# Exam P Practice Problem 90 – Insurance Benefits

Problem 90-A

A random loss follows an exponential distribution with mean 20. An insurance reimburses this random loss up to a benefit limit of 30.

When a loss occurs, what is the expected value of the benefit not paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 4.5$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 5.1$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 6.3$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 8.5$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 11.2$

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Problem 90-B

A random loss follows an exponential distribution with mean 100. An insurance reimburses this random loss up to a benefit limit of 200.

When a loss occurs, what is the expected value of the benefit not paid by this insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 12.6$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 13.5$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 24.6$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 40.6$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 40.7$

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$\copyright \ 2014 \ \ \text{ Dan Ma}$

# Exam P Practice Problem 89 – Finding Median

Problem 89-A

The random variables $X$ and $Y$ have the following joint density function.

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

Suppose that $m$ is the median of $X+Y$. Which of the following is true about $m$?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.5

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 2

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 3.5

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 3.5

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Problem 89-B

The random variable $X$ has the following density function.

$\displaystyle f(x)=\frac{3}{16000} \ (400-x^2) \ \ \ \ \ \ \ 0

Suppose that $m$ is the median of $X$. Which of the following is true about $m$?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 6

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 5.5

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 5.5

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 7

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 7

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$\copyright \ 2014 \ \ \text{ Dan Ma}$

# Exam P Practice Problem 87 – Modeling Insurance Payments

Problem 87-A

A business owner is facing a risk whose economic loss is modeled by the random variable $X$. The following is the density function of $X$.

$\displaystyle f(x)=\frac{1}{8} \ (4-x) \ \ \ \ \ \ \ \ 0

The business owner purchases an insurance policy to cover this potential loss. The insurance policy pays the business owner 80% of the amount of each loss.

Given that a loss has occurred, what is the probability that the amount of the insurance payment to the business owner is less than 2?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.25$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.36$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.64$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.75$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.86$

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Problem 87-B

An individual purchases an insurance policy to cover a loss $X$ whose density function is:

$\displaystyle f(x)=\frac{1}{1800} \ x \ \ \ \ \ \ \ \ 0

The insurance policy reimburses the policy owner 50% of each loss. Given that a loss has occurred, what is the median amount of the insurance payment made to the policy owner?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 15.00$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 18.65$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 21.21$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 23.63$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 42.43$

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$\copyright \ 2014 \ \ \text{ Dan Ma}$

# Exam P Practice Problem 82 – Estimating the Median Weight of Bears

Problem 82-A

A wildlife biologist wished to estimate the median weight of bears in Alaska. The weights of the bear population he studied follow a continuous distribution with an unknown median $M$. He captured a sample of 15 bears. Let $Y_5$ be the weight (in pounds) of the fifth smallest bear in the sample of 15 captured bears. Let $Y_{11}$ be the weight (in pounds) of the fifth largest bear in the sample.

Calculate the probability that the median $M$ is between $Y_5$ and $Y_{11}$, i.e., $P(Y_5.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.5000$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.7899$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.8218$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.8815$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.9232$

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Problem 82-B

The wildlife biologist in Problem 82-A also wishes to estimate $\tau_{75}$, the seventy fifth percentile of the weights of bear population he studied. Let $Y_{10}$ be the weight of the tenth smallest bear in the sample of 15 captured bears. Let $Y_{14}$ be the weight of the second largest bear in the sample of 15 bears.

Calculate the probability that $\tau_{75}$ is between $Y_{10}$ and $Y_{14}$, i.e., $P(Y_{10}<\tau_{75}.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.6155$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.7500$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.7715$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.8383$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.9232$

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$\copyright \ 2014 \ \text{ Dan Ma}$

# Exam P Practice Problem 66 – Median Cholesterol Level

Problem 66-A

The blood cholesterol levels of men aged 55 to 64 are normally distributed with mean 225 milligrams per deciliter (mg/dL) and standard deviation 39.5 mg/dL.

A medical researcher is planning a clinical study targeting men from the age group of 55 to 64 who have high blood cholesterol levels (above 240 mg/dL).

What is the median cholesterol level of the men in the target population of this medical study?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 225 \ \text{ mg/dL}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 249 \ \text{ mg/dL}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 262 \ \text{ mg/dL}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 284 \ \text{ mg/dL}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 294 \ \text{ mg/dL}$

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Problem 66-B

The blood cholesterol levels of women aged 55 to 64 are normally distributed with mean 190 milligrams per deciliter (mg/dL) and standard deviation 40 mg/dL.

A medical researcher is planning a clinical study targeting women from the age group of 55 to 64 who have borderline high blood cholesterol levels (between 200 and 240 mg/dL).

What is the median cholesterol level of the women in the target population of this medical study?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 214 \ \text{ mg/dL}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 216 \ \text{ mg/dL}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 218 \ \text{ mg/dL}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 220 \ \text{ mg/dL}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 225 \ \text{ mg/dL}$

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$\copyright \ 2013$

# Exam P Practice Problem 64 – Median Claim Size

Problem 64-A

The following is the density function of the claim size $X$ of an insurance policy from a block of property and casualty insurance policies:

$\displaystyle f(x)=\frac{1}{2} \biggl( 0.4 \ e^{-0.4 x} \biggr) + \frac{1}{2} \biggl( 0.2 \ e^{-0.2 x} \biggr) \ \ \ \ \ \ \ 0

Calculate the median claim size $X$.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.618$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.739$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.406$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.599$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.533$

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Problem 64-B

The following is the density function of the claim size $X$ of an insurance policy from a block of auto insurance policies:

$\displaystyle f(x)=\frac{3}{4} \biggl( \frac{1}{8} \ x \biggr) + \frac{1}{4} \biggl( \frac{1}{64} \ x^3 \biggr) \ \ \ \ \ \ \ 0

Calculate the median claim size $X$.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.2216$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.4972$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.8653$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.9622$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.9975$

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$\copyright \ 2013$

# Exam P Practice Problem 52 – Reliability of Refrigerators

Problem 52-A

The time from initial purchase to the time of the first major repair (in years) for a brand of refrigerators is modeled by the random variable $Y=e^X$ where $X$ is normally distributed with mean 1.2 and variance 2.25.

A customer just bought a brand new refrigerator of this particular brand. The refrigerator came with a two-year warranty. During the warranty period, any repairs, both minor and major, are the responsibilities of the manufacturer.

What is the probability that the newly purchased refrigerator will not require major repairs during the warranty period?

$\displaystyle (A) \ \ \ \ \ \ \ \ 0.2451$

$\displaystyle (B) \ \ \ \ \ \ \ \ 0.2981$

$\displaystyle (C) \ \ \ \ \ \ \ \ 0.3669$

$\displaystyle (D) \ \ \ \ \ \ \ \ 0.6331$

$\displaystyle (E) \ \ \ \ \ \ \ \ 0.7549$

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Problem 52-B

The time from initial purchase to the time of the first major repair (in years) for a brand of refrigerators is modeled by the random variable $Y=e^X$ where $X$ is normally distributed with mean 0.8 and standard deviation 1.5.

What is the median length of time (from initial purchase) that is free of any need for major repairs?

$\displaystyle (A) \ \ \ \ \ \ \ \ 0.80 \text{ years}$

$\displaystyle (B) \ \ \ \ \ \ \ \ 2.23 \text{ years}$

$\displaystyle (C) \ \ \ \ \ \ \ \ 3.50 \text{ years}$

$\displaystyle (D) \ \ \ \ \ \ \ \ 4.71 \text{ years}$

$\displaystyle (E) \ \ \ \ \ \ \ \ 6.86 \text{ years}$

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$\copyright \ 2013$