91Ó°ÊÓ

$X$: amount earned by a Life Insurance Company on a 5-year term life policy picked at random

Mean,

${\mathrm{μ}}_X=\$303.35$

Standard deviation,

${\mathrm{σ}}_X=\$9707.57$

Insurer's average income,

localid="1654240461251" $V=\frac{X_1+X_2+X_3+X_4}{2}$

Where,

$X_i$: Earnings from insuring just one person

The Property mean:

${\mathrm{μ}}^{\mathrm{μ}}\underset{t=0}{\overset{\mathrm{Î\mu }}{∑}}{a}_t{X}_t=\underset{i=0}{\overset{n}{∑}}{a}_i{\mathrm{μ}}_{X_t}$

Property standard deviation:

$\mathrm{σ}\underset{i=0}{\overset{s{a}_l{X}_i}{∑}}=\sqrt{\underset{i=0}{\overset{n}{∑}}{a}_i^2{\mathrm{σ}}_{X_i}^2}$

Because V indicates the average income of the insurer.

Then

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

So that

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

And

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

Thus,

We have

Mean of V,

Standard deviation of V,

$\begin{array}{rcl}{\mathrm{σ}}_V& =& {\mathrm{σ}}_{0.25x_1+0.25x_2+0.25x_3+0.25x_4}\\ & =& \sqrt{0.{25}^2{\mathrm{σ}}_{X_1}^2+0.{25}^2{\mathrm{σ}}_{X_2}^2+0.{25}^2{\mathrm{σ}}_{X_1}^2+0.{25}^2{\mathrm{σ}}_{X_4}^2}\\ & =& \sqrt{0.0625(9707.57{)}^2+0.0625(9707.57{)}^2+0.0625(9707.57{)}^2+0.0625(9707.57{)}^2}\\ & ≈& \$4853.79\end{array}$