Tag Archives: Standard Deviation

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<x<10 \\            \text{ } & \text{ } \\           0 &\ \ \ \ \ \ \text{otherwise}           \end{array} \right.

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<x<20 \\            \text{ } & \text{ } \\           0 &\ \ \ \ \ \ \text{otherwise}           \end{array} \right.

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 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<x<20

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<x<10

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 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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Answers

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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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\copyright \ 2013

Exam P Practice Problem 29 – Expected Insurance Payment

Problem 29A

When there is a loss due to fire damage to a house, the loss follows the distribution indicated below.

            \displaystyle \begin{bmatrix} \text{Loss Amount}&\text{ }&\text{Probability}  \\\text{ }&\text{ }&\text{ } \\ 5,000&\text{ }&0.5  \\ 10,000&\text{ }&0.3  \\ 50,000&\text{ }&0.1  \\ 75,000&\text{ }&0.05  \\ 100,000&\text{ }&0.03  \\ 150,000&\text{ }&0.015 \\ 200,000&\text{ }&0.005   \end{bmatrix}

The owner of the house purchases an insurance policy with a deductible of 5,000 to insure againse this loss. What is the expected payment to the owner under this policy.

      A. \ \ \ \ \ \text{15,000}
      B. \ \ \ \ \ \text{15,500}
      C. \ \ \ \ \ \text{20,500}
      D. \ \ \ \ \ \text{31,000}
      E. \ \ \ \ \ \text{30,000}

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Problem 29B
Use the same information as in Problem 29A. What is the standard deviation of the payment to the owner under this insurance policy?

      A. \ \ \ \ \ \text{29,541.49}
      B. \ \ \ \ \ \text{30450.18}
      C. \ \ \ \ \ \text{31,480.15}
      D. \ \ \ \ \ \text{35,597.34}
      E. \ \ \ \ \ \text{38,749.19}

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Answers

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