Question

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 $X_1$ and ${\mathrm{X}}_2$ are the insurer's income from the two insurance policies, the insurer's average income $W$ on the two policies is

$\mathrm{W}=\mathrm{X}1+\mathrm{X}{222}^W=\frac{X_1+X_2}{2}$

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

The required value for Mean of W,

${\mu }_W=\$303.35$

The required value for Standard deviation of W ,

${\sigma }_W=\$6864.29$

Step-by-step solution

Given Information

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

Mean,

${\mu }_X=\$303.35$

The standard deviation (SD)

${\sigma }_X=\$9707.57$

The insurer's income from two insurance policies is$X_1$and $X_2$

So that

The average income of the insurer on two policies,

$W=\frac{X_1+X_2}{2}$

Find the mean and standard deviation of W

When both X and Y are independent,

Property mean:

${\mu }_{aX+bY}=a_{{\mu }_X}+b_{{\mu }_Y}$

Property variance:

${\sigma }_{aX+bY}^2=a^2{\mu }_X^2+b^2{\mu }_Y^2$

Since W represents insurer's average income.

Then

$W=\frac{X_1+X_2}{2}=0.5X_1+0.5X_2$

Thus,

We have

Mean of W,

$\begin{array}{rcl}{\mu }_W& =& {\mu }_{0.5x_1+0.5x_2}\\ & =& 0.5{\mu }_X+0.5{\mu }_X\\ & =& 0.5(303.35)+0.5(303.35)\\ & =& \$303.35\end{array}$

Standard deviation of W,

$\begin{array}{rcl}{\sigma }_W& =& {\sigma }_{0.5x_1+0.5x_2}\\ & =& \sqrt{0.5^2{\sigma }_x^2+0.5^2{\sigma }_x^2}\\ & =& \sqrt{0.25(9707.57{)}^2+0.25(9707.57{)}^2}\\ & \approx & \$6864.29\end{array}$