Chapter 6.2, Problem 64E

To determine

Mean and standard deviation of V .

Answer to Problem 64E

Mean of V ,

  ${\mu }_V=\$303.35$

Standard deviation of V ,

  ${\sigma }_V=\$4853.79$

Explanation of Solution

Given information:

X : amount earned by Life Insurance Company on a randomly chosen 5 − year term life policy

Such that

Mean,

  ${\mu }_X=\$303.35$

Standard deviation,

  ${\sigma }_X=\$9707.57$

Insurer’s average income,

  $V=\frac{X_1+X_2+X_3+X_4}{2}$

Where,

X_i : Income from insuring one man

Calculations:

Property mean:

  ${\mu }_{{\displaystyle {\sum }_{i=0}^n{a}_i{X}_i}}={\displaystyle \sum _{i=0}^n{a}_i{\mu }_{X_i}}$

Property standard deviation:

  ${\sigma }_{{\displaystyle {\sum }_{i=0}^n{a}_i{X}_i}}=\sqrt{{\displaystyle \sum _{i=0}^n{a}_i^2{\sigma }_{X_i}^2}}$

Since V represents insurer’s average income.

Then

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

Such that

  ${\mu }_X={\mu }_{X_1}={\mu }_{X_2}={\mu }_{X_3}={\mu }_{X_4}=\$303.35$

And

  ${\sigma }_X={\sigma }_{X_1}={\sigma }_{X_2}={\sigma }_{X_3}={\sigma }_{X_4}=\$9707.57$

Thus,

We have

Mean of V ,

  $\begin{array}{l}{\mu }_V={\mu }_{0.25X_1+0.25X_2+0.25X_3+0.25X_4}\\ =0.25{\mu }_{X_1}+0.25{\mu }_{X_2}+0.25{\mu }_{X_3}+0.25{\mu }_{X_4}\\ ={\mu }_X=\$303.35\end{array}$

Standard deviation of V ,

  $\begin{array}{l}{\sigma }_V={\sigma }_{0.25X_1+0.25X_2+0.25X_3+0.25X_4}\\ =\sqrt{{0.25}^2{\sigma }_{X_1}^2+{0.25}^2{\sigma }_{X_2}^2+{0.25}^2{\sigma }_{X_3}^2+{0.25}^2{\sigma }_{X_4}^2}\\ =\sqrt{0.0625{\left(9707.57\right)}^2+0.0625{\left(9707.57\right)}^2+0.0625{\left(9707.57\right)}^2+0.0625{\left(9707.57\right)}^2}\\ \approx \$4853.79\end{array}$