Category Archives: Normal Distribution

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 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 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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\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

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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\copyright \ 2013

Exam P Practice Problem 45 – Heights of Male Students

Problem 45-A

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

Five male students from this university are randomly selected.

What is the probability that the shortest student among the five randomly selected students is taller than 5 feet 5 inches?

Note that one feet = 12 inches.

        \displaystyle A. \ \ \ \ \ \ \ \ \ \ (0.0764)^5

        \displaystyle B. \ \ \ \ \ \ \ \ \ \ 0.0764

        \displaystyle C. \ \ \ \ \ \ \ \ \ \ 0.3279

        \displaystyle D. \ \ \ \ \ \ \ \ \ \ 0.6721

        \displaystyle E. \ \ \ \ \ \ \ \ \ \ 0.9236

The answers are based on this normal table from SOA.

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Problem 45-B

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

Ten male students from this university are randomly selected.

What is the probability that the tallest student among the ten randomly selected students is shorter than 6 feet 2 inches?

Note that one feet = 12 inches.

        \displaystyle A. \ \ \ \ \ \ \ \ \ \ (0.0367)^{10}

        \displaystyle B. \ \ \ \ \ \ \ \ \ \ 0.0367

        \displaystyle C. \ \ \ \ \ \ \ \ \ \ 0.3120

        \displaystyle D. \ \ \ \ \ \ \ \ \ \ 0.6880

        \displaystyle E. \ \ \ \ \ \ \ \ \ \ 0.9633

The answers are based on this normal table from SOA.

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Answers

The answers are based on this normal table from SOA.

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