BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: An actuary compiles a sample of 100 auto insurance claims. The sizes of these sampled claims are independently and identically distributed with mean 1000 and standard deviation 400.

Calculate the approximate probability that the sum of the sizes of the 100 claims is less than 92,000.

Options:
A
0.023
B
0.050
C
0.421
D
0.579
E
0.977

Correct answer: A

Solution: Mean of sum \(=100(1000)=100,000\)
Standard deviation of sum \(=\sqrt{100(400)^{2}}=4,000\)
\(P(X<92,000)=P\left(Z<\frac{92,000-100,000}{4000}\right)=P(Z<-2)=0.0228\)
Question: The monthly commission that an agent earns is modeled by a random variable \(X\) with probability density function
\(f(x)= \begin{cases}\frac{1}{20} e^{-\frac{x}{20}}, & \text { for } x>0 \\ 0, & \text { otherwise. }\end{cases}\)
Calculate the probability that the commission the agent earns in a month is within 0.5 standard deviations of \(\mathrm{E}(X)\).

Options:
A
0.34
B
0.38
C
0.50
D
0.68
E
0.95

Correct answer: B

Solution: This is an exponential distribution with mean 20 and standard deviation 20. So, we want:
\(\int_{10}^{30} \frac{1}{20} e^{-\frac{1}{20} x} d x=\left.e^{-\frac{1}{20} x}\right|_{10} ^{30}=-e^{-1.5}+e^{-0.5}=0.3834\)
Question: Individual burglary claim amounts covered by policies of an insurance company are normally distributed with mean 2500 and standard deviation 500.

The probability that the mean of a random sample of 100 claims will exceed \(K\) is 0.01 .
Calculate \(K\).

Options:
A
2505
B
2512
C
2616
D
3663
E
4950

Correct answer: C

Solution: Average of 100 claims is normal with mean 2500 and standard deviation \(500 / \sqrt{100}=50\).
\[
\begin{aligned}
& P(\bar{X}<K)=0.99 \\
& P\left(Z<\frac{K-2500}{50}\right)=0.99 \\
& \frac{K-2500}{50}=2.32635 \\
& K=2616
\end{aligned}
\]
Question: The sales for a product can be modeled by \(Z=4 X-Y-3 . X\) and \(Y\) are independent random variables with \(\operatorname{Var}(X)=2\) and \(\operatorname{Var}(Y)=3\).

Calculate Var (Z).

Options:
A
5
B
11
C
29
D
32
E
35

Correct answer: E

Solution: \(\operatorname{Var}(Z)=\operatorname{Var}(4 X-Y-3)=(4)^{2} \operatorname{Var}(X)+(-1)^{2} \operatorname{Var}(Y)=(16)(2)+3=35\)
Question: A group of 17 people in a study on lung cancer consists of three heavy smokers, four light smokers, and ten non-smokers. Six people from the group are chosen at random for a new treatment.

Calculate the probability that three of those chosen are non-smokers.

Options:
A
0.176
B
0.284
C
0.300
D
0.339
E
0.588

Correct answer: D

Solution: Hypergeometric:
\(\frac{\binom{10}{3}\binom{7}{3}}{\binom{17}{6}}=\frac{75}{221}=0.33937\)
Question: Let \(X\) represent the number of defective parts in a shipment of five parts.
\(P[X \geq x]=\frac{1}{2}\left(1-\sqrt{\frac{x-1}{5}}\right), \quad x=1,2,3,4,5\)
Calculate \(\mathrm{E}(X)\).

Options:
A
0.9
B
1.1
C
2.1
D
2.3
E
3.9

Correct answer: B

Solution: \(\mathrm{P}[X \geq 0]=1\)
\(\mathrm{P}[X \geq 1]=0.5\)
\(\mathrm{P}[X \geq 2]=0.27639\)
\(\mathrm{P}[X \geq 3]=0.18377\)
\(\mathrm{P}[X \geq 4]=0.11270\)
\(\mathrm{P}[X \geq 5]=0.05279\)
\(p(x)= \begin{cases}0.5, & x=0 \\ 0.22361, & x=1 \\ 0.09262, & x=2 \\ 0.07107, & x=3 \\ 0.05991, & x=4 \\ 0.05279, & x=5\end{cases}\)
\(E(X)=\sum_{i=0}^{5} x_{i} p\left(x_{i}\right)=1.12565\)
Question: An actuary wishes to predict the size \(W\) of a claim using a predictor \(T\). Suppose that \(W\) and \(T\) are independent and normally distributed with the same mean and with variances 4 and 12 , respectively.

Calculate \(P[|W-T|<1]\).

Options:
A
0.20
B
0.23
C
0.38
D
0.60
E
0.68

Correct answer: A

Solution: \(W-T\) has a normal distribution with mean 0 and variance \(4+12=16\).
\(P(|W-T|<1)=P(-1<W-T<1)\)
\(=P(-1<N(0,16)<1)=P(-.25<N(0,1)<.25)=0.197\)
Question: Let \(X=1\) if termites are present in a building and \(X=0\) if they are not.

Let \(Y=1\) if a test indicates the presence of termites in that building and \(Y=0\) if it does not.
The joint probabilities of \(X\) and \(Y\) are:
\[
\begin{aligned}
& \mathrm{P}[X=0, Y=0]=0.90 \\
& \mathrm{P}[X=1, Y=0]=0.01 \\
& \mathrm{P}[X=0, Y=1]=0.05 \\
& \mathrm{P}[X=1, Y=1]=0.04
\end{aligned}
\]

Calculate the coefficient of variation for \(Y\).

Options:
A
0.31
B
0.91
C
0.95
D
3.18
E
4.36

Correct answer: D

Solution: \(p(y)= \begin{cases}0.91, & y=0 \\ 0.09, & y=1\end{cases}\)
\(E(Y)=(0)(0.91)+(1)(0.09)=0.09\)
\(E\left(Y^{2}\right)=\left(0^{2}\right)(0.91)+\left(1^{2}\right)(0.09)=0.09\)
\(V(Y)=0.09-0.09^{2}=0.0819\)
\(S D(Y)=\sqrt{0.0819}=0.28618\)
\(C V(Y)=\frac{0.28618}{.09}=3.179797\)
Question: In a group of four employees, two are high-risk and two are low-risk. This year, each high-risk employee has probability 0.6 of having no accidents; each low-risk employee has probability 0.9 of having no accidents.

The occurrences of accidents among employees are independent events.
Calculate the probability that at most one employee has one or more accidents this year.

Options:
A
0.4536
B
0.5184
C
0.6804
D
0.7084
E
0.7452

Correct answer: E

Solution: \(\mathrm{P}(0\) employees have accidents \()=(0.6)^{2}(0.9)^{2}=0.2916\)
P(1 employee has accidents)
\(=\mathrm{P}(1\) high-risk employee has accidents and 0 low-risk employees have accidents
OR 0 high-risk employees have accidents and1 low-risk employee has accidents)
\(=\mathrm{P}(1\) high-risk employee has accidents \() \mathrm{P}(0\) low-risk employees have accidents \()\)
\(+\mathrm{P}(0\) high-risk employees have accidents \() \mathrm{P}(1\) low-risk employee has accidents \()\)
[combining the addition rule for a union of mutually exclusive events and the multiplication rule for independent events]
\(=(2)(0.4)(0.6)(0.9)^{2}+(0.6)^{2}(2)(0.1)(0.9)=0.4536\)
\(\mathrm{P}(\) at most 1 employee has accidents \()=0.2916+0.4536=0.7452\).
Question: An insurance company surcharges a driver, based on the year of the driver's last accident, using the following table with the current year denoted by \(t\) :

\begin{tabular}{|l|c|c|c|c|}
\hline \begin{tabular}{l} 
Year of Last \\
Accident
\end{tabular} & \(\mathrm{t}-1\) & \(\mathrm{t}-2\) & \(\mathrm{t}-3\) & \(\mathrm{t}-4\) \\
\hline Surcharge & \(20 \%\) & \(15 \%\) & \(10 \%\) & \(5 \%\) \\
\hline
\end{tabular}

The probability that a driver has at least one accident in any given year is 0.10 , independent of the number of accidents in all other years.

Calculate the expected surcharge in year \(t\) for a driver who has been driving since the beginning of year \(t-4\).

Options:
A
\(\quad 4.5 \%\)
B
\(\quad 5.0 \%\)
C
\(\quad 8.6 \%\)
D
\(\quad 10.0 \%\)
E
\(19.4 \%\)

Correct answer: A

Solution: The probability of a \(20 \%\) surcharge is 0.1 , the probability of a \(15 \%\) surcharge is \(0.9(0.1)\), the probability of a \(10 \%\) surcharge is \(0.9(0.9)(0.1)\) and the probability of a \(5 \%\) surcharge is \(0.9(0.9)(0.9)(0.1)\).
Expected surcharge \(=0.20(0.1)+0.15(0.1)(0.9)+0.10(0.1)(0.9)^{2}+0.05(0.1)(0.9)^{3}=0.045245\)
Question: A two-part machine functions when at least one of its parts is working. Both parts are working today. The future lifetime of each part is exponentially distributed with mean five years. The lifetimes of the parts are independent.

The machine functions one year from now.
Calculate the probability that both parts will be working at that time.

Options:
A
0.003
B
0.409
C
0.670
D
0.693
E
0.819

Correct answer: D

Solution: Let \(X\) be the random variable that a part works. Then:
\[
\begin{aligned}
& P(X>x)=e^{-0.2 x} \\
& P(X<x)=1-e^{-0.2 x}
\end{aligned}
\]

One year from now, P [both parts will work | at least 1 part will work]
\[
\begin{aligned}
& =\frac{P[\text { both parts will work }]}{P[\text { at least } 1 \text { part will work }]} \\
& \frac{e^{-0.2} e^{-0.2}}{1-P(\text { neither part works })} \\
& =\frac{e^{-0.2} e^{-0.2}}{1-\left(1-e^{-0.2}\right)\left(1-e^{-0.2}\right)} \\
& =\frac{0.67032}{0.96714}=0.69310
\end{aligned}
\]
Question: The lifetime (in years) of a car is a random variable with probability density function
\[
f(x)= \begin{cases}0.0625 e^{-0.0625 x}, & \text { for } x>0 \\ 0, & \text { otherwise }\end{cases}
\]

Calculate the probability that the car's lifetime is less than 20 years, given that the car's lifetime is at least five years.

Options:
A
0.445
B
0.522
C
0.608
D
0.832
E
0.975

Correct answer: C

Solution: This is an exponential distribution with \(P[X<x]=1-e^{-x / 16}\).
\(P[X<20 \mid X>5]=\frac{P[5<X<20]}{P[X>5]}\)
\(=\frac{P[X<20]-P[X<5]}{P[X>5]}\)
\(=\frac{\left(1-e^{-\frac{20}{16}}\right)-\left(1-e^{-\frac{5}{16}}\right)}{e^{-\frac{5}{16}}}=0.60839\)
Question: Four men at a wedding party throw their hats into a big box. Later, each of them randomly selects a hat from the box and places it on his head.

Calculate the probability that none of the four men has his own hat on his head.

Options:
A
0.042
B
0.250
C
0.333
D
0.375
E
0.500

Correct answer: D

Solution: Let the hats (and men) be \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), and D . There are 24 orderings of the hats drawn. Consider the complement, where at least one man draws his hat.
If \(A\) is drawn first, all 6 subsequent combinations meet this event.
If B is drawn first, combinations \(\mathrm{ACD}, \mathrm{CAD}\), and DCA meet this event.
If C or D is drawn first, each also have three combinations that meet this event.
Hence there are \(6+3+3+3=15\) ways to meet the complementary event.
The desired probability is \((24-15) / 24=0.375\).
Question: An insurer's losses are modeled by a random variable \(X\), with density function, \(f\), where \(f(x)\) is proportional to \(\frac{1}{x^{2}}\), for \(x>100\), and 0 , otherwise.

Calculate the median of \(X\).

Options:
A
100
B
120
C
150
D
200
E
300

Correct answer: D

Solution: \(1=c \int_{100}^{\infty} x^{-2} d x=-\left.c x^{-1}\right|_{100} ^{\infty}=c 100^{-1} \Rightarrow c=100\)
\(0.5=100 \int_{100}^{M} x^{-2} d x=-\left.100 x^{-1}\right|_{100} ^{M}=1-100 M^{-1} \Rightarrow M=200\)
Question: A monthly state lottery picks five distinct integers from 1 to 30 and selects one of the five to be the bonus number. An entry consists of five distinct integers from 1 to 30 with one of the numbers designated as the bonus number.

Each month there are 100,000 entries. Each entry that matches the five distinct numbers is awarded 50,000 . If the bonus number is also matched, an additional 250,000 is awarded to that entry. Nothing is awarded for matching fewer than five numbers.

Calculate the expected payout from the lottery in a month.

Options:
A
\(\quad 70,172\)
B
\(\quad 77,190\)
C
\(\quad 84,731\)
D
100,000
E
\(\quad 175,431\)

Correct answer: A

Solution: The expected monthly payout is:
\(100,000 *[300,000 * \mathrm{P}(\) match all 5 and bonus number \()+50,000 * \mathrm{P}\) (match all 5 but no bonus number) \(+0 * \mathrm{P}\) (match fewer than 5 numbers)].
\(P(\) Match all five and Bonus \()=\frac{1}{\binom{30}{5}} * \frac{1}{5}=\frac{1}{712,530}\)
\(P(\) Match all five and No Bonus \()=\frac{1}{\binom{30}{5}} * \frac{4}{5}=\frac{4}{712,530}\)
Thus, expected monthly payout equals:
\(100,000\left[300,000 \frac{1}{712,530}+50,000 \frac{4}{712,530}\right]=70,172\).
Question: An auditor is examining insurance policies for fraud. A policyholder can only file one claim. The probability of any given policy having a claim is 0.90 , and the probability of a claim being fraudulent is 0.20 . The auditor picks five policies at random and examines them in order until he finds two fraudulent claims. He then stops examining policies. If he doesn't find two fraudulent claims, he stops after examining the fifth policy.

Calculate the expected number of policies he will examine.

Options:
A
4.68
B
4.73
C
4.78
D
4.83
E
4.88

Correct answer: B

Solution: This is a negative binomial distribution. The probability of finding fraud is \(0.9(0.2)=0.18\). Let \(p(x)\) be the probability of examining \(x\) policies. Then, \(p(1)=0\)
\(p(2)=0.18^{2}=0.0324\)
\(p(3)=2(0.82)(0.18)^{2}=0.0531\)
\(p(4)=3(0.82)^{2}(0.18)^{2}=0.0654\)
\(p(5)=1-0.0324-0.0531-0.0654=0.8491\)
\(E(X)=2(0.0324)+3(0.0531)+4(0.0654)+5(0.8491)=4.7312\).
Question: A policyholder purchases car insurance for two years. In a given year, the policyholder's number of car accidents is zero with probability 0.9 , exactly one with probability 0.08 , and exactly two with probability 0.02 . The number of accidents in the first year is independent of the number in the second.

Calculate the probability that the policyholder has one accident in each year, given that the policyholder has a total of exactly two accidents.

Options:
A
0.006
B
0.042
C
0.151
D
0.262
E
0.960

Correct answer: C

Solution: Let \(M=\#\) of accidents in first year, \(N=\#\) of accidents in second year.
\(P(M=1, N=1 \mid M+N=2)=\frac{P(M=1, N=1)}{P(M+N=2)}=\frac{0.08(0.08)}{0.9(0.02)+0.08(0.08)+0.02(0.9)}\)
\(=\frac{0.0064}{0.0424}=0.151\)
Question: The lifetimes of televisions of a certain model are exponentially distributed with a median of 2.7 years.

Calculate the \(87.5^{\text {th }}\) percentile of the lifetimes for these televisions.

Options:
A
3.08
B
4.73
C
8.10
D
10.80
E
19.68

Correct answer: C

Solution: Let \(X=\) lifetime of a television of this brand:
\(0.5=P[X<2.7]=1-e^{-2 / 7 / \beta}\)
\(2.7 / \beta=0.6931\)
\(\beta=3.8953\)
\(0.875=P[X<P]=1-e^{-p / 3.8953}\)
\(p / 3.8953=2.0794\)
\(p=8.10\).
Question: This year, the number of tooth fillings a policyholder undergoes is Poisson distributed. The probability that the policyholder undergoes no tooth fillings this year is 0.18 .

Calculate the mode of the number of tooth fillings the policyholder undergoes this year.

Options:
A
0
B
1
C
2
D
5
E
6

Correct answer: B

Solution: Let \(X=\) number of fillings, which has a Poisson distribution with mean \(\lambda\)
\[
0.18=P[X=0]=\frac{e^{-\lambda} \lambda^{0}}{0!}=e^{-\lambda}
\]
\(\lambda=1.7147\)
Since the mean is 1.7147 , the mode will be 1 as it can be shown that when the mean is not a whole number, the mode is the largest whole number less than the mean. When the mean is a whole number, there are two modes, the mean and one less than the mean.
Question: The time, in years, until replacement for a new telephone pole has probability density function
\[
f(t)= \begin{cases}k t, & \text { for } 0<t<50 \\ 0, & \text { otherwise }\end{cases}
\]
where \(k\) is a constant.
Calculate the probability that a new telephone pole will be replaced within 25 years given that it is not replaced within 20 years.

Options:
A
0.09
B
0.11
C
0.16
D
0.17
E
0.84

Correct answer: B

Solution: \(1=\int_{0}^{50} k t d t=1250 k \Rightarrow k=\frac{1}{1250}\)
\(P[T \leq 25 \mid T \geq 20]=\frac{P[20<T<25]}{P[T>20]}=\frac{\int_{20}^{25} \frac{t}{1250} d t}{\int_{20}^{50} \frac{t}{1250} d t}=\frac{(625-400) / 2500}{(2500-400) / 2500}=\frac{225}{2100}=0.10714\)
Question: The working lifetime of a master computer chip that regulates the electronic components of an automobile engine is exponentially distributed with a mean of 7.2 years. Under a warranty, the chip manufacturer will replace any chip that fails within \(t\) years. It is expected that \(5 \%\) of all chips will be replaced under this warranty.

Calculate \(t\).

Options:
A
0.007
B
0.369
C
0.416
D
0.501
E
0.720

Correct answer: B

Solution: Let \(X=\) working lifetime of chips, in years. Then
\(0.05=P[X \leq t]=1-e^{-t / 7.2}\)
\(e^{-t / 7.2}=0.95\)
\(-t / 7.2=-0.05129\)
\(t=0.3693\)
Question: A baseball-pitching machine is used for batting practice. The machine is out of adjustment such that every pitched baseball arrives at the batter's box between 0 and 2 feet higher than intended.

Let \(X\) equal the difference, in feet, between the actual arrival height and the intended arrival height of a pitched baseball.

The density of \(X, f(x)\), is proportional to \(x\).
Calculate the \(80^{\text {th }}\) percentile for \(X\).

Options:
A
0.40
B
0.89
C
1.26
D
1.60
E
1.79

Correct answer: E

Solution: \(1=\int_{0}^{2} f(x) d x=\int_{0}^{2} c x d x=2 c \Rightarrow c=0.5\)
\(0.8=\int_{0}^{p} 0.5 x d x=0.25 p^{2} \Rightarrow p=\sqrt{3.2}=1.7889\)
Question: An insurance company sells flood and fire insurance. This year, the company's profit from selling flood insurance is normally distributed, and its profit from selling fire insurance is normally distributed with three times the mean and three times the standard deviation as from flood insurance.

The profits from the two types of insurance are independent. The probability that the company earns a positive profit from selling flood insurance this year is 0.67 .

Calculate the probability that the insurance company earns an overall positive profit this year.

Options:
A
\(\quad 0.71\)
B
0.73
C
0.81
D
0.92
E
\(\quad 0.96\)

Correct answer: A

Solution: Let \(X\) be the profit from flood insurance and \(Y\) be the profit from fire insurance and let \(S=X+Y\) be the total profit. Let \(\mu\) and \(\sigma\) be the mean and standard deviation of \(X\).
\(P[X>0]=P\left[Z>\frac{0-\mu}{\sigma}\right]=0.67 \Rightarrow \frac{0-\mu}{\sigma}=-0.44 \Rightarrow 0.44 \sigma=\mu\)
\(S\) has a normal distribution with mean \(E[X]+E[Y]=\mu+3 \mu=4 \mu\) and standard deviation \(\sqrt{\operatorname{Var}(X)+\operatorname{Var}(Y)}=\sqrt{\sigma^{2}+9 \sigma^{2}}=\sqrt{10} \sigma\).
\(P[S>0]=P\left[Z>\frac{0-4 \mu}{\sqrt{10} \sigma}=\frac{-4(0.44 \sigma)}{\sqrt{10} \sigma}=-0.56\right]=0.71\)
Question: An amusement park has two roller coasters. This year, the numbers of accidents occurring on the first and second roller coasters are Poisson distributed with means \(\lambda_{1}=0.5\) and \(\lambda_{2}\), respectively. The probability that at least one accident occurs on the second roller coaster is twice the probability for the first roller coaster.

Calculate \(\lambda_{2}\).

Options:
A
1.00
B
1.19
C
1.23
D
1.55
E
2.00

Correct answer: D

Solution: Let \(X_{j}=\) number of accidents occurring on the \(j\) th roller coaster, for \(j=1,2\), which is Poisson distributed with mean \(\lambda_{j}\).
The probability that at least one accident occurs on roller coaster \(j\) equals
\(1-\mathrm{P}\left[X_{j}=0\right]=1-\frac{e^{-\lambda_{j}}\left(\lambda_{j}\right)^{0}}{0!}=1-e^{-\lambda_{j}}\).
Since the probability that at least one accident occurs on the second roller coaster is twice that for the first roller coaster, \(1-e^{-\lambda_{2}}=2\left(1-e^{-\lambda_{1}}\right)\). Then
\(e^{-\lambda_{2}}=2 e^{-\lambda_{1}}-1 \Rightarrow \lambda_{2}=-\ln \left(2 e^{-\lambda_{1}}-1\right)=-\ln \left(2 e^{-0.5}-1\right)=1.546\).
Question: A patient must undergo hospitalization and surgery. The hospitalization and surgery charges are uniformly distributed on the intervals \([0, b]\) and \([0,2 b-6]\), respectively, where \(b\) is a constant larger than 3 .

The standard deviation of the hospitalization charge is 9.60 .
Calculate the standard deviation of the surgery charge.

Options:
A
13.2
B
15.7
C
17.5
D
19.2
E
\(\quad 19.9\)

Correct answer: C

Solution: The variance of a uniform distribution is \(1 / 12\) of the square of the interval length. The standard deviation of the hospitalization charge is \(\frac{b}{\sqrt{12}}=9.6 \Rightarrow b=33.255\). The standard deviation of the surgery charge is \(\frac{2 b-6}{\sqrt{12}}=\frac{2(33.255)-6}{\sqrt{12}}=17.4677\).
Question: Claims under a product liability policy have the following characteristics:
i) The number of claims does not exceed two.
ii) The probability that the number of claims is exactly one is 0.08 .
iii) The probability that the number of claims is exactly two is 0.02 .
iv) For \(n=1\) or 2 claims, the total claim amount under the policy is a random variable \(X\) with cumulative distribution function
\[
F(x)=1-\left(\frac{500 n}{x}\right), \quad \text { for } x \geq 500 n
\]

Calculate the probability that there is exactly one claim, given that there is at least one claim and the total claim amount under the policy is less than or equal to 2000.

Options:
A
\(\quad 1 / 2\)
B
\(\quad 3 / 5\)
C
\(\quad 2 / 3\)
D
\(\quad 3 / 4\)
E
\(6 / 7\)

Correct answer: E

Solution: \(P(N=1, X \leq 2000)=0.08 P(X \leq 2000 \mid N=1)=0.08(3 / 4)=0.06\)
\(P(N=2, X \leq 2000)=0.08 P(X \leq 2000 \mid N=2)=0.02(1 / 2)=0.01\)
\(P(N>0, X \leq 2000)=0.06+0.01=0.07\)
\(P(N=1 \mid N>0, X \leq 2000)=\frac{P(N=1, X \leq 2000)}{P(N>0, X \leq 2000)}=\frac{0.06}{0.07}=6 / 7\)
Question: Drivers are classified as either high-risk or low-risk. Ten percent of drivers are classified as high-risk. The risk classification of each driver remains the same from this year to next year.

The probability that a driver classified as high-risk is involved in an accident is 0.12 for this year and, independently, 0.12 for next year. The probability that a driver classified as low-risk is involved in an accident is 0.05 for this year and, independently, 0.05 for next year.

Calculate the probability that a driver is involved in an accident next year, given that the driver is involved in an accident this year.

Options:
A
0.004
B
0.014
C
0.057
D
0.065
E
0.099

Correct answer: D

Solution: Let \(H\) be the event that a driver is high-risk and \(L\) be the event that a driver is low-risk.
Let \(T\) be the event that a driver has an accident this year and \(N\) be the event that the same driver has an accident next year. Then,
\[
\begin{aligned}
& P[N \mid T]=\frac{P[N \cap T]}{P[T]}=\frac{P[N \cap T \mid H] P[H]+P[N \cap T \mid L] P[L]}{P[T \mid H] P[H]+P[T \mid L] P[L]} \\
& \quad=\frac{P[N \mid H] P[T \mid H] P[H]+P[N \mid L] P[T \mid L] P[L]}{P[T \mid H] P[H]+P[T \mid L] P[L]} \\
& \quad=\frac{(0.12)^{2}(0.10)+(0.05)^{2}(0.90)}{(0.12)(0.10)+(0.05)(0.90)}=0.6474
\end{aligned}
\]
Question: Let \(X\) represent the number of defective parts in a shipment of four.
\[
\mathrm{P}[X \geq x]=\frac{1}{4}\left(1-\frac{x-2}{3}\right)^{2}, \text { for } x=1,2,3,4
\]

Calculate \(\mathrm{E}(X)\).

Options:
A
0.83
B
\(\quad 0.96\)
C
1.39
D
1.81
E
1.83

Correct answer: A

Solution: \(p(x)=\left\{\begin{array}{l}1-P(X \geq 1)=1-0.444=0.556, \text { for } x=0 \\ P(X \geq 1)-P(X \geq 2)=0.444-0.250=0.194, \text { for } x=1 \\ P(X \geq 2)-P(X \geq 3)=0.250-0.111=0.139, \text { for } x=2 \\ P(X \geq 3)-P(X \geq 4)=0.111-0.028=0.083, \text { for } x=3 \\ P(X \geq 4)=0.028, \text { for } x=4\end{array}\right.\)
\(\mathrm{E}(X)=0(0.556)+1(0.194)+2(0.139)+3(0.083)+4(0.028)=0.833\).
Question: The time, in years, until replacement for a new telephone pole has probability density function
\[
f(t)= \begin{cases}k t, & \text { for } 0<t<20 \\ 0, & \text { otherwise }\end{cases}
\]
where \(k\) is a constant.
Calculate the probability that a new telephone pole will be replaced within ten years given that it is not replaced within five years.

Options:
A
0.19
B
0.20
C
0.25
D
0.33
E
\(\quad 0.94\)

Correct answer: B

Solution: \(P[T \leq 10 \mid T \geq 5]=\mathrm{P}[5 \leq T \leq 10] / \mathrm{P}[T \leq 5]\)
\(\frac{\int_{5}^{10} k t d t}{\int_{5}^{20} k t d t}=\frac{\left.t^{2}\right|_{5} ^{10}}{\left.t^{2}\right|_{5} ^{20}}=\frac{75}{375}=0.20\)
Question: A flight is delayed due to bad weather. The delay time is modeled by a random variable with a continuous uniform distribution. The expected delay time is three hours, and the standard deviation of the delay time is one hour.

Calculate the shortest possible delay time, in hours.

Options:
A
0.58
B
1.27
C
1.73
D
2.31
E
\(\quad 2.42\)

Correct answer: B

Solution: \(E(X)=\frac{a+b}{2}=3\)
\(a+b=6\)
\(b=6-a\)
\(V(X)=\frac{(b-a)^{2}}{12}=1^{2}=1\)
\(\frac{(6-a-a)^{2}}{12}=1\)
\(a=1.26795\).
Question: A study of fire damage claims is conducted. Claim amounts are mutually independent and each claim amount exceeds 1000 with probability 0.20 .

Calculate the probability that at least one of three randomly selected claim amounts exceeds 1000.

Options:
A
0.472
B
0.488
C
0.512
D
0.528
E
0.600

Correct answer: B

Solution: The number of such claims, \(N\), is binomial with \(n=3\) and \(p=0.2\).
\(P(N>0)=1-P(N=0)=1-0.8^{3}=0.488\)