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Chapter 6: Q. 63 (page 392)

Life insurance The risk of insuring one person's life is reduced if we insure many people. Suppose that we randomly select two insured 21-year-old males, and that their ages at death are independent. If X1 and X2 are the insurer's income from the two insurance policies, the insurer's average income W on the two policies is

W=X1+X222W=X1+X22

Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy, but the standard deviation is less.)

Short Answer

Expert verified

The required value for Mean of W,

渭W=$303.35

The required value for Standard deviation of W ,

蟽W=$6864.29

Step by step solution

01

Given Information

X:amount earned by Life Insurance Company on a 5-year term life insurance chosen at random.

Mean,

渭X=$303.35

The standard deviation (SD)

蟽X=$9707.57

The insurer's income from two insurance policies isX1and X2

So that

The average income of the insurer on two policies,

W=X1+X22

02

Find the mean and standard deviation of W

When both X and Y are independent,

Property mean:

渭aX+bY=a渭X+b渭Y

Property variance:

蟽aX+bY2=a2渭X2+b2渭Y2

Since W represents insurer's average income.

Then

W=X1+X22=0.5X1+0.5X2

Thus,

We have

Mean of W,

渭W=渭0.5x1+0.5x2=0.5渭X+0.5渭X=0.5(303.35)+0.5(303.35)=$303.35

Standard deviation of W,

蟽W=蟽0.5x1+0.5x2=0.52蟽x2+0.52蟽x2=0.25(9707.57)2+0.25(9707.57)2鈮�$6864.29

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