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

# Exam P Practice Problem 86 – Finding Mean and Variance

The following is the cumulative distribution function of the random variable $X$.

$\displaystyle F(x)=\left\{\begin{matrix} \displaystyle 0&\ \ \ \ \ \ x < 0 \\{\text{ }}& \\{\displaystyle \frac{(x+2)^2}{100}}&\ \ \ \ \ \ 0 \le x <6 \\{\text{ }}& \\{\displaystyle 1}&\ \ \ \ \ \ 6 \le x <\infty \end{matrix}\right.$

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Problem 86-A

Calculate the expected value of $X$.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 2.16$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 3.35$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 4.32$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 6.00$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 6.67$

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Problem 86-B

Calculate the variance of $X$.

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 3.240$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 3.658$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 3.957$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 4.694$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 5.556$

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

# Exam P Practice Problem 83 – Claim Size of Auto Insurance Policies

Problem 83-A

An insurance company has a block of auto insurance policies. The claim size (in thousands) for a policy in this block of auto insurance policies is modeled by the random variable $Y=X^2$ where $X$ has a normal distribution with mean 0 and variance 1.5.

What is the expected claim size for such an auto insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 1250$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 1500$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 1750$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 2250$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 2500$

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Problem 83-B

An insurance company has a block of auto insurance policies. The claim size (in thousands) for a policy in this block of auto insurance policies is modeled by the random variable $Y=X^2$ where $X$ has a normal distribution with mean 0 and variance 3.

What is the standard deviation of the claim size for such an auto insurance policy?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 1732$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 3000$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 4243$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 4987$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 5732$

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

# Exam P Practice Problem 77 – Estimating Random Claim Sizes

Problem 77-A

The probability distribution of the claim size from an auto insurance policy randomly selected from a large pool of policies is described by the following density function.

$\displaystyle f(x)=\frac{3}{1000} \ (50-5x+\frac{1}{8} \ x^2), \ \ \ \ \ \ \ \ \ \ 0

What is the probability that a randomly selected claim from this insurance policy is within 120% of the mean claim size?

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

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.85$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.88$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.91$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.95$

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Problem 77-B

The probability distribution of the claim size from an auto insurance policy randomly selected from a large pool of policies is described by the following density function.

$\displaystyle f(x)=\frac{3}{2500} \ (100x-20x^2+ x^3), \ \ \ \ \ \ \ \ \ \ 0

What is the probability that a randomly selected claim from this insurance policy is within one-half of a standard deviation of the mean claim size?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.34$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.37$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.60$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.62$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.64$

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$\copyright \ 2013 \ \ \text{Dan Ma}$

# Exam P Practice Problem 71 – Estimating Claim Frequency

Problem 71-A

An auto insurer issued policies to a large group of drivers under the age of 40. These drivers are classified into five distinct groups by age. These groups are equal in size.

The annual claim count distribution for any driver being insured by this insurer is assumed to be a binomial distribution. The following table shows more information about these drivers.

$\displaystyle \begin{bmatrix} \text{Age}&\text{ }&\text{ }&\text{Mean} &\text{ }&\text{ }&\text{Variance} \\\text{Group}&\text{ }&\text{ }&\text{Of Claim Count} &\text{ }&\text{ }&\text{Of Claim Count} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{16-17}&\text{ }&\text{ }&\displaystyle \frac{5}{2}&\text{ }&\text{ }&\displaystyle \frac{5}{4} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{18-24}&\text{ }&\text{ }&\displaystyle 2&\text{ }&\text{ }&\displaystyle 1 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{25-29}&\text{ }&\text{ }&\displaystyle \frac{3}{2}&\text{ }&\text{ }&\displaystyle \frac{3}{4} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{30-34}&\text{ }&\text{ }&\displaystyle 1&\text{ }&\text{ }&\displaystyle \frac{1}{2} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{35-39}&\text{ }&\text{ }&\displaystyle \frac{1}{2} &\text{ }&\text{ }&\displaystyle \frac{1}{4} \end{bmatrix}$

An insured driver is randomly selected from this large pool of insured and is observed to have one claim in the last year.

What is the probability that the mean number of claims in a year for this insured driver is greater than 1.5?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{67}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{57}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{3}{5}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{51}{67}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{50}{64}$

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Problem 71-B

An auto insurer issued policies to a large group of drivers under the age of 40. These drivers are classified into five distinct groups by age. These groups are equal in size.

The annual claim count distribution for any driver being insured by this insurer is assumed to be a geometric distribution. The following table shows more information about these drivers.

$\displaystyle \begin{bmatrix} \text{Age}&\text{ }&\text{ }&\text{Mean} &\text{ }&\text{ }&\text{Variance} \\\text{Group}&\text{ }&\text{ }&\text{Of Claim Count} &\text{ }&\text{ }&\text{Of Claim Count} \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{35-39}&\text{ }&\text{ }&\displaystyle 1 &\text{ }&\text{ }&\displaystyle 2 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{30-34}&\text{ }&\text{ }&\displaystyle 2&\text{ }&\text{ }&\displaystyle 6 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{25-29}&\text{ }&\text{ }&\displaystyle 3&\text{ }&\text{ }&\displaystyle 12 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{18-24}&\text{ }&\text{ }&\displaystyle 4&\text{ }&\text{ }&\displaystyle 20 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ \text{16-17}&\text{ }&\text{ }&\displaystyle 5&\text{ }&\text{ }&\displaystyle 30 \end{bmatrix}$

An insured driver is randomly selected from this large pool of insured and is observed to have one claim in the last year.

What is the probability that the mean number of claims in a year for this insured driver is greater than 2.5?

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.49$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.51$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.55$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.57$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.60$

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$\copyright \ 2013 \ \ \text{Dan Ma}$

# Exam P Practice Problem 58 – Dental Care and Vision Care Expenses

Problem 58-A

A health plan offers dental care and vision care benefits. Let $X$ represents the total annual amount (in millions) paid in dental care benefits. Let $Y$ represents the total annual amount (in millions) paid in vision care benefits.

The health plan determined that

• $X=K^2$ where $K$ follows a normal distribution with mean 0 and variance 1,
• $Y=L^2$ where $L$ follows a normal distribution with mean 0 and variance 2, and
• $K$ and $L$ are independent.

Given that the total annual vision care benefits paid by the health plan exceeds 2.5 millions, what is the probability that the total annual dental care benefits paid by the health plan exceeds 2 millions?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0228$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0793$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1586$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8416$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.9207$

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Problem 58-B

A health plan offers dental care and vision care benefits. Let $X$ represents the total annual amount (in millions) paid in dental care benefits. Let $Y$ represents the total annual amount (in millions) paid in vision care benefits.

The health plan determined that

• $X=2.5 K^2$ where $K$ follows a normal distribution with mean 0 and variance 1,
• $Y=5 L^2$ where $L$ follows a normal distribution with mean 0 and variance 1, and
• $K$ and $L$ are independent.

What is the probability that the total annual dental care benefits exceeds 3 millions and that the total annual vision care benefits exceeds 4 millions?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1013$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.4565$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.6266$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.7286$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.7881$

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# Exam P Practice Problem 57 – Lifetimes of Machines

Problem 57-A

A factory owner purchased two identical machines for her factory. Let $X$ and $Y$ be the lifetimes (in years) of these two machines. These lifetimes are modeled by the following joint probability density function.

$\displaystyle f(x,y)=\frac{0.01}{\sqrt{x} \ \sqrt{y}} \ e^{-0.2 \sqrt{x}} \ e^{-0.2 \sqrt{y}} \ \ \ \ \ \ \ 0

The machine whose lifetime is modeled by the random variable $Y$ came online 2 years after the beginning of operation of the machine that is modeled by the random variable $X$.

Given that $X$ exceeds 2, that is the probability that $Y$ exceeds 3?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2928$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.4670$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.5330$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.7072$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.7536$

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Problem 57-B

A company purchased two machines for its factory. Let $X$ and $Y$ be the lifetimes (in years) of these machines. The following is the joint density function of their lifetimes.

$\displaystyle f(x,y)=\frac{3}{125} \ y \ e^{-0.3 x} \ \ \ \ \ \ \ 0

The machine whose lifetime is modeled by the random variable $Y$ came online after the failure of the machine whose lifetime is modeled by $X$.

What is the variance of the total time of operation of these two machines?

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 12.50$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 13.60$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 17.20$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 19.85$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 23.61$

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# 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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# Exam P Practice Problem 49 – Aggregate Claim Costs

Problem 49-A

The aggregate claim amount (in millions) in a year for a block of fire insurance policies is modeled by a random variable $Y=e^X$ where $X$ has a normal distribution with mean 4 and variance 2. What is the expected annual aggregate claim amount?

$\displaystyle (A) \ \ \ \ \ \ \ 403.43$

$\displaystyle (B) \ \ \ \ \ \ \ 244.69$

$\displaystyle (C) \ \ \ \ \ \ \ 148.41$

$\displaystyle (D) \ \ \ \ \ \ \ 90.02$

$\displaystyle (E) \ \ \ \ \ \ \ 54.60$

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Problem 49-B

The aggregate claim amount (in millions) in a year for a block of fire insurance policies is modeled by a random variable $Y=e^X$ where $X$ has a normal distribution with mean 1.15 and variance 1.2. What is the probability that the annual aggregate claim amount will be less than the expected annual aggregate claim amount?

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

$\displaystyle (B) \ \ \ \ \ \ \ 0.6915$

$\displaystyle (C) \ \ \ \ \ \ \ 0.7088$

$\displaystyle (D) \ \ \ \ \ \ \ 0.8749$

$\displaystyle (E) \ \ \ \ \ \ \ 0.9599$

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# Exam P Practice Problem 46 – Finding Moment of a Sum

Problem 46-A

Suppose that $X$ and $Y$ are random losses that are jointly distributed with the following density function:

$\displaystyle f(x,y)=\frac{1}{40} \ (4-x) \ \ \ \ \ \ 0

Find the second moment of the sum of the two losses.

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Problem 46-B

Suppose that $X$ and $Y$ are random losses that are jointly distributed with the following density function:

$\displaystyle f(x,y)=32 \ x^2 \ y \ e^{-2 \ (x \ + \ y)} \ \ \ \ \ \ 0

Find the second moment of the sum of the two losses.

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