Exam P Practice Problem 7

Problem 7a
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by

      \displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0

The property owner purchased an insurance policy that pays the amount of the damage in full during the next year.

  1. What is the probability that the insurer’s payment to the owner will exceed $17,500?
  2. What is the mean payment made by the insurer to the owner of the property?
  3. What is the variance of the amount of payment made by the insurer?

Problem 7b
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by

      \displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0

The property owner purchased an insurance policy with a coinsurance provision that pays 80% of the amount of the damage during the next year. The remaining 20% of the amount of the damage is retained by the property owner.

  1. What is the probability that the insurer’s payment to the owner will exceed $17,500?
  2. What is the mean payment made by the insurer to the owner of the property?
  3. What is the variance of the amount of payment made by the insurer?

Solution is found below.

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Solution to Problem 7a
Let X be the loss amount (i.e. the pdf is the one given in the problem). One important thing to keep in mind is that a loss to the property is not certain. So the answer for #1 is not P(X>17.5). Let Y be the payment made by the insurer to the property owner. The answer for #1 is P(Y>17.5). One way to look at the problem is through the law of total probability.

\displaystyle \begin{aligned}P(Y>17.5)&=P(Y>17.5 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +P(Y>17.5 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+e^{-0.05 (17.5)} \times 0.2 \\&=0.2 \times e^{-0.875} \\&=0.08337  \end{aligned}

The following provides the answers to the rest of the problem:

\displaystyle \begin{aligned}E(Y)&=E(Y \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{1}{0.05} \times 0.2 \\&=4 \\&=\$4000  \end{aligned}

\displaystyle \begin{aligned}E(Y^2)&=E(Y^2 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y^2 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{2}{0.05^2} \times 0.2 \\&=160  \end{aligned}

\displaystyle \begin{aligned}Var(Y)&=E(Y^2)-E(Y)^2 \\&=160-4^2 \\&=144  \end{aligned}

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Answer to Problem 7b
\displaystyle 7b.1 \ \ \ \ \ \ 0.2 e^{-1.09375}=0.0669916086

\displaystyle 7b.2 \ \ \ \ \ \ \$3200

\displaystyle 7b.3 \ \ \ \ \ \ 92.16

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