Question

Life insurance If four $21$-year-old men are insured, the insurer’s average income is

$V=\frac{X_1+X_2+X_3+X_4}{4}=0.25X_1+0.25X_2+0.25X_3+0.25X_4$

where $X_i$is the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of V. (If you compare with the results of Exercise $57$, you should see that averaging over more insured individuals reduces risk.)

Short answer

The mean and standard deviation,

${\mu }_V=303.35$

${\sigma }_V=4853.79$

Step-by-step solution

Given Information

Given in the question that the average income of the insures is

$V=\frac{X_1+X_2+X_3+X_4}{4}=0.25X_1+0.25X_{{}_2}+0.25X_2+0.25X_4$

Calculate the mean and standard deviation of V

Let's compute the mean of $V$

Here,

${\mu }_x=303.35\text{and}{\sigma }_y=9707.57$

Therefore,

$$\begin{gathered} {\mu }_Y=0.25\mu X_1+0.25\mu X_2+0.25\mu X_3+0.25\mu X_4 \\ =0.25\times 303.35+0.25\times 303.25+0.25\times 303.25+0.25\times 303.25 \\ =303.35 \end{gathered}$$

Compute the standard deviation of$V$

$$\begin{gathered} {\sigma }_V=\sqrt{0.0625\times 9707.{57}^2+0.0625\times 9707.{57}^2+0.0625\times 9707.{57}^2} \\ =\sqrt{23559228.83} \\ =4853.79 \end{gathered}$$