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18. Life insurance

(a) It would be quite risky for you to insure the life of a $21$-year-old friend under the terms of Exercise $14$. There is a high probability that your friend would live and you would gain $\backslash (1250$in premiums. But if he were to die, you would lose almost $\backslash )100,000$. Explain carefully why selling insurance is not risky for an insurance company that insures many thousands of $21$-year-old men.

(b) The risk of an investment is often measured by the standard deviation of the return on the investment. The more variable the return is, the riskier the
investment. We can measure the great risk of insuring a single person鈥檚 life in Exercise $14$by computing the standard deviation of the income Y that the insurer will receive. Find 蟽Y using the distribution and mean found in Exercise 14.

Step by step solution

01

Part (a) Step 1:Given information 

Given in the question that the high probability would gain $\$1250$in premiums. If die, would lose almost $\$100,000$. Selling insurance is not risky for an insurance company that insures many thousands of $21$-year-old men.

02

Part (a) Step 2: Explanation 

According to the information, the gain around is$\$1250.$
Amount would lost in case of death is$\$100,000$
The expected value with its probability is:
$$\begin{gathered} E\left(X\right)=鈭�x脳P\left(x\right) \\ =(鈭�99750)脳0.00183+鈥�.+\left(1250\right)脳0.99058 \\ =303.35 \end{gathered}$$

03

Part (b) Step 1: Given information

The standard deviation of the investment return is used to determine the risk of an investment. The riskier the investment is, the more varied the return is.

04

Part (b) Step 2: Explanation 

The standard deviation of$\mathrm{Y}$ can be determined as:
$$\begin{gathered} 蟽=\sqrt{鈭�x^2脳P\left(x\right)-{\left(鈭�x脳P\left(x\right)\right)}^2} \\ =\sqrt{(鈭�99750鈭�303.3525{)}^2脳0.00183+鈥�鈥�鈥�鈥�鈥�..+(1250鈭�303.3525{)}^2脳0.99058} \\ =9708 \end{gathered}$$

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