# Exam P Practice Problem 110 – likelihood of auto accidents

Problem 110-A

An actuary studied the likelihood of accidents in a one-year period among a large group of insured drivers. The following table gives the results.

Age Group Percent of Drivers Probability of 0 Accidents Probability of 1 Accident
16-20 15% 0.20 0.25
21-30 25% 0.35 0.40
31-50 35% 0.60 0.30
51-70 20% 0.67 0.23
71+ 5% 0.50 0.35

Suppose that a randomly selected insured driver in the studied group had at least 2 accidents in the past year. Calculate the probability that the insured driver is in the age group 21-30.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 7$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 4$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 3 \bold 0$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 0$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5$

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Problem 110-B

An auto insurance company performed a study on the frequency of accidents of its insured drivers in a one-year period. The following table gives the results of the study.

Age Group Percent of Drivers Probability of At Least 1 Accident
16-20 10% 0.30
21-40 20% 0.20
41-65 35% 0.10
66+ 35% 0.12

A randomly selected insured driver from the study was found to have no accidents in the one-year period.

Calculate the probability that the insured driver is from the age group 16-20.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 0 \bold 8$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 2$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 5$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 9$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 0$

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# Exam P Practice Problem 109 – counting insurance payments

Problem 109-A

Amounts of damages due to auto collision accidents follow a probability distribution whose density function is given by the following.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{8000} \ (400-40x+x^2) &\ \ \ \ \ \ 0 < x < 20 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

When occurred, the collision damages are reimbursed by an insurance coverage subject to a deductible of 4.

Fifteen unrelated auto collision accidents have been reported. Determine the probability that exactly nine or ten of the accidents will be reimbursed by the insurance coverage.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 4 \bold 2$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 6 \bold 3$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 2$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 6 \bold 6$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 8 \bold 9$

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Problem 109-B

Amounts of damages due to auto collision accidents follow a probability distribution whose density function is given by the following.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{4000} \ (400-80x+4 x^2) &\ \ \ \ \ \ 0 < x < 10 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

When occurred, the damages are reimbursed by an insurance coverage subject to a deductible of 2.

Twelve unrelated auto collision accidents have been reported. Determine the probability that exactly six or seven of the accidents will not be reimbursed by the insurance coverage.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 8 \bold 4$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 2 \bold 2 \bold 5$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 0 \bold 8$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 2 \bold 7$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 0$

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# Exam P Practice Problem 108 – random selection of balls

Both 108-A and 108-B use the following information.

Bowl One contains 1 blue ball and 4 orange balls. Bowl Two contains 3 blue balls and 2 orange balls. A bowl is chosen at random. Balls are randomly chosen one at a time from the chosen bowl, with each chosen ball returning to the bowl.

.

Problem 108-A

What is the probability that four of the first six selections are blue ball?

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 4 \bold 8 \bold 6}{\bold 3 \bold 1 \bold 2 \bold 5}$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 0 \bold2}{\bold 6 \bold 2 \bold 5}$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 3 \bold 4}{\bold 2 \bold 0 \bold 5}$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 2 \bold 1}{\bold 1 \bold 2 \bold 5}$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 0 \bold 2 \bold 0}{\bold 3 \bold 1 \bold 2 \bold 5}$

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Problem 108-B

If four of the first six selections are blue balls, what is the probability that the balls are selected from Bowl One?

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 9}{\bold 2 \bold 5 \bold 5}$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 2}{\bold 4 \bold 3}$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 4}{\bold 8 \bold 5}$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1 \bold 5}{\bold 2 \bold 5 \bold 5}$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \frac{\bold 1}{\bold 2}$

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# Exam P Practice Problem 107 – wait time at a busy restaurant

Both 107-A and 107-B use the following probability density function.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{1}{450} \ (30-x) &\ \ \ \ \ \ 0 < x < 30 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

Problem 107-A

The wait time (in minutes) for a table at a busy restaurant on the weekend is distributed according to the density function $f(x)$ given above.

A customer plans to dine in this restaurant on two different weekends.

Determine the expected value of the longest wait of these two visits to the restaurant.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 0 \bold . \bold 0$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 1 \bold . \bold 0$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 2 \bold . \bold 8$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 3 \bold . \bold 5$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 4 \bold . \bold 0$

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Problem 107-B

The wait time (in minutes) for a table at a busy restaurant on the weekend is distributed according to the density function $f(x)$ given above.

A customer plans to dine in this restaurant on two different weekends.

Determine the expected value of the shortest wait of these two visits to the restaurant.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 4 \bold . \bold 5$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 6 \bold . \bold 0$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 7 \bold . \bold 0$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 8 \bold . \bold 6$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 1 \bold 0 \bold . \bold 0$

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# Exam P Practice Problem 106 – average height of students

Problem 106-A

Heights of male students in a large university follow a normal distribution with mean 69 inches and standard deviation 2.8 inches.

Four male students from this university are randomly selected.

Determine the probability that the average height of the selected students is between 5 feet 7 inches and 5 feet 11 inches.

Note that one feet = 12 inches.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 1 \bold 0 \bold 4$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 5 \bold 2 \bold 2 \bold 2$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 4 \bold 7 \bold 2$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 8 \bold 9 \bold 6$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 9 \bold 0 \bold 7 \bold 5$

The answers are based on this normal table from SOA.

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Problem 106-B

Heights of female students in a large university follow a normal distribution with mean 65 inches and standard deviation 2.2 inches.

Sixteen female students are randomly selected.

Determine the probability that the average height of the selected students is greater than 5 feet 6 inches.

Note that one feet = 12 inches.

$\displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 0 \bold 3 \bold 4 \bold 4$

$\displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 1 \bold 7 \bold 8 \bold 2$

$\displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 3 \bold 2 \bold 6 \bold 4$

$\displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 7 \bold 2$

$\displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 7 \bold 3 \bold 6$

The answers are based on this normal table from SOA.

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# Exam P Practice Problem 104 – two random insurance losses

Problem 104-A

Two random losses $X$ and $Y$ are jointly modeled by the following density function:

$\displaystyle f(x,y)=\frac{1}{32} \ (4-x) \ (4-y) \ \ \ \ \ \ 0

Suppose that both of these losses had occurred. Given that $X$ is exactly 2, what is the probability that $Y$ is less than 1?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{7}{24}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{11}{24}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{12}{24}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{24}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{24}$

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Problem 104-B

Two random losses $X$ and $Y$ are jointly modeled by the following density function:

$\displaystyle f(x,y)=\frac{1}{96} \ (x+2y) \ \ \ \ \ \ 0

Suppose that both of these losses had occurred. Determine the probability that $Y$ exceeds 2 given that the loss $X$ is known to be 2.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{36}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{24}{36}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{26}{36}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{28}{36}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{29}{36}$

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# Exam P Practice Problem 103 – randomly selected auto collision claims

Problem 103-A

The size of an auto collision claim follows a distribution that has density function $f(x)=2(1-x)$ where $0.

Two randomly selected claims are examined. Compute the probability that one claim is at least twice as large as the other.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{10}{36}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{15}{36}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{20}{36}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{21}{36}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{23}{36}$

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Problem 103-B

Auto collision claims follow an exponential distribution with mean 2.

For two randomly selected auto collision claims, compute the probability that the larger claim is more than four times the size of the smaller claims.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.4$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.5$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.6$

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# Exam P Practice Problem 101 – auto collision claims

Problem 101-A

The amount paid on an auto collision claim by an insurance company follows a distribution with the following density function.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{1}{96} \ x^3 \ e^{-x/2} &\ \ \ \ \ \ x > 0 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

The insurance company paid 64 claims in a certain month. Determine the approximate probability that the average amount paid is between 7.36 and 8.84.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.8320$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.8376$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.8435$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.8532$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.8692$

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Problem 101-B

The amount paid on an auto collision claim by an insurance company follows a distribution with the following density function.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{1}{1536} \ x^3 \ e^{-x/4} &\ \ \ \ \ \ x > 0 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$

The insurance company paid 36 claims in a certain month. Determine the approximate 25th percentile for the average claims paid in that month.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 15.11$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 15.43$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 15.75$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 16.25$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 16.78$

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# Exam P Practice Problem 100 – find the variance of loss in profit

Problem 100-A

The monthly amount of time $X$ (in hours) during which a manufacturing plant is inoperative due to equipment failures or power outage follows approximately a distribution with the following moment generating function.

$\displaystyle M(t)=\biggl( \frac{1}{1-7.5 \ t} \biggr)^2$

The amount of loss in profit due to the plant being inoperative is given by $Y=12 X + 1.25 X^2$.

Determine the variance of the loss in profit.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,927.20}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,608.20}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \text{475,693.76}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \text{583,358.20}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \text{601,769.56}$

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Problem 100-B

The weekly amount of time $X$ (in hours) that a manufacturing plant is down (due to maintenance or repairs) has an exponential distribution with mean 8.5 hours.

The cost of the downtime, due to maintenance and repair costs, is modeled by $Y=15+5 X+1.2 X^2$.

Determine the variance of the cost of the downtime.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \text{130,928.05}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \text{149,368.45}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \text{181,622.05}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \text{188,637.67}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \text{195,369.15}$

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Wording revised November 17, 2019

# Exam P Practice Problem 99 – When Random Loss is Doubled

Problem 99-A

A business owner faces a risk whose economic loss amount $X$ follows a uniform distribution over the interval $0. In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.

Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 0.5 is the responsibility of the business owner and any loss amount that is greater than 0.5 is paid by the insurer in full. When a loss occurs next year, determine the expected payment made by the insurer to the business owner.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{8}{16}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{9}{16}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{16}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{15}{16}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{17}{16}$

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Problem 99-B

A business owner faces a risk whose economic loss amount $X$ has the following density function:

$\displaystyle f(x)=\frac{x}{2} \ \ \ \ \ \ 0

In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.

Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 1 is the responsibility of the business owner and any loss amount that is greater than 1 is paid by the insurer in full. When a loss occurs next year, what is the expected payment made by the insurer to the business owner?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.6667$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 1.5833$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 1.6875$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 1.7500$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 2.6250$

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