BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: The amount of loss under an insurance policy is uniformly distributed on \([10,60]\). The insurer will reimburse the full amount of the loss up to a maximum benefit. The expected benefit per loss is equal to 31 .

Calculate the maximum benefit under the policy.

Options:
A
35
B
38
C
40
D
52
E
56

Correct answer: C

Solution: Let \(b=\) maximum benefit. Then,
\(31=\int_{10}^{b} x / 50 d x+\int_{b}^{60} b / 50 d x=\frac{b^{2}-10^{2}}{100}+b \frac{60-b}{50}=\frac{-b^{2}+120 b-100}{100}\)
\(3100=-b^{2}+120 b-100\)
\(b^{2}-120 b+3200=0\)
\((b-40)(b-80)=0\).
The only feasible solution is \(b=40\).
Question: An insurer sells a group life and disability policy.

The joint probability distribution for death and disability is given in the table below.

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \multicolumn{2}{|c|}{} & \multicolumn{5}{c|}{ Number of Deaths } \\
\cline { 2 - 7 } \multicolumn{2}{|c|}{} & 0 & 1 & 2 & 3 & 4 \\
\hline \multirow{4}{*}{\begin{tabular}{c} 
Number of \\
Disabilities
\end{tabular}} & 0 & 0.51 & 0.09 & 0.04 & 0.01 & 0.01 \\
\cline { 2 - 7 } & 1 & 0.08 & 0.06 & 0.02 & 0.01 & 0.01 \\
\cline { 2 - 7 } & 2 & 0.04 & 0.03 & 0.01 & 0.01 & 0.00 \\
\cline { 2 - 7 } & 3 & 0.03 & 0.02 & 0.01 & 0.01 & 0.00 \\
\hline
\end{tabular}

Calculate the probability of at least two disabilities given no more than one death.

Options:
A
0.12
B
0.14
C
0.15
D
0.25
E
\(\quad 0.75\)

Correct answer: B

Solution: \(\mathrm{P}[\) Dis \(\geq 2 \mid\) Death \(\leq 1]=\frac{0.04+0.03+0.03+0.02}{0.51+0.09+0.08+0.06+0.04+0.03+0.03+0.02}=\frac{0.12}{0.86}=0.1395\).
Question: Three identical machines are purchased simultaneously and are still operating after one year. The manufacturer then buys an insurance policy that pays 1000 if any of the three machines fails during the second year of use. The three machines operate independently.

The age at failure, in years, for each of these machines has cumulative distribution function
\[
F(x)= \begin{cases}1-\left(\frac{1}{x}\right)^{\frac{1}{4}}, & \text { for } x \geq 1 \\ 0, & \text { otherwise. }\end{cases}
\]

Note that \(x \geq 1\) because each machine has already successfully completed one year of operation.

Calculate the expected claim payment for this policy.

Options:
A
4
B
159
C
405
D
595
E
996

Correct answer: C

Solution: The probability that a machine fails in the second year is \(F(2)=1-(1 / 2)^{1 / 4}=0.1591\).
The probability that at least one machine fails is the complement of the probability of no machines failing, \(1-(1-0.1591)^{3}=0.4054\). The expected payment is \(1000(0.4054)=405\).
Question: In a large state, drivers are divided into three distinct age groups. The following table provides the probabilities that a one-car accident, involving a driver in a given age group, occurs in an urban, suburban or rural location in that state.

\begin{tabular}{|l||c|c|c|}
\cline { 2 - 4 } \multicolumn{1}{c|}{} & \multicolumn{3}{c|}{ Probability of accident location by age group } \\
\hline Driver's age & Urban & Suburban & Rural \\
\hline Under 25 & \(7 / 10\) & \(1 / 5\) & \(1 / 10\) \\
\hline \(25-65\) & \(1 / 4\) & \(3 / 5\) & \(3 / 20\) \\
\hline Over 65 & \(1 / 20\) & \(7 / 20\) & \(3 / 5\) \\
\hline
\end{tabular}

Fifty-five percent of one-car accidents involve a driver who is under 25, and fifteen percent involve a driver who is over 65.

A one-car accident occurs in a suburban location.
Calculate the probability that the driver is under 25.

Options:
A
0.11
B
0.17
C
0.32
D
0.55
E
\(\quad 0.76\)

Correct answer: C

Solution: \(P(\) suburban \()=0.55(1 / 5)+0.30(3 / 5)+0.15(7 / 20)=0.3425\).
\(P(\) suburban and under 25) \(=0.55(1 / 5)=0.11\).
\(P(\) under \(25 \mid\) suburban \()=0.11 / 0.3425=0.3212\).
Question: A medical insurance policyholder just experienced a hospitalization, resulting in an undetermined financial loss that is exponentially distributed. The policyholder has Plan A, which reimburses the first two units for each hospitalization. The policyholder could have chosen Plan B, which reimburses the first ten units for each hospitalization.

The probability that this hospitalization is completely covered under Plan A is 0.15
Calculate the probability that this hospitalization would have been completely covered had the policyholder chosen Plan B.

Options:
A
0.200
B
0.351
C
0.443
D
0.556
E
0.750

Correct answer: D

Solution: Let \(X=\) loss due to hospitalization, which is exponentially distributed with parameter \(\lambda\).
\(0.15=P(X \leq 2)=\int_{0}^{2} \lambda e^{-\lambda x} d x=1-e^{-2 \lambda}\)
\(0.85=e^{-2 \lambda}\).
\(P(X \leq 10)=1-e^{-10 \lambda}=1-0.85^{5}=0.55629\).
Question: A construction worker experiences a back injury resulting in a loss on an insurance policy. Losses under the policy are uniformly distributed on the interval [2, 10]. The insurer reimburses the worker for \(60 \%\) of this loss.

Calculate the \(30^{\text {th }}\) percentile of the construction worker's unreimbursed loss from the back injury.

Options:
A
\(\quad 1.76\)
B
1.80
C
2.64
D
3.04
E
\(\quad 4.40\)

Correct answer: A

Solution: Let \(X\) be the loss, which is uniform [2,10].
Let \(Y\) be the unreimbursed loss. Then, \(Y=0.4 X\). Then,
\(0.3=P[Y \leq p]=P[0.4 X \leq p]=P[X \leq 2.5 p]=\frac{2.5 p-2}{10-2}=5 p / 16-1 / 4\)
\(5 p / 16=3 / 10+1 / 4=11 / 20\)
\(p=\frac{16(11)}{5(20)}=1.76\).
Question: A college office has two student assistants, Bob and Ann, who work on alternate days. Each day, the assistant draws a morning and an afternoon assignment from a jar containing four assignments, one outdoor and three indoor.

Bob always makes his selection with replacement, while Ann makes hers without replacement.

On a particular day, one outdoor and one indoor assignment are selected.
Calculate the probability that Bob made the selections.

Options:
A
\(\quad 3 / 16\)
B
\(\quad 3 / 8\)
C
\(3 / 7\)
D
\(\quad 7 / 16\)
E
\(1 / 2\)

Correct answer: C

Solution: The probability of getting one of each type when Bob is selecting is \(\frac{3}{4} \times \frac{1}{4} \times 2=\frac{3}{8}\). The same probability when Ann is selecting is \(\frac{3}{4} \times \frac{1}{3} \times 2=\frac{1}{2}\). Thus, the desired probability is \(\frac{\frac{3}{8}}{\frac{1}{2}+\frac{3}{8}}=\frac{3}{7}\).
Question: A random variable, \(X\), denotes the number of airplanes landing each hour at a certain airport. \(X\) is modeled by a Poisson distribution. On average, 20 airplanes are expected to land each hour.

Calculate \(\mathrm{E}\left(X^{2}\right)+\mathrm{E}(X)\).

Options:
A
40
B
60
C
420
D
440
E
820

Correct answer: D

Solution: \(\mathrm{E}(X)=\mathrm{V}(X)=\lambda=20\)
\(\mathrm{V}(X)=\mathrm{E}\left(X^{2}\right)-\mathrm{E}(X)^{2}\)
\(\mathrm{E}\left(X^{2}\right)=20+20^{2}=420\)
\(\mathrm{E}\left(X^{2}\right)+\mathrm{E}(X)=420+20=440\)
Question: A tornado occurs in each of two cities, A and B . The durations of the tornadoes in cities \(A\) and \(B\) are independent and exponentially distributed with means 5 minutes and 4 minutes, respectively.

The tornadoes in cities A and B result in damages valued at 1.00 and \(c\) per minute, respectively, where \(c\) is a constant.

The standard deviation of the total damage to the two cities is 5.20.
Calculate \(c\).

Options:
A
0.05
B
0.13
C
0.36
D
2.35
E
5.51

Correct answer: C

Solution: Let \(X\) represent the number of minutes that city A's tornado lasts.
Let \(Y\) represent the number of minutes that city B's tornado lasts.
\(X\) and \(Y\) are independent and exponentially distributed with means 5 and 4, respectively.
The variance of an exponential distribution is the square of its mean, the variance of a sum of independent random variables is the sum of the variances and multiplying a random variable by a constant results in multiplying the variance by the square of the constant. Therefore, the variance of the monetary value of the damage is
\[
(5.2)^{2}=\operatorname{Var}(X+c Y)=\operatorname{Var}(X)+\operatorname{Var}(c Y)=\operatorname{Var}(X)+c^{2} \operatorname{Var}(Y)=5^{2}+c^{2}(4)^{2}=25+16 c^{2} .
\]

Solving for \(c\) gives \(c=\sqrt{\frac{(5.2)^{2}-25}{16}}=0.35707\).
Question: The random vector ( \(W, X, Y\) ) has the following properties:
i) \(\quad W\) assumes values of only 0 and 1 .
ii) \(\quad X\) and \(Y\) each assume values of only 1 and 2 .
iii) \(\quad \mathrm{P}[W=w, X=x, Y=y]\) is proportional to \(x+y-w\).

Calculate \(\operatorname{Var}(W \mid Y=1)\).

Options:
A
\(\quad 51 / 400\)
B
\(15 / 64\)
C
\(\quad 6 / 25\)
D
\(\quad 3 / 8\)
E
\(\quad 2 / 5\)

Correct answer: B

Solution: Here are the possible vectors with their probabilities:

\begin{tabular}{|c|c|c|}
\hline\((w, x, y)\) & \(x+y-w\) & probability \\
\hline\((0,1,1)\) & 2 & \(2 / 20\) \\
\hline\((0,1,2)\) & 3 & \(3 / 20\) \\
\hline\((0,2,1)\) & 3 & \(3 / 20\) \\
\hline\((0,2,2)\) & 4 & \(4 / 20\) \\
\hline\((1,1,1)\) & 1 & \(1 / 20\) \\
\hline\((1,1,2)\) & 2 & \(2 / 20\) \\
\hline\((1,2,1)\) & 2 & \(2 / 20\) \\
\hline\((1,2,2)\) & 3 & \(3 / 20\) \\
\hline
\end{tabular}

Given that \(Y=1, W\) is 0 with probability \((2+3) /(2+3+1+2)=5 / 8\) and 1 with probability \(3 / 8\). Then,
\(E(W)=\frac{3}{8}\)
\(E\left(W^{2}\right)=\frac{3}{8}\)
\(V(W)=\frac{3}{8}-\left(\frac{3}{8}\right)^{2}=\frac{15}{64}\).
Question: In a certain region, \(5 \%\) of all banks will fail in the next five years. The FDIC insures \(80 \%\) of the banks in the region. Only \(3 \%\) of the banks in the region insured by the FDIC will fail in the next five years.

Calculate the probability that a bank in the region that fails in the next five years is insured by the FDIC.

Options:
A
0.02
B
0.04
C
0.06
D
0.48
E
0.60

Correct answer: D

Solution: \(\mathrm{P}(F)=0.05, \mathrm{P}(F D I C)=0.80\)
\(\mathrm{P}(F \mid F D I C)=0.03=\frac{\mathrm{P}(F \cap F D I C)}{\mathrm{P}(F D I C)}=\frac{\mathrm{P}(F \cap F D I C)}{0.80} \Rightarrow \mathrm{P}(F \cap F D I C)=0.024\).
We want: \(\mathrm{P}(F D I C \mid F)=\frac{\mathrm{P}(F \cap F D I C)}{\mathrm{P}(F)}=\frac{0.024}{0.05}=0.48\).
Question: An actuary accepts an assignment to live and work in Key West, Florida for three years.

The probability that there are no hurricanes in Key West is 0.75 per year. If there is at least one hurricane in Key West in a year, then the probability is 0.35 that the actuary will need to evacuate in that year. There is no other reason for the actuary to evacuate.

The year-to-year occurrences of hurricanes in Key West are independent events.
Calculate the probability that the actuary will need to evacuate at least once from Key West before the assignment is complete.

Options:
A
0.088
B
0.219
C
0.240
D
0.422
E
0.578

Correct answer: C

Solution: In a given year, the probability of evacuating is the probability of evacuating given there is at least one hurricane times the probability of at least one hurricane, \(0.35(0.25)=0.0875\).
The probability of never evacuating in three years is \(0.9125^{3}=0.7598\).
The probability of being evacuated at least once is \(1-0.7598=0.2402\).
Question: Each of the 76 employees of a company is either a low-risk or a high-risk employee. This year, each low-risk employee's number of accidents has standard deviation 0.50 and each high-risk employee's number of accidents has standard deviation 5.50. The standard deviation of the total number of accidents for all the employees is 43.00 . The frequencies of accidents among the employees are independent.

Calculate the number of low-risk employees.

Options:
A
1
B
15
C
38
D
61
E
75

Correct answer: B

Solution: Let \(l=\) number of low-risk employees; therefore, there are \(76-l\) high-risk employees. For independent random variables, the variance of the sum equals the sum of the variances. Therefore: \(43^{2}=1849=\) Var[Total \# of Accidents] \(=l \times \operatorname{Var}[\#\) of Accidents of Low Risk Ee] \(+(76-l) \times \operatorname{Var}[\#\) of Accidents of High Risk Ee] \(=l(0.50)^{2}+(76-l)(5.50)^{2} =2299-30 l 450=30 l l=15\).
Question: Losses under a boat insurance policy are exponentially distributed with mean \(\frac{400}{\ln 2}\).

For each loss, the claim payment is the amount of the loss, up to a maximum of 1000.
Calculate the expected value of a claim payment.

Options:
A
298
B
367
C
400
D
475
E
577

Correct answer: D

Solution: Let \(\beta=400 / \ln 2\) be the mean. The expected claim payment is
\(\int_{0}^{\infty} \min (x, 1000) \beta^{-1} e^{-x / \beta} d x=\int_{0}^{1000} \beta^{-1} x e^{-x / \beta} d x+\int_{1000}^{\infty} 1000 \beta^{-1} e^{-x / \beta} d x=-\left.(x+\beta) e^{-x / \beta}\right|_{0} ^{1000}-\left.1000 e^{-x / \beta}\right|_{1000} ^{\infty} =\beta-(1000+\beta) e^{-1000 / \beta}+1000 e^{-1000 / \beta}=\beta\left(1-e^{-1000 / \beta}\right)=\frac{400}{\ln 2}\left(1-e^{-2.5 \ln 2}\right)=475.06\).
Question: A company tests five batteries. Each battery has a probability \(p\) of being defective, where \(p\) is a constant. The events of different batteries being defective are mutually independent.

The number of defective batteries out of the five batteries is a random variable, \(N\), where \(\mathrm{E}\left(N^{2}\right)=1.80\).

Calculate \(\operatorname{Var}(N)\).

Options:
A
0.80
B
0.98
C
1.15
D
1.67
E
1.80

Correct answer: A

Solution: The second moment is the variance plus the square of the mean. For a binomial distribution, we have (knowing that \(p\) must be positive)
\(1.80=5 p(1-p)+(5 p)^{2}=5 p+20 p^{2}\)
\(0=20 p^{2}+5 p-1.80\)
\(p=\frac{-5 \pm \sqrt{25+144}}{40}=\frac{-5+13}{40}=0.20\).
The variance is \(5 p(1-p)=5(0.2)(1-0.2)=0.80\).
Question: A homeowner hires a moving company to transport ten pieces of electronic equipment to a new home. Each piece is equally likely to be damaged during the move, and the event of damage to any one piece is independent of the events of damage to the other pieces.

The homeowner fully insures two of these pieces, partially insures three of them, and leaves the other five pieces uninsured.

During the move, three pieces are damaged.
Calculate the probability that at least one damaged piece is fully insured and at least one damaged piece is partially insured.

Options:
A
0.242
B
0.250
C
0.325
D
0.378
E
\(\quad 0.400\)

Correct answer: C

Solution: There are \({ }_{10} \mathrm{C}_{3}=120\) possible and equally likely combinations of three damaged pieces.
There are three events to add up, with the indicated number of possibilities.
i) 1 damaged piece insured, 1 damaged piece partially insured, 1 damaged piece uninsured;
\({ }_{2} \mathrm{C}_{1}\left({ }_{3} \mathrm{C}_{1}\right)\left({ }_{5} \mathrm{C}_{1}\right)=2(3)(5)=30\).
ii) 1 damaged piece insured, 2 damaged pieces partially insured, 0 damaged pieces uninsured;
\(\left({ }_{2} \mathrm{C}_{1}\right)\left({ }_{3} \mathrm{C}_{2}\right)\left({ }_{5} \mathrm{C}_{0}\right)=2(3)(1)=6\).
iii) 2 damaged pieces insured, 1 damaged piece partially insured, 0 damaged pieces uninsured; \(\left({ }_{2} \mathrm{C}_{2}\right)\left({ }_{3} \mathrm{C}_{1}\right)\left({ }_{5} \mathrm{C}_{0}\right)=1(3)(1)=3\).
The solution is \((30+6+3) / 120=39 / 120=0.325\).
Question: The number of thefts in any year a warehouse owner experiences is modeled by a Poisson distribution. In each year, the probability of at least one theft is 0.10 , and the yearly number of thefts are independent.

Calculate the variance of the total number of thefts the owner experiences in the next 15 years.

Options:
A
0.11
B
0.41
C
1.35
D
1.50
E
1.58

Correct answer: E

Solution: Let \(X\) be number of thefts in one year and \(S\) be the total number of thefts in 15 years. The sum of 15 independent Poisson distributions with mean \(\lambda\) is Poisson with mean \(15 \lambda\).
\(0.10=\mathrm{P}[X \geq 1]=1-\mathrm{P}[X=0]=1-\frac{e^{-\lambda} \lambda^{0}}{0!}=1-e^{-\lambda}\)
\(0.9=e^{-\lambda}\)
\(\lambda=0.105361\)
\(\operatorname{Var}(S)=15 \lambda=1.580408\).
Question: The check-out times for customers at a grocery store are modeled by a continuous uniform distribution between 1.1 and 8.6 minutes. The store would like to advertise that "no more than \(5 \%\) of customers require more than \(y\) minutes to check out."

Calculate the minimum value for \(y\).

Options:
A
1.475
B
7.125
C
8.170
D
8.225
E
8.545

Correct answer: D

Solution: Let \(X=\) checkout time, in minutes. Then \(X\) has a Uniform distribution on the interval [1.1, 8.6].
\(\mathrm{P}[X<y]=\frac{y-1.1}{8.6-1.1}=0.95\)
\(y=8.225\).
Question: A property-casualty insurer sells only homeowners insurance. Among the policies that the company sold, \(40 \%\) belong to policyholders who have a security system for their houses and \(60 \%\) belong to those who did not.

Assume in any particular year:
i) For any house without a security system, the number of break-ins follows a Poisson distribution, with mean 3.
ii) For any house with a security system, the number of break-ins follows a Poisson distribution, and the probability of no break-ins is three times that for a house without a security system.

Calculate the probability that a house has a security system, given that there are exactly two break-ins in a year.

Options:
A
0.11
B
0.27
C
0.40
D
0.45
E
0.55

Correct answer: D

Solution: Let \(S\) be the number of break-ins with a security system and \(N\) be the number without. Let \(E(S)=\lambda\). Then,
\(P(S=0)=e^{-\lambda}=3 P(N=0)=3 e^{-3}\)
\(-\lambda=\ln \left(3 e^{-3}\right)=(\ln 3)-3\)
\(\lambda=3-\ln 3=1.90139\).
Let \(A\) be the event that the house has a security system and \(X\) be the number of break-ins. Then, \(P(A \mid X=2)=\frac{P(X=2 \mid A) P(A)}{P(X=2 \mid A) P(A)+P\left(X=2 \mid A^{c}\right) P\left(A^{c}\right)}\)
\(=\frac{P(S=2) P(A)}{P(S=2) P(A)+P(N=2) P\left(A^{c}\right)}=\frac{\frac{1.90139^{2} e^{-1.90139}}{2!}(0.4)}{\frac{1.90139^{2} e^{-1.90139}}{2!}(0.4)+\frac{3^{2} e^{-3}}{2!}(0.6)}\)
\(=\frac{0.10800}{0.10800+0.13443}=0.44549\).
Question: A taxicab company purchases an insurance policy that covers damages from car accidents. The number of accidents on each day is Poisson distributed with mean \(\lambda\), and the number of accidents on different days are mutually independent. The expected number of accidents each week is 63 .

Calculate the probability that the number of accidents in one day is two or more standard deviations below the expected number of accidents in one day.

Options:
A
0.006
B
0.021
C
0.116
D
0.156
E
0.207

Correct answer: B

Solution: Let \(D\) be the number of accidents in one day. Due to independence the expected number of accidents in one day is \(\mathrm{E}(D)=63 / 7=9\). For a Poisson distribution the variance is equal to the mean. Thus, \(\operatorname{Var}(D)=\mathrm{E}(D)=9, \mathrm{SD}(D)=3\).

The probability of being two more standard deviations below the mean is \(\mathrm{P}[D \leq 9-2 \times 3]=\mathrm{P}[D \leq 3]\)
\(=\mathrm{P}[D=0]+\mathrm{P}[D=1]+\mathrm{P}[D=2]+\mathrm{P}[D=3]\)
\(=\frac{e^{-9} 9^{0}}{0!}+\frac{e^{-9} 9^{1}}{1!}+\frac{e^{-9} 9^{2}}{2!}+\frac{e^{-9} 9^{3}}{3!}\)
\(=0.000123+0.001111+0.004998+0.014994=0.021226\).
Question: Two random number generators, labeled \(C\) and \(D\), are used to select numbers between 0 and 10. Generator C uses the continuous uniform distribution on \([0,10]\). Generator D uses the discrete uniform distribution on the integers 0 through 10 .

Calculate the absolute value of the difference between the standard deviation of the selection using \(C\) and the standard deviation of the selection using \(D\).

Options:
A
0.00
B
0.02
C
0.28
D
0.50
E
1.67

Correct answer: C

Solution: For Random Number Generator C:
\(\operatorname{Var}(C)=\frac{(10-0)^{2}}{12}=8.3333, \mathrm{SD}(C)=\sqrt{8.3333}=2.8868\)
For Random Number Generator D :
\(\mathrm{E}(D)=\frac{1}{11}[0+1+2+3+4+5+6+7+8+9+10]=5\)
\(\mathrm{E}\left(D^{2}\right)=\frac{1}{11}\left[0^{2}+1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}\right]=35\)
\(\operatorname{Var}(D)=35-5^{2}=10, S D(D)=\sqrt{10}=3.1623\).
The absolute difference is: \(|2.8868-3.1623|=0.2755\).
Question: A policyholder owns two properties and has hurricane insurance on both properties. One property is located in Texas. The other property is located in Florida.

Let \(T\) be the number of hurricane claims in a 5 -year period for the Texas property, and let \(F\) be the number of hurricane claims in a 5 -year period for the Florida property. The joint probability function for \(T\) and \(F\) is given in the following table.

\begin{tabular}{|c|r|r|r|r|}
\hline \multicolumn{5}{|c|}{ Number of Hurricane Claims in 5-year period } \\
\hline \multirow{2}{*}{ Texas } & 0 & 1 & 2 & 3 \\
\cline { 2 - 5 } & & 0.14 & 0.04 & 0.02 \\
0 & 0.40 & 0.08 & 0.04 & 0.02 \\
1 & 0.16 & 0.03 & 0.02 & 0.01 \\
2 & 0.04 & & & \\
\hline
\end{tabular}

Calculate \(\operatorname{Var}(T \mid F=1)\).

Options:
A
0.18
B
0.24
C
0.45
D
0.49
E
0.80

Correct answer: D

Solution: \(\operatorname{Var}(T \mid F=1)=\mathrm{E}\left(T^{2} \mid F=1\right)-\{\mathrm{E}(T \mid F=1)\}^{2}\). To calculate these two expected values, we must first find the conditional probability mass function \(p(t \mid F=1)\), where \(t=0,1,2\).
\(p(t \mid F=1)= \begin{cases}14 / 25=0.56, & t=0 \\ 8 / 25=0.32, & t=1 \\ 3 / 25=0.12, & t=2\end{cases}\)
\(\mathrm{E}(T \mid F=1)=0(0.56)+1(0.32)+2(0.12)=0.56\)
\(\mathrm{E}\left(T^{2} \mid F=1\right)=0^{2}(0.56)+1^{2}(0.32)+2^{2}(0.12)=0.80\)
\(\operatorname{Var}(T \mid F=1)=0.80-0.56^{2}=0.4864\).
Question: The time in days, \(X\), that an individual suffers from a common cold is continuous and uniformly distributed on \([a, b]\). The mean of \(X\) is 8.50, and the variance of \(X\) is 0.75 .

Calculate the probability that a person who has suffered from a common cold for the past 7.5 days will continue to suffer from that cold for at least an additional 1.5 days.

Options:
A
0.33
B
0.40
C
0.50
D
0.60
E
0.83

Correct answer: B

Solution: \(\frac{a+b}{2}=8.5, a=17-b\)
\(\frac{(b-a)^{2}}{12}=0.75,(b-a)^{2}=9, b-a=3\)
\(b-(17-b)=3,2 b=20, b=10\)
\(a=17-10=7\)
\(\mathrm{P}[X>9 \mid X>7.5]=\frac{10-9 / 10-7}{10-7.5 / 10-7}=\frac{1.0}{2.5}=0.40\).
Question: The numbers of hits received by a website during the \(1^{\text {st }}, 2^{\text {nd }}, \ldots, 1440^{\text {th }}\) minutes of a given day are mutually independent Poisson random variables with common mean. The probability that there are exactly four hits during the \(1^{\text {st }}\) minute is 54 times the probability that there are no hits during the \(1440^{\text {th }}\) minute.

Calculate the number of hits the website is expected to receive between in a 60-minute period.

Options:
A
120
B
180
C
240
D
300
E
360

Correct answer: E

Solution: \(\frac{e^{-\lambda} \lambda^{4}}{4!}=\frac{54 e^{-\lambda} \lambda^{0}}{0!}\)
\(\frac{\lambda^{4}}{4!}=54\)
\(\lambda^{4}=1296, \lambda=6\)
The expected number of hits in a 60 -minute period is \(60(6)=360\).
Question: A policyholder owns a car and a truck. An auto insurer provides coverage on both vehicles.

Let \(X\) be the number of claims incurred over a five-year period on the car.
Let \(Y\) be the number of claims incurred over a five-year period on the truck.
The probability function of \(X\) and \(Y\) is shown in the table below:

\begin{tabular}{|c|r|c|c|c|c|}
\cline { 2 - 6 } \multicolumn{2}{c|}{} & \multicolumn{4}{c|}{\(Y\)} \\
\cline { 2 - 6 } \multicolumn{2}{c|}{} & 0 & 1 & 2 & 3 \\
\hline \multirow{3}{*}{\(X\)} & 0 & 0.06 & 0.10 & 0.12 & 0.02 \\
\cline { 1 - 6 } & 1 & 0.08 & 0.11 & 0.15 & 0.06 \\
\cline { 1 - 6 } & 2 & 0.05 & 0.07 & 0.05 & 0.03 \\
\cline { 1 - 6 } & 3 & 0.01 & 0.02 & 0.03 & 0.04 \\
\hline
\end{tabular}

Calculate \(\mathrm{E}(Y \mid X<2)\).

Options:
A
1.30
B
1.37
C
1.41
D
1.45
E
1.53

Correct answer: C

Solution: \[
\begin{aligned}
& p(y \mid X<2)= \begin{cases}0.14 / 0.70=0.2000, & y=0 \\
0.21 / 0.70=0.3000, & y=1 \\
0.27 / 0.70=0.3857, & y=2 \\
0.08 / 0.70=0.1143, & y=3\end{cases} \\
& E(Y \mid X<2)=0(0.2000)+1(0.3000)+2(0.3857)+3(0.1143)=1.4143
\end{aligned}
\]
Question: Let \(X\) be a continuous random variable with density function
\(f(x)= \begin{cases}\frac{p-1}{x^{p}}, & x>1 \\ 0, & \text { otherwise }\end{cases}\)
Calculate the value of \(p\) such that \(E(X)=2\).

Options:
A
1
B
2.5
C
3
D
5
E
There is no such \(p\).

Correct answer: C

Solution: \(E(X)=\int_{1}^{\infty} x \frac{p-1}{x^{p}} d x=(p-1) \int_{1}^{\infty} x^{1-p} d x\)
\(\left.(p-1) \frac{x^{2-p}}{2-p}\right|_{1} ^{\infty}=\frac{p-1}{p-2}=2\)
\(p-1=2(p-2)=2 p-4\)
\(p=3\)