to reach conclusions about the population from which it
was drawn is known as inferential statistics. The purpose of this chapter is to
examine how sample means and proportions from a population will tend to be
distributed, and to lay the groundwork for the inferential statistics applications
in the chapters that follow.
To underscore the pivotal importance of this chapter, let’s specifically consider it in terms of the chapters that immediately precede and follow:
• In Chapter 7, we had access to the population mean, and we made probability
statements about individual x values taken from the population.
• In this chapter, we again have access to the population mean, but we will be
making probability statements about the means of samples taken from the
population. (With this chapter, the sample mean itself is now considered as a
random variable!)
• In Chapter 9, we will lack access to the population mean, but we will begin
using sample data as the basis from which to make probability statements
about the true (but unknown) value of the population mean.
In essence, this chapter is a turning point where we begin the transition from (a)
knowing the population and making probability statements about elements taken
from it, to (b) not knowing the population, but making probability statements
about the population based on sample data taken from it.
A PREVIEW OF SAMPLING DISTRIBUTIONS
As suggested in Section 8.1, we’re now going to escalate to a new level and consider the sample mean itself as a random variable. If we were to take a great many
samples of size n from the same population, we would end up with a great many
different values for the sample mean. The resulting collection of sample means
could then be viewed as a new random variable with its own mean and standard
(8.1)
(8.2)246 Part 3: Sampling Distributions and Estimation
deviation. The probability distribution of these sample means is called the distribution of sample means, or the sampling distribution of the mean.
For any specified population, the sampling distribution of the mean is defined
as the probability distribution of the sample means for all possible samples of
that particular size. Table 8.1 provides an introduction to the concept of the sampling distribution of the mean. Referring to parts A–C of Table 8.1:
• Part A The population. Each of the four members of the population has a
given number of bottles of Diet Pepsi in his or her refrigerator. Bill has 1,
Carl has 1, Denise has 3, and Ed has 5. The probability distribution of
the discrete random variable, x 5 number of bottles of Diet Pepsi in the
refrigerator, is shown in this portion of Table 8.1, and  5 (1 1 1 1 3 1 5)y4,
or  5 2.5 bottles of Diet Pepsi.
• Part B All possible simple random samples of n 5 2 from this population,
along with the probability that each possible sample might be selected. With
the simple random sampling technique, each sample has the same probability
A. The Population, Its Probability Distribution, and Its Mean
Person x
Bill 1
Carl 1
Denise 3
Ed 5
B. All possible simple random samples of n 5 2
Sample Mean of this Sample
Probability of
Selecting this Sample
Bill, Carl x 5 (1 1 1)y2 5 1.0 1y6
Bill, Denise x 5 (1 1 3)y2 5 2.0 1y6
Bill, Ed x 5 (1 1 5)y2 5 3.0 1y6
Carl, Denise x 5 (1 1 3)y2 5 2.0 1y6
Carl, Ed x 5 (1 1 5)y2 5 3.0 1y6
Denise, Ed x 5 (3 1 5)y2 5 4.0 1y6
C. The Probability Distribution of the Sample Means (the Sampling Distribution of
the Mean)
2/6
1/6
0
0 1 2 3 4 5
x
mx
= 2.5
The mean of the sample means, x, is calculated as
x 5
1
__
6
(1.0) 1 __ 2
6
(2.0) 1 __ 2
6
(3.0) 1 __ 1
6
(4.0) 5 2.5 bottles
2
2
TABLE 8.1
Although the population is
very small, parts (A) through
(C) show the basic principles
underlying the sampling
distribution of the mean. The
discrete random variable
is x 5 the number of bottles
of Diet Pepsi in the person’s
refrigerator.
0.50
0.25
0.00
0 1 2 3 4 5
x
Mean: m = (1 + 1 + 3 + 5)/4 = 2.5Chapter 8: Sampling Distributions 247
of selection. The number of combinations of 4 elements taken 2 at a time is
6, the number of different simple random samples possible for n 5 2. In this
portion of the table, we also have the mean of each possible sample.
• Part C The probability distribution of the sample means (i.e., the sampling
distribution of the mean), along with _
x
5 the mean of the sampling distribution. Note that _
x
is the same as , or 2.5 bottles.
The sampling distribution of the mean has four important characteristics:
1. The sampling distribution of the mean will have the same mean as the original population from which the samples were drawn [i.e., E(x_) 5 _
x
5 ].
2. The standard deviation of the sampling distribution of the mean is referred
to as the standard error of the mean, or _
x
. For the sampling distribution in
Table 8.1, _
x
can be calculated in the same way as the standard deviations in
Chapter 6, as the positive square root of its variance:
2_
x
5 E[(x_ 2 )2] 5 (1.0 2 2.5)2s__ 1 6+ 1 (2.0 2 2.5)2s__ 2 6+
1 (3.0 2 2.5)2s__ 2 6+ 1 (4.0 2 2.5)2s__ 16+ 5 0.917
and
_
x
5 Ïwww 0.917 5 0.958 bottles
3. If the original population is normally distributed, the sampling distribution of
the mean will also be normal.
4. If the original population is not normally distributed, the sampling distribution of the mean will be approximately normal for large sample sizes and will
more closely approach the normal distribution as the sample size increases.
Known as the central limit theorem, this result will be an especially important
topic in the next section.
When a discrete random variable is the result of a Bernoulli process, with
x 5 the number of successes in n trials, the result can be expressed as a sample
proportion, p:
p 5
Number of successes
___________________
Number of trials
5
x
__
n
Like the sample mean, the sample proportion can also be considered a random
variable that will take on different values as the procedure leading to the sample
proportion is repeated a great many times. The resulting probability distribution
is called the sampling distribution of the proportion, and it will also have
an expected value (, the probability of success on any given trial) and a standard
deviation (p). These will be discussed in Section 8.4.
8.1 What is the difference between a probability
distribution and a sampling distribution?
8.2 What is the difference between a standard deviation
and a standard error?
8.3 In a simple random sample of 1000 households,
150 households happen to own a barbecue grill. Based
on the characteristics of the population, the expected
E X E R C I
S E S248 Part 3: Sampling Distributions and Estimation
THE SAMPLING DISTRIBUTION OF THE MEAN
When the Population Is Normally Distributed
When a great many simple random samples of size n are drawn from a population
that is normally distributed, the sample means will also be normally distributed.
This will be true regardless of the sample size. In addition, the standard error of
the distribution of these means will be smaller for larger values of n. In summary:
(8.3)
number of grill owners in the sample was 180. What are
the values of , p, and n?
8.4 For a population of five individuals, television ownership is as follows:
x 5 Number of
Television Sets Owned
Allen 2
Betty 1
Chuck 3
Dave 4
Eddie 2
a. Determine the probability distribution for the discrete random variable, x 5 number of television sets
owned. Calculate the population mean and standard
deviation.
b. For the sample size n 5 2, determine the mean for
each possible simple random sample from the five
individuals.
c. For each simple random sample identified in part (b),
what is the probability that this particular sample
will be selected?
d. Combining the results of parts (b) and (c), describe
the sampling distribution of the mean.
e. Repeat parts (b)–(d) using n 5 3 instead of n 5 2.
What effect does the larger sample size have on the
mean of the sampling distribution? On the standard
error of the sampling distribution?
8.5 Given the following probability distribution for an
infinite population with the discrete random variable, x:
x 1 2
P(x) 0.5 0.5
a. Determine the mean and the standard deviation of x.
b. For the sample size n 5 2, determine the mean
for each possible simple random sample from this
population.
c. For each simple random sample identified in part (b),
what is the probability that this particular sample
will be selected?
d. Combining the results of parts (b) and (c), describe
the sampling distribution of the mean.
e. Repeat parts (b)–(d) using a sample size of n 5 3
instead of n 5 2. What effect does the larger sample
size have on the mean of the sampling distribution?
On the standard error of the sampling distribution?
8.6 The operator of a museum exhibit has found that 30%
of the visitors donate the dollar that has been requested to
help defray costs, with the others leaving without paying.
Three visitors are on their way into the exhibit. Assuming
that they will be making independent decisions on whether
to donate the dollar that has been requested:
a. Determine the mean and standard deviation for
x 5 the amount paid by individual visitors.
b. Determine the mean and the standard deviation for
the sampling distribution of x} for samples consisting
of 3 visitors each.
Sampling distribution of the mean, simple random samples from a normally
distributed population:
Regardless of the sample size, the sampling distribution of the mean will be
normally distributed, with
• Mean:
E(} x) 5 x} 5 
• Standard error:
 }
x 5

____
Ïwn
where  5 population mean
 5 population standard deviation
n 5 sample sizeChapter 8: Sampling Distributions 249
The standard deviation (referred to as the standard error) of the sample
means will be smaller than that of the original population. This is because each
sample mean is the average of several values from the population, and relatively
large or small values of x are being combined with values that are less extreme.
EXAMPLE
Sampling Distribution
In Chapter 7, we reported that the average annual number of hours flown by
general-aviation aircraft was 120 hours.1 For purposes of our discussion here,
we will continue to assume that the flying hours for these aircraft are normally
distributed with a standard deviation of 30 hours.
SOLUTION
Assuming a normal distribution with  5 120.0 hours and  5 30.0 hours, part (a)
of Figure 8.1 shows the shape of the population distribution, while part (b) shows
the sampling distribution of the mean for simple random samples of n 5 9 aircraft
each. Notice how the sampling distribution of the mean is much narrower than
the distribution of flying hours for the original population. While the population x
values have a standard deviation of 30.0 hours, the sample means have a standard
error of just 30.0yÏ
__
9, or 10.0 hours.
In part (c) of Figure 8.1, we see how the sampling distribution of the mean
becomes even narrower when the sample size is increased. For simple random
samples of size n 5 36, the sample means have a standard error of only 30.0yÏ
___
36,
or 5.0 hours.
EXAMPLEEXAMPLEEXAMPLEEXAMPL
Just as we can use the z-score to convert a normal distribution to a standard
normal distribution, we can also convert a normally distributed sampling distribution of means to a standard normal distribution. The z-score formula for doing
so is very similar to the one introduced in the previous chapter:
z-score for the sampling distribution of the mean, normally distributed
population:
z 5
}
x 2 
______
}
x
where z 5 distance from the mean, measured
in standard error units
}
x 5 value of the sample mean in which
we are interested
 5 population mean
}
x 5 standard error of the sampling
distribution of the mean, or yÏ__ n
1Source: General Aviation Manufacturers Association, General Aviation Statistical Databook, 2008
Edition, pp. 28–29.250 Part 3: Sampling Distributions and Estimation
FIGURE 8.1
For a normal population, the
sampling distribution of } x will
also be normal regardless of
the sample size. As n increases,
the standard error of the
sampling distribution will
become smaller.
30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210
30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210
30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210
(a) The population
m = 120 hours
s = 30 hours
x x x
(b) The sampling distribution of x when n = 9
(c) The sampling distribution of x when n = 36
mx
= m = 120 hours
sx
= s/√n = 30/√9
= 10 hours
mx
= m = 120 hours
sx
= s/√n = 30/√36
= 5 hours
EXAMPLE
z-Score
For this example, we have a population of general-aviation aircraft, with the
assumption of  5 120.0 flying hours and  5 30.0 flying hours.
SOLUTION
Two important notes:
1. This is the same population of aircraft we discussed in Chapter 7. In Chapter 7,
we were making probability statements about individual planes selected from
EXAMPLEEXAMPLEEXAMPLChapter 8: Sampling Distributions 251
FIGURE 8.2
For a simple random sample
of n 5 36 from the aircraft
flying-hours population, the
probability is just 0.0548
that the sample mean will
be 128 hours or more.
mx
= 120 hours
sx
= 5 hours
P(x ≥ 128) = P(z ≥ 1.60)
or 1.0000 (Cumulative area to z = +∞)
–0.9452 (Cumulative area to z = 1.60)
= 0.0548
z:
x:
0.00
120
1.60
128
(s
x
units)
(hours)
z =
x – m
–––––
sx
128 – 120
––––––––
5
For x = 128 hours, = = 1.60
the population. Here we are making probability statements about the means
of samples of planes taken from the original population.
2. In Chapter 7, we dealt with a population consisting of individual planes. Here
we are considering a population consisting of sample means—this is what is
meant by the sampling distribution of the mean.
For a Simple Random Sample of 36 Aircraft, What Is the Probability That
the Average Flying Time for the Aircraft in the Sample Was at Least
128.0 Hours?
As calculated previously, the standard error of the sampling distribution of the
mean for samples of this size is x} 5 30.0yÏ
___
36, or 5.0 hours. We can now
proceed to calculate the z-score corresponding to a sample mean of 128.0 hours:
z 5
}
x 2 
______
}
x
5
128.0 2 120.0
_____________
5.0
5
8.0
___
5.0
5 1.60
Referring to the standard normal table, the cumulative area to z 5 1.60 is
0.9452. Since the total area beneath the curve is 1.0000, the probability that
}
x will be at least 128.0 hours is 1.0000 2 0.9452, or 0.0548, as shown in
Figure 8.2.
As with our use of the z-score in finding areas beneath the standard normal
curve in the preceding chapter, many other possibilities are available for identifying
probabilities associated with the sample mean. For example, you may wish to verify
on your own that P(123.0 # } x # 129.0) 5 0.2384 for a simple random sample
of n 5 36 aircraft.
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE252 Part 3: Sampling Distributions and Estimation
When the Population Is Not Normally Distributed
The assumption of normality for a population isn’t always realistic, since in many
cases the population is either not normally distributed or we have no knowledge
about its actual distribution. However, provided that the sample size is large (i.e.,
n $ 30), the sampling distribution of the mean can still be assumed to be normal.
This is because of what is known as the central limit theorem:
Central limit theorem. For large, simple random samples from a population
that is not normally distributed, the sampling distribution of the mean will be
approximately normal, with the mean } x 5  and the standard error } x 5 yÏ__ n.
As the sample size (n) is increased, the sampling distribution of the mean will
more closely approach the normal distribution.
As shown in Figure 8.3, regardless of how a population happens to be
distributed, the central limit theorem enables us to proceed as though the samples
FIGURE 8.3
When populations are
not normally distributed,
the central limit theorem
states that the sampling
distribution of the mean will be
approximately normal for large
simple random samples.
x
x
x
x x
Any population having a nonnormal distribution
For large simple random samples, the sample means
will be approximately normally distributed
mx
= m
sx
= s/√nChapter 8: Sampling Distributions 253
were being drawn from a population that is normally distributed.2 The only
requirement here is that the sample size be sufficiently large. In Chapter 9, we will
consider the constraint that  is known. As a practical matter, when the sample
size is at least 30, the “sufficiently large” requirement is assumed to have been
met. As the definition of the central limit theorem indicates, the approximation
to the normal distribution becomes even better for larger samples. Refer to Seeing
Statistics Applets 7 and 8, at the end of the chapter, for visual demonstrations of
the sampling distribution of the mean for fair and unfair dice, respectively.
8.7 A random variable is normally distributed with mean
 5 $1500 and standard deviation  5 $100. Determine
the standard error of the sampling distribution of the
mean for simple random samples with the following
sample sizes:
a. n 5 16 b. n 5 100
c. n 5 400 d. n 5 1000
8.8 For a random variable that is normally distributed,
with  5 200 and  5 20, determine the probability
that a simple random sample of 4 items will have a
mean that is
a. greater than 210.
b. between 190 and 230.
c. less than 225.
8.9 For a random variable that is normally distributed,
with  5 80 and  5 10, determine the probability that
a simple random sample of 25 items will have a mean
that is
a. greater than 78.
b. between 79 and 85.
c. less than 85.
8.10 Employees in a large manufacturing plant worked
an average of 62.0 hours of overtime last year, with a
standard deviation of 15.0 hours. For a simple random
sample of n 5 36 employees and x 5 the number of
overtime hours worked last year, determine the z-score
corresponding to each of the following sample means:
a. } x 5 55.0 hours b. } x 5 60.0 hours
c. } x 5 65.0 hours d. } x 5 47.0 hours
8.11 For the manufacturing plant discussed in
Exercise 8.10, the union president and the human
resources director jointly select a simple random sample
of 36 employees to engage in a discussion with regard to
the company’s work rules and overtime policies. What
is the probability that the average number of overtime
hours last year for members of this sample would have
been less than 65.0 hours? Between 55.0 and 65.0 hours?
8.12 Shirley Johnson, a county political leader, is looking
ahead to next year’s election campaign and the possibility
that she might be forced to support a property tax
increase to help combat projected shortfalls in the county
budget. For the dual purpose of gaining voter support
and finding out more about what her constituents think
about taxes and other important issues, she is planning to
hold a town meeting with a simple random sample of 40
homeowners from her county. For the more than 100,000
homes in Shirley’s county, the mean value is $190,000 and
the standard deviation is $50,000. What is the probability
that the mean value of the homes owned by those invited
to her meeting will be greater than $200,000?
E X E R C I
S E S
2The central limit theorem applies to any distribution that has a nonzero variance. If there is zero
variance in the population, sample means will all have the same value, hence they cannot be normally
distributed under this (rare) condition.
The central limit theorem is basic to the concept of statistical inference because
it permits us to draw conclusions about the population based strictly on sample
data, and without having any knowledge about the distribution of the underlying
population. Such inferences are important to the topics of the next three chapters, and they also play a key role in the application of statistics to total quality
management. This chapter’s opening vignette is a process-control application to
be examined in more detail in Chapter 20.254 Part 3: Sampling Distributions and Estimation
8.13 An industrial crane is operated by four electric
motors working together. For the crane to perform properly, the four motors have to generate a total of at least
380 horsepower. Engineers from the plant that produces
the electric motors have stated that motors off their
assembly line average 100 horsepower, with a standard
deviation of 10 horsepower. The four electric motors on
the crane have just been replaced with four brand new
motors from the motor manufacturer. A safety engineer
is concerned that the four motors might not be powerful
enough to properly operate the crane. Stating any additional assumptions you are using, does the engineer’s
concern appear to be justified?
8.14 From past experience, an airline has found the
luggage weight for individual air travelers on its
trans-Atlantic route to have a mean of 80 pounds and
a standard deviation of 20 pounds. The plane is consistently fully booked and holds 100 passengers. The pilot
insists on loading an extra 500 pounds of fuel whenever
the total luggage weight exceeds 8300 pounds. On what
percentage of the flights will she end up having the
extra fuel loaded?
8.15 A hardware store chain has just received a truckload of 5000 electric drills. Before accepting the shipment, the purchasing manager insists that 9 of the drills
be randomly selected for testing. He intends to measure
the maximum power consumption of each drill and
reject the shipment if the mean consumption for the
sample is greater than the 300 watts listed on the
product label. Unknown to the purchasing manager, the
drills on the truck require an average of 295 watts, with
a standard deviation of 12 watts. Stating any additional
assumptions you are using, find the probability that the
truckload of drills will be rejected.
8.16 The tensile strength of spot welds produced by a
robot welder is normally distributed, with a mean of
10,000 pounds per square inch and a standard deviation
of 800 pounds per square inch. For a simple random
sample of n 5 4 welds, what is the probability that the
sample mean will be at least 10,200 pounds per square
inch? Less than 9900 pounds per square inch?
8.17 At a department store catalog orders counter, the
average time that a customer has to wait before being
served has been found to be approximately exponentially
distributed, with a mean (and standard deviation) of
3.5 minutes. For a simple random sample of 36 recent
customers, invoke the central limit theorem and determine the probability that their average waiting time was
at least 4.0 minutes.
8.18 It has been reported that the average U.S. teenager
sends 80 text messages per day. For purposes of this
exercise, we will assume the daily number of text messages
sent is normally distributed with a standard deviation
of 15.0 messages. For a randomly selected group of 10
teenagers, and considering these persons to be a simple
random sample of all U.S. teens, what is the probability
that the group will send at least 900 text messages (i.e.,
a sample mean of at least 90 messages per teen) next
Wednesday? Source: healthnews.com, May 27, 2009.
8.19 The average length of a hospital stay in general or
community hospitals in the United States is 5.5 days.
Assuming a population standard deviation of 2.5 days
and a simple random sample of 50 patients, what is the
probability that the average length of stay for this group
of patients will be no more than 6.5 days? If the sample
size had been 8 patients instead of 50, what further
assumption(s) would have been necessary in order to
solve this problem? Source: Bureau of the Census, Statistical
Abstract of the United States 2009, p. 115.
8.20 According to the Newspaper Association of America,
the average visitor to online newspaper sites spends
45 minutes per month reading online news content.
Assuming a population standard deviation of 10 minutes
and a simple random sample of 30 online newspaper readers, what is the probability that members of this group
will average at least 40 minutes reading online newspapers
during the coming month? Source: niemanlab.org, July 17, 2009.
THE SAMPLING DISTRIBUTION OF THE PROPORTION
As mentioned earlier, the proportion of successes in a sample consisting of n trials
will be
Sample proportion 5 p 5 Number of successes ___________________
Number of trials
5
x
__
n
Whenever both n and n(1 2 ) are $5, the normal distribution can be
used to approximate the binomial, an application examined in Chapter 7. When
these conditions are satisfied and the procedure leading to the sample outcome is
(8.4)Chapter 8: Sampling Distributions 255
repeated a great many times, the sampling distribution of the proportion (p) will
be approximately normally distributed, with this approximation becoming better
for larger values of n and for values of  that are closer to 0.5.
Like the sampling distribution of the mean, the sampling distribution of
the proportion has an expected value and a standard error. Also, as the size of the
sample becomes larger, the standard error becomes smaller. For simple random
samples, the sampling distribution of the proportion can be described as follows:
EXAMPLE
Sampling Distribution, Proportion
According to the Bureau of the Census, 85.2% of adult residents of Oklahoma
have completed high school.3
SOLUTION
What Is the Probability That No More Than 80% of the Persons in a Simple
Random Sample of 200 Adult Residents of Oklahoma Have Finished High
School (i.e., the Probability That p # 0.800)?
The expected value of the sampling distribution of the proportion is  5 0.852.
This, along with the sample size of n 5 200, is used in determining the standard
error of the sampling distribution and the z-score corresponding to a sample
proportion of p 5 0.800:
Standard error 5 
p 5 Ïwwwww _________ (1 n2 ) 5 Ïwwwwwwww 0.852(1 ________________ 200 2 0.852) 5 0.0251
z 5
p 2 
______
p
5
0.800 2 0.852
_____________
0.0251
5 22.07
Referring to the standard normal distribution table, the cumulative area to
z 5 22.07 is 0.0192. As shown in Figure 8.4, the probability that no more than
80% of these 200 people are high school graduates is 0.0192.
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLE
3Source: Bureau of the Census, Statistical Abstract of the United States 2009, p. 149.
• Sampling distribution of the proportion, p:
Mean 5 E(p) 5 
Standard error 5 
p 5 Ïwwwww _________ (1n2 ) where  5 population proportion
n 5 sample size
• z-score for a given value of p:
z 5
p 2 
______

p
where p 5 the sample proportion value of interest256 Part 3: Sampling Distributions and Estimation
Although the binomial distribution is discrete and the normal continuous, we have not
used the correction for continuity in reaching this solution. As Chapter 7 explained,
such correction makes little difference in the results when sample sizes are large and
the number of successes can take on a large number of possible values. Because
of the sample sizes typically involved when dealing with the sampling distribution
of the proportion, this correction is generally not employed in these applications and
will not be necessary in the exercises that follow.
NOTE
FIGURE 8.4
The population proportion
() of adult residents of
Oklahoma who have finished
high school is 0.852. For a
simple random sample of
200 adults from that state, the
probability is 0.0192 that the
proportion of sample members
who are high school graduates
will be no more than 0.800.
p = 0.852
sp
= 0.0251
P(p ≤ 0.800) = P(z ≤ –2.07)
= 0.0192 (Cumulative area to z = –2.07)
z:
p:
–2.07 0.00
0.800 0.852
(s
p
units)
z =
p – p
–––––
sp
0.800 – 0.852
–––––––––––
0.0251
For p = 0.800, = = –2.07
8.21 According to the Federal Reserve System, 46% of
U.S. households have credit-card debt. For simple
random samples of size n, with n . 30, describe the
sampling distribution of p 5 the proportion of households in the sample who have credit-card debt. Source:
Statistical Abstract of the United States 2009, p. 719.
8.22 Shirley Johnson, the county political leader discussed
in Exercise 8.12, is a Republican. According to voter
registration records, 60% of the homeowners in her county
are also Republicans. For the simple random sample
of 40 homeowners invited to attend her town meeting,
describe the sampling distribution of p 5 the proportion
of Republicans among those invited to the meeting.
8.23 A simple random sample with n 5 300 is drawn
from a binomial process in which  5 0.4. Determine the
following probabilities for p 5 the proportion of successes:
a. P(p $ 0.35)
b. P(0.38 # p # 0.42)
c. P(p # 0.45)
8.24 A simple random sample with n 5 200 is drawn
from a binomial process in which  5 0.6. Determine
the following probabilities for p 5 the proportion of
successes:
a. P(0.50 # p)
b. P(0.65 # p # 0.70)
c. P(p $ 0.70)
8.25 It has been reported that 40% of U.S. workers
employed as purchasing managers are females. In a simple
random sample of U.S. purchasing managers, 70 out of
the 200 are females. Given this information: Source: Bureau
of the Census, Statistical Abstract of the United States 2009, p. 384.
a. What is the population proportion, ?
b. What is the sample proportion, p?
c. What is the standard error of the sample proportion?
d. In the sampling distribution of the proportion, what
is the probability that a sample of this size would
result in a sample proportion at least as large as that
found in part (b)?
E X E R C I
S E SChapter 8: Sampling Distributions 257
8.26 According to the Investment Company Institute,
40% of U.S. households have an Individual Retirement
Account (IRA). Assuming the population proportion to
be  5 0.40 and that a simple random sample of 400
households has been selected: Source: Investment Company
Institute, 2008 Investment Company Fact Book, p. 87.
a. What is the expected value of p 5 the proportion in
the sample having an IRA?
b. What is the standard error of the sampling distribution of the proportion?
c. What is the probability that at least 35% of those in
the sample will have an IRA?
d. What is the probability that between 38% and 45%
of those in the sample will have an IRA?
8.27 A mail-order discount outlet for running shoes has
found that 20% of the orders sent to new customers end
up being returned for refund or exchange because of size or
fit problems experienced by the customer. The 100-member
Connor County Jogging Society has just found out about
this mail-order source and each member has ordered a
pair of running shoes. The group-orders manager of the
mail-order firm has promised the club a 15% discount on
its next order if 10 or more of the pairs ordered must be
returned because of size or fit problems. What is the probability that the club will get 15% off on its next order?
8.28 The Temployee Company supplies temporary
workers to industry and has found that 58% of the
positions it is asked to fill require a knowledge of
Microsoft Excel. The firm currently has 25 individuals
available who know Microsoft Excel. If Temployee
Company receives 50 inquiries for temporary workers
next week, how likely is it that the company will not
have enough Microsoft Excel-skilled people on hand?
8.29 The absentee rate for bus drivers in a large school
district has been 5% in recent years. During the past few
months, the school district and the bus drivers’ union
have been engaged in negotiations for a new contract.
Negotiations are at an impasse over several key issues,
and the drivers already have worked 6 months past the
termination of their previous labor contract with the
district. Yesterday’s negotiating session was particularly
contentious, with members of both sides engaging in
heated debate and name-calling. Of the 400 bus drivers
who serve the school district, 60 of them called in sick
this morning, an absentee rate of 15%. The school
district has accused the union of collusion in carrying
out a work stoppage and the union has responded that
the school district is just trying to make a big deal out
of more people than usual happening to be sick on a
given day. Given the historical absentee rate, what is the
probability that a daily absentee rate would happen to be
15% or more? Comment on the competing claims of the
school district and the bus drivers’ union.
8.30 According to the Federal Trade Commission (FTC),
about 3.4% of items come up at the wrong price when
scanned at the retail checkout counter. Freda Thompson,
director of investigations for her state’s consumer affairs
office, sets up a study in which she has undercover
shoppers (sometimes known as “mystery shoppers”)
randomly select and purchase 600 items from Wal-Mart
department stores across her state. Suppose that researchers find 15 of the 600 items have been incorrectly priced
by the checkout scanner, an error rate of just 2.5%. If
Wal-Mart’s overall rate of scanning errors is the same as
the rate cited by the FTC, what is the probability that
no more than 2.5% of the items purchased by Freda’s
investigators would be incorrectly priced when they are
scanned? Given the assumed sample result, comment on
whether Wal-Mart might be typical of other companies
in terms of their error rate for scanning retail items.
Source: nist.gov, July 17, 2009.
SAMPLING DISTRIBUTIONS WHEN
THE POPULATION IS FINITE
When sampling is without replacement and from a finite population, the techniques we’ve discussed must be modified slightly. The purpose is to arrive at a
corrected (reduced) value of the standard error for the sampling distribution.
To better appreciate why this is necessary, consider the possibility of selecting a
sample of 990 persons from a population of just 1000. Under these conditions,
there would be practically no sampling error because the sample would include
almost everyone in the population.
Whether we are dealing with the sampling distribution of the mean (} x) or the
sampling distribution of the proportion (p), the same correction factor is applied.
This factor depends on the sample size (n) versus the size of the population (N)
and, as a rule of thumb, should be applied whenever the sample is at least 5% as
large as the population. When n , 0.05N, the correction will have very little effect.
(8.5)258 Part 3: Sampling Distributions and Estimation
Note that the first term in each of the preceding formulas is the same as
we would have used earlier in calculating the standard error. The second term
in each case is the finite population correction factor, and its purpose is to reduce the standard error according to how large the sample is compared to the
population.
• Standard error for the sample mean when sampling without replacement
from a finite population:
}
x 5

____
Ïwn ? Ïwww ______ NN 2 1 n where n 5 sample size
N 5 population size
 5 population standard deviation
• Standard error for the sample proportion when sampling without
replacement from a finite population:

p 5 Ïwwwww _________ (1n2 ) ? Ïwww ______ NN 22 1 n where n 5 sample size
N 5 population size
 5 population proportion
EXAMPLE
Finite Populations
Of the 629 passenger vehicles imported by a South American country in a
recent year, 117 were Volvos. A simple random sample of 300 passenger vehicles
imported during that year is taken.
SOLUTION
What Is the Probability That at Least 15% of the Vehicles in This Sample
Will Be Volvos?
For the 629 vehicles imported, the population proportion of Volvos is 5 117y629,
or 0.186. The sample size (n 5 300) is more than 5% of the size of the population
(N 5 629), so the finite population correction factor is included in the calculation
of the standard error of the proportion:
This is the same as
would have been
calculated in
Section 8.4.
The finite population correction
term, this will be smaller than 1.0 so
the standard error will be less than if
an infinite population were involved.

p 5 Ïwwwww _________ (1 n2 ) ? Ïwww N______ N 22 1 n
5 Ïwwwwwwww 0.186(1 ________________ 300 2 0.186) ? Ïwwwww 629 __________ 62922300 1 5 0.0225 ? 0.7238 5 0.0163
EXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLEEXAMPLChapter 8: Sampling Distributions 259
For problems where we happen to be dealing with a sample mean, the same
finite population correction approach is used. As with the sample proportion, the
correction term will be less than 1.0 in order to reduce the value of the standard
error. Again, whenever n , 0.05N, the correction will have little effect. For example,
at the extreme for this rule of thumb, if n 5 100 and N 5 2000, the correction term
will have a value of 0.975. The resulting standard error will be very close to that
which would have been determined under the assumption of an infinite population.
Note that the correction factor has done its job. The resulting standard error
is 72.38% as large as it would have been for an infinite population. Having obtained 
p 5 0.0163, we can now determine the z-score that corresponds to the
sample proportion of interest, p 5 0.15:
z 5
p 2 
______
p
5
0.15 2 0.186
____________
0.0163
5 22.21
Using the standard normal distribution table, we see the cumulative area to
z 5 22.21 is 0.0136. The total area beneath the standard normal curve is 1.0000,
so the probability that at least 15% of the cars in the sample will be Volvos is
1.0000 2 0.0136, or 0.9864.
XAMPLEEXAMPLEEXAMPLE
8.31 What is meant by a finite population and how
can this affect the sampling distribution of a mean or a
proportion?
8.32 Under what circumstances is it desirable to use
the finite population correction factor in describing the
sampling distribution of a mean or a proportion?
8.33 Compared to the situation in which the population
is either extremely large or infinite in size, what effect
does the use of the finite population term tend to have
on the resulting value of the standard error?
8.34 A population of 500 values is distributed such that
 5 $1000 and  5 $400. For a simple random sample
of n 5 200 values selected without replacement, describe
the sampling distribution of the mean.
8.35 A civic organization includes 200 members, who have
an average income of $58,000, with a standard deviation
of $10,000. A simple random sample of n 5 30 members
is selected to participate in the annual fund-raising drive.
What is the probability that the average income of the
fund-raising group will be at least $60,000?
8.36 Of the 217 airports available for public use in
Belgium, 132 are paved. For a simple random sample of
n 5 50, what is the probability that at least 22 of the
airports in the sample will be paved? Source: General
Aviation Manufacturers Association, General Aviation Statistical Databook,
2008 Edition, p. 48.
8.37 A firm’s receiving department has just taken in a
shipment of 300 generators, 20% of which are defective.
The quality control inspector has been instructed to
reject the shipment if, in a simple random sample of 40,
15% or more are defective. What is the probability that
the shipment will be rejected?
E X E R C I
S E S
COMPUTER SIMULATION OF
SAMPLING DISTRIBUTIONS
Using the computer, we can simulate the collection of simple random samples from
a population or a probability distribution. For example, Computer Solutions 8.1
describes the procedures for using Excel or Minitab to generate a simple random
sample of observations from a Poisson probability distribution with a mean of 4.
(8.6)260 Part 3: Sampling Distributions and Estimation
Sampling Distributions and Computer Simulation
These procedures simulate 200 samples from a Poisson distribution with a mean of 4.0. In the first
simulation, the size of each sample is n 5 5. In the second simulation, the size of each sample is n 5 30.
The Excel results are displayed in Figure 8.5 and the Minitab results would be similar.
EXCEL
1. Start with a blank worksheet. From the Data ribbon, click Data Analysis in the rightmost menu section. Click
Random Number Generation. Click OK.
2. Enter 5 into the Number of Variables box. Enter 200 into the Number of Random Numbers box. Within the
Distribution box, select Poisson. Enter 4.0 into the Lambda box. Select Output Range and enter A2 into the
dialogue box. Click OK. The 200 simulated samples will be located in A2:E201, with each row representing one
of the 200 samples.
3. Click on cell F2 and enter the formula 5AVERAGE(A2:E2). Click at the lower right corner of cell F2 and drag
downward to fill the remaining cells in column F. These values are the 200 sample means. Click on cell F1 and enter
the title for the F column as Means.
4. Determine the mean of the sample means by clicking on cell G2 and entering the formula 5AVERAGE(F2:F201).
Determine the standard deviation of the sample means by clicking on cell H2 and entering the formula
5STDEV(F2:F201).
5. Generate the histogram of means when n 5 5: Click on cell J1 and enter the title for the J column as Bin. Enter
1.0 into J2 and 1.2 into J3. Select J2:J3, then click at the lower right corner of cell J3 and drag downward to fill,
so that 7.0 will be in cell J32.
6. From the Data ribbon, click Data Analysis in the rightmost menu section. Within Analysis Tools, click
Histogram. Click OK. Enter F1:F201 into Input Range. Enter J1:J32 into Bin Range. Select Labels. Select Output
Range and enter K1 into the dialogue box. Select Chart Output. Click OK. In fine-tuning the histogram, eliminate
the gaps between the bars by right-clicking on one of the bars, selecting the Format Data Series menu, then
clicking on the Gap Width slider and moving it leftward to the No Gap position, and clicking Close. The histogram
is shown in part (b) of Figure 8.5.
7. For samples with n 5 30, the procedure for generating 200 samples, their means, and the histogram of the sample
means will be similar to the steps shown above. For example, in step 2, we would enter 30 into the Number of
Variables box and enter 200 into the Number of Random Numbers box. For each sample, the sample values
would be in columns A through AD and the sample means would be in column AE.
MINITAB
1. Click Calc. Select Random Data. Click Poisson. Enter 200 into the Number of rows of data to generate box.
Enter C1–C5 (the destination columns) into the Store in columns box. Enter 4.0 into the Mean box. Click OK.
The 200 simulated samples will be located in columns C1–C5, with each row constituting a separate sample.
2. Determine the sample means: Click Calc. Click Row Statistics. Select Mean. Enter C1–C5 into the Input
Variables box. Enter C6 into the Store result in box. Click OK. Enter Means as the label of column C6.
3. Generate the histogram for the 200 sample means: Click Graph. Click Histogram. Select Simple and Click OK.
Enter C6 into the Graph variables box. Click OK.
4. For samples with n 5 30, the procedure for generating 200 samples, their means, and the histogram of the sample
means will be similar to the steps shown above. For example, the destination in step 1 will be columns C1–C30,
and the means will be located in column C31.
COMPUTER 8.1 SOLUTIONSChapter 8: Sampling Distributions 261
FIGURE 8.5
Although the population is
a Poisson distribution with a
mean of 4.0, the sampling
distribution of the mean
still approaches the normal
distribution as the Excelsimulated samples become
larger. The simulations
represented in parts (b) and
(c) were generated using the
Excel procedures in Computer
Solutions 8.1. The Minitab
results would be similar.
(a) The population from which the Excel-simulated samples were drawn
(b) Distribution of means for 200 simulated samples, each with n 5 5
(c) Distribution of means for 200 simulated samples, each with n 5 30262 Part 3: Sampling Distributions and Estimation
In part (a) of Figure 8.5, it is evident that the Poisson distribution from which the
simulated samples are to be taken is neither continuous nor symmetrical. Part (b)
of Figure 8.5 shows the distribution of sample means when Excel generated 200
simulated samples, each with a sample size of n 5 5. In part (c) of Figure 8.5, each
of the 200 simulated samples is larger, with n 5 30. Note that the distribution of
sample means not only becomes narrower, but it also more closely resembles a
normal distribution.
The probability distribution in part (a) of Figure 8.5 has a mean of 4.0. Because
it is a Poisson distribution, its variance is also 4.0, and its standard deviation is
the positive square root of the variance, or 2.0. According to the central limit
theorem, samples of n 5 30 from this population should have a mean of 4.0 and
a standard error of yÏwn 5 2.0yÏww 30 5 0.365. For the 200 samples described
in part (c) of Figure 8.5, the mean was 4.009 and the standard error was 0.3759,
both very close to the values predicted by the theorem.
Note: These exercises require Excel, Minitab, or other
statistical software capable of simulating simple random
samples from a population or probability distribution.
8.38 Using the procedures in Computer Solutions 8.1
as a general guide, simulate 150 simple random samples
of size n 5 2 from an exponential distribution with
a mean of 6. Determine the mean and the standard
deviation of the 150 sample means, then compare these
with their theoretically expected values. Given the
nature of the population distribution, would we expect
the distribution of sample means to be at least approximately normally distributed? Generate a histogram of
the sample means and comment on the shape of the
distribution.
8.39 Repeat Exercise 8.38, but with n 5 30 observations
in each sample.
8.40 Using the procedures in Computer Solutions 8.1 as a
general guide, simulate 150 simple random samples of size
n 5 4 from a normal distribution with a mean of 80 and
a standard deviation of 20. Determine the mean and the
standard deviation of the 150 sample means, then compare
these with their theoretically expected values. Given the
nature of the population distribution, would we expect the
distribution of sample means to be at least approximately
normally distributed? Generate a histogram of the sample
means and comment on the shape of the distribution.
8.41 Repeat Exercise 8.40, but with n 5 30 observations
in each sample.
E X E R C I
S E S
SUMMARY
• The sample mean or sample proportion considered as a random variable
The two preceding chapters dealt with populations representing discrete and
continuous random variables, and with the probability distributions for these
variables. The purpose of this chapter is to examine how sample means and
proportions from a population will tend to be distributed, and to lay the foundation for the inferential statistics of the chapters that follow. Inferential statistics
involves the use of sample information to draw conclusions about the population
from which the sample was drawn.
• Sampling distribution of the sample mean or the sample proportion
If we were to take a great many samples of a given size from the same population,
the result would be a great many different values for the sample mean or sample
proportion. Such a process involves what is referred to as a sampling distribution
(8.7)Chapter 8: Sampling Distributions 263
of the sample mean or the sample proportion, and the sample mean or proportion is the random variable of interest. The standard deviation of the sampling
distribution of the mean or proportion is referred to as the standard error of the
sampling distribution. As the sample size becomes larger, the standard error of
the sampling distribution will decrease.
• The central limit theorem
If a population is normally distributed, the sampling distribution of the mean will
also be normal. Regardless of how the population values are distributed, the central
limit theorem holds that the sampling distribution of the mean will be approximately normal for large samples (e.g., n $ 30) and will more closely approach the
normal distribution as the sample size increases. Whenever both n and n(1 2 )
are $ 5, the normal distribution can be used to approximate the binomial, and the
sampling distribution of the proportion will be approximately normally distributed, with the approximation improving for larger values of n and for values of 
that are closer to 0.5.
Just as we can use the z-score to convert a normal distribution to the standard
normal distribution, the same approach can be used to convert a normal sampling
distribution to the standard normal distribution. In the process, the z-score can be
used to determine probabilities associated with the sample mean or the sample
proportion.
• When the sample is large compared to the population
When sampling is without replacement and from a finite population, calculation
of the standard error of the sampling distribution may require multiplying by a
correction term called the finite population correction factor. The purpose of the
correction is to reduce the value of the standard error according to how large
the sample is compared to the population from which it was drawn. As a rule of
thumb, the finite population correction is made whenever the sample is at least
1y20th the size of the population.
• Using the computer to demonstrate the central limit theorem
Some computer statistical packages can generate simple random samples from
a specified distribution. The chapter uses Excel and Minitab to demonstrate the
central limit theorem by showing how the distribution of means of simulated
samples from a nonnormal population tends to approach the normal distribution
as the sample size is increased.
Sampling Distribution of the Mean: Simple Random
Samples from a Normally Distributed Population
• Regardless of the sample size, the sampling distribution of the mean will be normally
distributed, with
• Mean:
E(x}) 5 x} 5 
E Q U A T I O N S264 Part 3: Sampling Distributions and Estimation
• Standard error:
}
x 5

____
Ïwn
where  5 population mean
 5 population standard deviation
n 5 sample size
• z-score for the sampling distribution of the mean:
z 5
}
x 2 
______
}
x
where z 5 distance from the mean, measured
in standard error units
}
x 5 value of the sample mean in which
we are interested
}
x 5 standard error of the sampling
distribution of the mean, or yÏwn
• According to the central limit theorem, the sampling distribution of the mean will be
approximately normal even when the samples are drawn from a nonnormal population, provided that the sample size is sufficiently large. It is common practice to
apply the above formulas when n $ 30.
Sampling Distribution of the Proportion, p
• Mean 5 E(p) 5 
• Standard error:
p
5 Ïwwwww _________ (1 n2 ) where  5 population proportion
n 5 sample size
• z-score for a given value of p:
z 5
p 2 
______
p
where p 5 sample proportion value of interest
Standard Error of the Sampling Distribution: Sampling
without Replacement from a Finite Population
• Standard error for the sample mean:
}
x 5

_____
Ïwn
? Ïwww N______ N 22 1 n where n 5 sample size
N 5 population size
 5 population standard deviation
• Standard error for the sample proportion:
p
5 Ïwwwww _________ (1 n2 ) ? Ïwww N______ N 22 1 n where n 5 sample size
N 5 population size
 5 population proportion
8.42 The campaign manager for a political candidate
claims that 55% of registered voters favor the candidate
over her strongest opponent. Assuming that this claim
is true, what is the probability that in a simple random
sample of 300 voters, at least 60% would favor the
candidate over her strongest opponent?
8.43 For the situation described in Exercise 8.42, suppose
that only 49% of the registered voters in the sample favor
the candidate over her strongest opponent. Given this result:
a. If the campaign manager’s claim is correct, what is
the probability that the sample proportion would be
no more than 0.49 for a sample of this size?
CHAPTER EXERCISESChapter 8: Sampling Distributions 265
b. Based on your answer to part (a), speculate on
whether the campaign manager’s claim might be
mistaken.
8.44 Given the following probability distribution for an
infinite population with the discrete random variable, x:
x 0 1 2 3
P(x) 0.2 0.1 0.3 0.4
a. Determine the mean and the standard deviation
of x.
b. For the sample size n 5 2, determine the mean
for each possible simple random sample from this
population.
c. For each simple random sample identified in part (b),
what is the probability that this particular sample
will be selected?
d. Combining the results of parts (b) and (c), describe
the sampling distribution of the mean.
8.45 In 2009, the average fee paid by H&R Block
tax preparation customers was $187. Assume that the
standard deviation of fees was $60 but that we have no
idea regarding the shape of the population distribution.
Source: hrblock.com, July 14, 2009.
a. What additional assumption about the population
would be needed in order to use the standard normal
table in determining the probability that the mean fee
for a simple random sample of 5 customers was less
than $170?
b. What is the probability that the mean fee for a
simple random sample of 36 customers was less
than $170? What role does the central limit
theorem play in making it possible to determine
this probability?
8.46 Based on past experience, a telemarketing firm has
found that calls to prospective customers take an average
of 2.0 minutes, with a standard deviation of 1.5 minutes.
The distribution is positively skewed, since persons who
actually become customers require more of the caller’s
time than those who are not home, who simply hang
up, or who say they’re not interested. Albert has been
given a quota of 220 calls for tomorrow, and he works
an 8-hour day. Assuming that his list represents a simple
random sample of those persons who could be called,
what is the probability that Albert will meet or exceed
his quota?
8.47 When a production machine is properly calibrated,
it requires an average of 25 seconds per unit produced,
with a standard deviation of 3 seconds. For a simple
random sample of n 5 36 units, the sample mean is
found to be }x 5 26.2 seconds per unit.
a. What z-score corresponds to the sample mean of
} x 5 26.2 seconds?
b. When the machine is properly calibrated, what is
the probability that the mean for a simple random
sample of this size will be at least 26.2 seconds?
c. Based on the probability determined in part (b), does
it seem likely that the machine is properly calibrated?
Explain your reasoning.
8.48 For a given sample size, the standard error of
the sample mean is 10 grams. With other factors
unchanged, to what extent must the sample size be
increased if the standard error of the sample mean is
to be just 5 grams?
8.49 Employees at a corporate headquarters own an
average of 240 shares of the company’s common stock,
with a standard deviation of 40 shares. For a simple
random sample consisting of 5 employees, what
assumption would be required if we are to use the standard normal distribution in determining P(}x $ 220)?
8.50 The manufacturer of a travel alarm clock claims
that, on the average, its clocks deviate from perfect time
by 30 seconds per month, with a standard deviation of
10 seconds. Engineers from a consumer magazine
purchase 40 of the clocks and find that the average
clock in the sample deviated from perfect accuracy by
34 seconds in one month.
a. If the manufacturer’s claim is correct (i.e.,  5 30
seconds,  5 10 seconds), what is the probability
that the average deviation from perfect accuracy
would be 34 seconds or more?
b. Based on your answer to part (a), speculate on the
possibility that the manufacturer’s claim might not be
correct.
8.51 Based on past experience, 20% of the contacts
made by a firm’s sales representatives result in a sale
being made. Charlie has contacted 100 potential
customers but has made only 10 sales. Assume that
Charlie’s contacts represent a simple random sample
of those who could have been called upon. Given this
information:
a. What is the sample proportion, p 5 proportion of
contacts that resulted in a sale being made?
b. For simple random samples of this size, what is the
probability that p # 0.10?
c. Based on your answer to (b), would you tend to
accept Charlie’s explanation that he “just had a
bad day”?
8.52 As a project for their high school mathematics
class, four class members purchase packages of a leading
breakfast cereal for subsequent weighing and comparison
with the 12-ounce label weight. The students are surprised when they weigh the contents of their four packages and find the average weight to be only 11.5 ounces.
When they write to the company, its consumer relations
representative explains that “the contents of the ‘12-
ounce’ packages average 12.2 ounces and have a standard deviation of 0.4 ounces. The amount of cereal will
vary from one box to the next, and your sample is very
small—this explains why your average weight could be266 Part 3: Sampling Distributions and Estimation
just 11.5 ounces.” If the spokesperson’s information is
correct, how unusual would it be for these students to
get a sample mean as small as the one they obtained?
Be sure to state any assumptions you have made in
obtaining your answer.
8.53 Of 500 vehicles sold by an auto dealership last
month, 200 have a defect that will require a factory
recall for dealer correction. For a simple random
sample of n 5 50 of last month’s customers, what is
the probability that no more than 15 will be affected
by the recall?
8.54 Seventy percent of the registered voters in a large
community are Democrats. The mayor will be selecting
a simple random sample of 20 registered voters to serve
on an advisory panel that helps formulate policy for the
community’s parks and recreation facilities. What is the
probability that the constituency of the committee will
be no more than 50% Democrats?
8.55 In the 2000 census, the so-called “long form”
received by one of every six households contained
52 questions, ranging from your occupation and income
all the way to whether you had a bathtub. According to
the U.S. Census Bureau, the mean completion time for
the long form is 38 minutes. Assuming a standard deviation of 5 minutes and a simple random sample of 50 persons who filled out the long form, what is the probability
that their average time for completion of the form was
more than 45 minutes? Source: Haya El Nasser, “Census Forms
Can Be Filed by Computer,” USA Today, February 10, 2000, p. 4A.
8.56 In performing research on extrasensory perception,
researchers sometimes place two individuals in different
locations, then have person A (the “sender”) view each
item in a series of stimuli while person B (the “receiver”)
tries to correctly identify the image person A is viewing. In carrying out such a study, a researcher enlists the
help of Amy and Donald as volunteers to “send” and
“receive,” respectively. The researcher randomly shows
Amy one of two pictures—either a sunny sky or a stormy
sky—and this procedure is carried out for a total of 200
exposures at 20-second intervals. If Donald is simply
guessing, he will tend to be correct 50% of the time. In
this particular study, Donald was successful in 108 out of
200 attempts to identify which picture Amy was viewing.
If someone has absolutely no extrasensory capabilities
and is simply guessing, what is the probability of guessing correctly on at least 108 out of 200 attempts? Given
the probability you’ve calculated, would you suggest that
Donald drop his plan to list himself in the yellow pages
as a psychic phenomenon?
8.57 A lighting vendor has described its incandescent
bulbs as having normally distributed lifetimes with
a mean of 2000 hours and a standard deviation of
100 hours. The vendor is faced with an especially
demanding industrial customer who insists on conducting
a longevity test before committing to a very large order.
The customer’s purchasing department plans to test a
simple random sample of 16 bulbs and to place the large
order only if the sample mean is at least 2050 hours.
Given the longevity of the vendor’s bulbs and the test the
potential customer will be carrying out, what is the probability that the vendor will get the contract?
8.58 The overall pass rate for law school graduates
taking the Maryland bar exam has been reported as
76%. Assume that a certain Maryland law school has
had 400 of its most recent graduates take the Maryland
bar exam, but only 60% passed. When asked about
these results, the dean of this university’s law school
claims his graduates are just as good as others who have
taken the Maryland bar exam, and the low pass rate for
the recent graduates was just “one of those statistical
fluctuations that happen all the time.” If this university’s
overall pass rate for the Maryland bar exam were really
76%, what would be the probability of a simple random
sample of 400 having a pass rate of 60% or less? Based
on this probability, comment on the dean’s explanation.
Source: U.S. News & World Report, Best Graduate Schools, 2010 Edition,
p. 22.
As he does on the first business day of each month, Ted has
just opened up the suggestion box that Luke placed in the
plant to attract ideas from the employees. Along with the
usual monthly suggestions that old Luke attempt activities
that are either inappropriate for his age or physiologically
impossible, Ted finds a message from a production worker
who is very concerned about the company’s racquetball
racquet machine:
Dear Thorndike management:
I am getting very worried about the racquetball
racquet machine. Some of the older workers say it was
Thorndike Sports Equipment
INTEGRATED C
ASEChapter 8: Sampling Distributions 267
here 15 years ago when they started, and I think it is
just about worn out. I know new machines are expensive, so I thought I would collect some numbers that
might help you out in making a decision to replace it.
I know that the weight of the racquets coming
off the machine is a big deal, since all of the racquets
are weighed and then you decide which ones will be
“Light,” “Regular,” and whatever you’ve decided to
call the heavier model. Anyway, I’ve been doing some
research. On my own time, honest. From time to time
during the day, I take the 30 racquets that have just
come from the machine and weigh them myself.
You’ve told us that some racquets weigh more
than others, so I can understand that one racquet
might weigh 245 grams, the very next one might weigh
230 grams, and then the next one off the line might
be all the way up to 250 grams. Like I said before, I
think the machine is worn out. The first 30 racquets
I weighed yesterday weighed an average of 236.5
grams, the next 30 weighed an average of 243.1 grams.
I think the machine is going crazy, and you should either get it fixed or replace it.
My friends would get pretty upset if they knew I
wrote this suggestion, since buying a brand-new machine could mean that nobody gets a Christmas bonus
this year. But I’m a young guy and I’d like for you to
be successful so that I can have some job security and
take care of my family. If you end up having a problem
with the weights of your racquets, I might end up being
out of a job.
I’d like to find out if maybe I’m missing something
here, and if the machine really is OK or not. So my
friends don’t find out about this suggestion, would you
please put together a reply and duct-tape it to the underside of the pay phone next to the candy machine?
I’ll plan on picking it up after the first work shift on the
15th of next month.
Discarding the suggestions that Luke would probably
find offensive, Ted mentions the young worker’s letter
regarding the racquetball racquet machine. Luke asks if
the machine is still working the same as when the decision was made to offer racquets in three different weight
categories. Ted replies that it seems to be. The average
weight is still 240 grams, the standard deviation is still
10 grams, although he isn’t too sure whether the weights
are still normally distributed. Luke asks Ted to generate a
reply to the worker’s suggestion and tape it to the underside of the pay phone as requested.268 Part 3: Sampling Distributions and Estimation
Distribution of Means: Fair Dice
SEEING STATISTICS: APPLET 7
Try out the Applets: http://www.cengage.com/bstatistics/weiers
This applet simulates the rolling of dice when the dice are
fair—i.e., for each die, the outcomes (integers 1 through
6) are equally likely. You can select the size of each sample
(n 5 1, 3, or 12 dice at a time), and the applet will update
the relative frequencies with which the sample means have
occurred. The three versions of this applet are shown and
described here.
SAMPLE SIZE 5 1
Each time you click the Roll One Set button, you are rolling a single die. N shows the number of samples you have
taken so far, and the chart shows the relative frequency with
which each of the possible sample means has occurred.
Naturally, since each sample is only n 5 1, the result in each
sample is also the mean of each sample. You can make
things happen much faster by clicking either the Roll 10
Sets button or the Roll 100 Sets button. In the latter cases,
you are shown the means for the 10 most recent samples.
SAMPLE SIZE 5 3
Each time you click the Roll One Set button, you are now
rolling three dice. N shows the number of samples (or sets)
you have taken so far, and the chart is updated to show the
relative frequency with which each of the possible sample
means has occurred.
SAMPLE SIZE 5 12
Each time you click the Roll One Set button, you are rolling
twelve dice. The procedure is the same as above, except
each sample mean is now the mean for a set of 12 dice.
Applet Exercises
7.1 Generate 3000 rolls of a single die: Select the Sample
Size 5 1 applet version, and click the Roll 100 Sets button 30 times. Describe the shape of the distribution of the
sample results in the relative frequencies chart. Is the shape
similar to what you might expect?
7.2 Generate 3000 samples, each one representing a set
of three dice: Select the Sample Size 5 3 applet version
and click the Roll 100 Sets button 30 times. Comment on
how the distribution of sample means has changed from
the distribution obtained in Applet Exercise 7.1.
7.3 Generate 3000 samples, each one representing a set
of twelve dice: Select the Sample Size 5 12 applet version and click the Roll 100 Sets button 30 times. How
has the increase in sample size affected the shape of the
distribution of sample means compared to those obtained
in Applet Exercises 7.1 and 7.2?
7.4 If there were an additional applet version that allowed
each sample to consist of 100 dice, and we generated
2000 of these samples, what do you think the distribution
of sample means would tend to look like? In general, as
the sample sizes become larger, what shape will the distribution of sample means tend to approach?Chapter 8: Sampling Distributions 269
SEEING STATISTICS: APPLET 8
Distribution of Means: Loaded Dice
Try out the Applets: http://www.cengage.com/bstatistics/weiers
This applet simulates the rolling of loaded dice—i.e., for
each die, the outcomes (integers 1 through 6) are most
definitely NOT equally likely, and you should never gamble
with people who use these things. As with the “honest”
dice in Applet 7, you can select the size of each sample
(n 5 1, 3, or 12 dice at a time), and the applet will update
the relative frequencies with which the sample means have
occurred. The same features and procedures apply as in
Applet 7, but you will find the results to be quite different.
Examples of the three versions of the “loaded” applet are
shown here:
Applet Exercises
8.1 Generate 3000 rolls of a single die: Select the “Sample
Size 5 1” applet version, and click the “Roll 100 Sets”
button 30 times. Describe the shape of the distribution
of the sample results in the relative frequencies chart. Does
the shape look a little peculiar? Can you tell which faces of
the dice might be “loaded”?
8.2 Generate 3000 samples, each one representing a set of
three dice: Select the “Sample Size 5 3” applet version,
and click the “Roll 100 Sets” button 30 times. Comment
on how the distribution of sample means has changed
from the distribution obtained in Applet Exercise 8.1.
8.3 Generate 3000 samples, each one representing a set
of twelve dice: Select the “Sample Size 5 12” applet version, and click the “Roll 100 Sets” button 30 times. How
has the increase in sample size affected the shape of the
distribution of sample means compared to those obtained
in Applet Exercises 8.1 and 8.2?
8.4 If there were an additional applet version that allowed
each sample to consist of 100 dice, and we generated
2000 of these samples, what do you think the distribution
of sample means would tend to look like? In general, as
the sample sizes become larger, what shape will the distribution of sample means tend to approach?