Author Archive: Dan Ma

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.

\text{ }

\text{ }

\text{ }

\text{ }

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.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

\copyright 2019 – Dan Ma

Advertisements

Exam P Practice Problem 105 – testing electronic devices

Problem 105-A

The length of operation (in years) for an electronic device follows an exponential distribution with mean 4. Ten such devices are being observed for one year for a quality control study.

The lengths of operation for these devices are independent.

Determine the probability that no more than three of the devices stop working before the end of the study.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \  \bold 0 \bold . \bold 2 \bold 2 \bold 5 \bold 7

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 1 \bold 3 \bold 2

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 7 \bold 5 \bold 6 \bold 8

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 3 \bold 8 \bold 9

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 8 \bold 5 \bold 6 \bold 0

\text{ }

\text{ }

\text{ }

\text{ }

Problem 105-B

Twelve patients are randomly selected from a population of patients with history of heart disease to be tracked in a health study. The study begins with an initial assessment of health status. The participants are instructed to return for a follow up visit one year after the initial assessment.

For these patients, the time (in years) from the initial assessment to the next heart attack has an exponential distribution with mean 6.25 years. The times to the next heart attack for these patients are independent.

Determine the probability that ten or more patients experience no heart attack prior to the one-year follow up visit.

      \displaystyle \bold ( \bold A \bold ) \ \ \ \ \ \ \ \ \ \ \ \  \bold 0 \bold . \bold 2 \bold 9 \bold 1 \bold 3

      \displaystyle \bold ( \bold B \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 4 \bold 5 \bold 1 \bold 9

      \displaystyle \bold ( \bold C \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 5 \bold 4 \bold 8 \bold 1

      \displaystyle \bold ( \bold D \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 6 \bold 4 \bold 5 \bold 5

      \displaystyle \bold ( \bold E \bold ) \ \ \ \ \ \ \ \ \ \ \ \ \bold 0 \bold . \bold 7 \bold 4 \bold 3 \bold 2

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

daniel ma actuarial science

Daniel Ma actuarial

\copyright 2019 – Dan Ma

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<x<4, \ 0<y<x

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}

\text{ }

\text{ }

\text{ }

\text{ }

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<x<4, \ 0<y<4

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}

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

Daniel Ma actuarial

\copyright 2018 – Dan Ma

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<x<1.

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

\text{ }

      \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}

\text{ }

\text{ }

\text{ }

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.

\text{ }

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

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

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

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

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

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers can be found in this page.

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

Daniel Ma actuarial

\copyright 2018 – Dan Ma

Exam P Practice Problem 102 – estimating claim costs

Problem 102-A

Insurance claims modeled by a distribution with the following cumulative distribution function.

    \displaystyle  F(x) = \left\{ \begin{array}{ll}           \displaystyle  0 &\ \ \ \ \ \ x \le 0 \\            \text{ } & \text{ } \\          \displaystyle  \frac{1}{1536} \ x^4 &\ \ \ \ \ \ 0 < x \le 4 \\           \text{ } & \text{ } \\          \displaystyle  1-\frac{2}{3} x+\frac{1}{8} x^2- \frac{1}{1536} \ x^4 &\ \ \ \ \ \ 4 < x \le 8 \\           \text{ } & \text{ } \\          \displaystyle  1 &\ \ \ \ \ \ x > 8 \\                      \end{array} \right.

The insurance company is performing a study on all claims that exceed 3. Determine the mean of all claims being studied.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 4.8

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 4.9

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 5.0

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 5.1

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 5.2

\text{ }

\text{ }

\text{ }

Problem 102-B

Insurance claims modeled by a distribution with the following cumulative distribution function.

    \displaystyle  F(x) = \left\{ \begin{array}{ll}           \displaystyle  0 &\ \ \ \ \ \ x \le 0 \\            \text{ } & \text{ } \\          \displaystyle  \frac{1}{50} \ x^2 &\ \ \ \ \ \ 0 < x \le 5 \\           \text{ } & \text{ } \\          \displaystyle  -\frac{1}{50} x^2+\frac{2}{5} x- 1 &\ \ \ \ \ \ 5 < x \le 10 \\           \text{ } & \text{ } \\          \displaystyle  1 &\ \ \ \ \ \ x > 10 \\                      \end{array} \right.

The insurance company is performing a study on all claims that exceed 4. Determine the mean of all claims being studied.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 5.9

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 6.0

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 6.1

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 6.2

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 6.3

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers can be found in this page.

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

Daniel Ma actuarial

\copyright 2018 – Dan Ma

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.

\text{ }

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

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

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

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

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

\text{ }

\text{ }

\text{ }

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.

\text{ }

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

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

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

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

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

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers can be found in this page.

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

Daniel Ma actuarial

\copyright 2017 – Dan Ma

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.

\text{ }

      \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}

\text{ }

\text{ }

\text{ }

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 lost production and maintenance and repair costs, is modeled by Y=15+5 X+1.2 X^2.

Determine the variance of the cost of the downtime.

\text{ }

      \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}

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers can be found in this page.

probability exam P

actuarial exam

math

Daniel Ma

mathematics

dan ma actuarial science

Daniel Ma actuarial

\copyright 2017 – Dan Ma

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<x<1. 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.

\text{ }

      \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}

\text{ }

\text{ }

\text{ }

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<x<2

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?

\text{ }

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

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

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

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

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

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

Answers can be found in this page.

probability exam P

actuarial exam

math

Daniel Ma

mathematics

expected insurance payment

deductible

\copyright 2017 – Dan Ma

Exam P Practice Problem 98 – flipping coins

Problem 98-A

Coin 1 is an unbiased coin, i.e. when flipping the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when flipping the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained.

Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2856

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3060

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.3295

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.3564

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.3690

\text{ }

\text{ }

\text{ }

Problem 98-B

Box 1 contains 3 red balls and 1 white ball while Box 2 contains 2 red balls and 2 white balls. The two boxes are identical in appearance. One of the boxes is chosen at random. A ball is sampled from the chosen box with replacement until a white ball is obtained.

Determine the probability that the chosen box is Box 1 if the first white ball is observed on the 6th draw.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.7530

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.7632

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.7825

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.7863

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.7915

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

probability exam P

actuarial exam

math

Daniel Ma

mathematics

geometric distribution

Bayes

Answers

\copyright 2017 – Dan Ma

Exam P Practice Problem 97 – Variance of Claim Sizes

Problem 97-A

For a type of insurance policies, the following is the probability that the size of claim is greater than x.

    \displaystyle  P(X>x) = \left\{ \begin{array}{ll}           \displaystyle  1 &\ \ \ \ \ \ x \le 0 \\            \text{ } & \text{ } \\          \displaystyle  \biggl(1-\frac{x}{10} \biggr)^6 &\ \ \ \ \ \ 0<x<10 \\           \text{ } & \text{ } \\           0 &\ \ \ \ \ \ x \ge 10           \end{array} \right.

Calculate the variance of the claim size for this type of insurance policies.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \frac{10}{7}

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \frac{75}{49}

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \frac{95}{49}

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \frac{15}{7}

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \frac{25}{7}

\text{ }

\text{ }

\text{ }

Problem 97-B

For a type of insurance policies, the following is the probability that the size of a claim is greater than x.

    \displaystyle  P(X>x) = \left\{ \begin{array}{ll}           \displaystyle  1 &\ \ \ \ \ \ x \le 0 \\            \text{ } & \text{ } \\          \displaystyle  \biggl(\frac{250}{x+250} \biggr)^{2.25} &\ \ \ \ \ \ x>0 \\                             \end{array} \right.

Calculate the expected claim size for this type of insurance policies.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 200.00

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 203.75

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 207.67

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 217.32

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 232.74

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

probability exam P

actuarial exam

math

Daniel Ma

mathematics

Answers

\copyright 2017 – Dan Ma