# 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.

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

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

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

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

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

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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.

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

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

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

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

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

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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 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.

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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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# 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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# 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.

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

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

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

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

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

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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.

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

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

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

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

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

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# 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

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

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

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

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

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

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

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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.

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

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

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

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

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

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# Exam P Practice Problem 96 – Expected Insurance Payment

Problem 96-A

An insurance policy is purchased to cover a random loss subject to a deductible of 1. The cumulative distribution function of the loss amount $X$ is:

$\displaystyle F(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ x<0 \\ \text{ } & \text{ } \\ \displaystyle \frac{3}{25} \ x^2 - \frac{2}{125} \ x^3 &\ \ \ \ \ \ 0 \le x<5 \\ \text{ } & \text{ } \\ 1 &\ \ \ \ \ \ 5

Given a random loss $X$, determine the expected payment made under this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.50$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 1.54$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 1.72$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 4.63$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 6.26$

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Problem 96-B

An insurance policy is purchased to cover a random loss subject to a deductible of 2. The density function of the loss amount $X$ is:

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{8} \biggl(1- \frac{1}{4} \ x + \frac{1}{64} \ x^2 \biggr) &\ \ \ \ \ \ 0

Given a random loss $X$, what is the expected benefit paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.51$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.57$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.63$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 1.60$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 2.00$

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# Exam P Practice Problem 95 – Measuring Dispersion

Problem 95-A

The lifetime (in years) of a machine for a manufacturing plant is modeled by the random variable $X$. The following is the density function of $X$.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{2500} \ (100x-20x^2+ x^3) &\ \ \ \ \ \ 0

Calculate the standard deviation of the lifetime of such a machine.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.0$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.7$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.0$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.0$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.9$

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Problem 95-B

The travel time to work (in minutes) for an office worker has the following density function.

$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{1000} \ (50-5x+\frac{1}{8} \ x^2) &\ \ \ \ \ \ 0

Calculate the variance of the travel time to work for this office worker.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.87$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5.00$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6.50$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.75$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15.00$

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# Exam P Practice Problem 94 – Tracking High School Students

Problem 94-A

A researcher tracked a group of 900 high school students taking standardized tests in math and chemistry. Some of the students were given after-school tutoring before the tests (in both subjects) and the rest of the students received no tutoring. The following information is known about the test results:

• 510 of the students passed math test and 475 of the students passed chemistry test.
• Of the students who failed both subjects, there were 20% more students who did not receive tutoring than there were students who received tutoring.
• Of the students who failed chemistry and had tutoring, there were 99 more students who failed math than there were students who passed math.
• Of the students who failed chemistry and had no tutoring, there were 4 more students who failed math than there were students who passed math.
• There were 126 students who failed math and passed chemistry.

Determine the probability that a randomly selected student from this group had tutoring given that the student passed both subjects.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.6810$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.6828$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.6859$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.6877$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.6989$

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Problem 94-B

An insurance company tracked a group of 800 insureds for 2 years. It was found that 560 of the insureds had no claims in year 1 and 380 of the insureds had no claims in year 2. Of the insureds who had no claims in both years, there were four times as many male insureds than there were female insureds. Furthermore, there were 230 male insureds who had no claims in year 2 and there were 53 females insureds who had claims in both years. It is also known that there were 85 male insureds who had claims in year 1.

Determine the number of insureds who had no claims in year 1 but had claims in year 2.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 320$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 347$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 369$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 420$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 560$

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$\copyright \ 2016 \ \ \text{ Dan Ma}$

# Exam P Practice Problem 93 – Determining Average Claim Frequency

Problem 93-A

An actuary performs a claim frequency study on a group of auto insurance policies. She finds that the probability function of the number of claims per week arising from this set of policies is $P(N=n)$ where $n=1,2,3,\cdots$. Furthermore, she finds that $P(N=n)$ is proportional to the following function:

$\displaystyle \frac{e^{-2.9} \cdot 2.9^n}{n!} \ \ \ \ \ \ \ n=1,2,3,\cdots$

What is the weekly average number of claims arising from this group of insurance policies?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.900$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 3.015$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3.036$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 3.069$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 3.195$

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Problem 93-B

Let $N$ be the number of taxis arriving at an airport terminal per minute. It is observed that there are at least 2 arrivals of taxis in each minute. Based on a study performed by a traffic engineer, the probability $P(N=n)$ is proportional to the following function:

$\displaystyle \frac{e^{-2.9} \cdot 2.9^n}{n!} \ \ \ \ \ \ \ n=2,3,4,\cdots$

What is the average number of taxis arriving at this airport terminal per minute?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.740$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 2.900$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3.339$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 3.489$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 3.692$

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