Chapter 6: Q. 64 (page 392)

Life insurance If we randomly select four insured 21-year-old men, the insurer's average income is
$V = X 1 + X 2 + X 3 + X 444^{V} = \frac{X_{1} + X_{2} + X_{3} + X_{4}}{4}$
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 63, you should see that averaging over more insured individuals reduces risk.)

Short Answer

Expert verified

The required value for Mean of $V$ ,

μ_{V} = $ 303.35

Standard deviation of $V$ ,

σ_{V} = $ 4853.79

Step by step solution

Given information

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

Mean,

$μ_{X} = $ 303.35$

Standard deviation,

$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

Calculation

The Property mean:

$μ^{μ} \sum_{t = 0}^{ε} a_{t} X_{t} = \sum_{i = 0}^{n} a_{i} μ_{X_{t}}$

Property standard deviation:

$σ \sum_{i = 0}^{s a_{l} X_{i}} = \sqrt{\sum_{i = 0}^{n} a_{i}^{2} σ_{X_{i}}^{2}}$

Because V indicates the average income of the insurer.

Then

$V = \frac{X_{1} + X_{2} + X_{3} + X_{4}}{4} = 0.25 X_{1} + 0.25 X_{2} + 0.25 X_{3} + 0.25 X_{4}$

So that

$μ_{X} = μ_{X_{1}} = μ_{X_{2}} = μ_{X_{3}} = μ_{X_{4}} = $ 303.35$

And

$σ_{X} = σ_{X_{1}} = σ_{X_{2}} = σ_{X_{3}} = σ_{X_{4}} = $ 9707.57$

Thus,

We have

Mean of V,

\begin{array}{rcl} μ_{V} & = & μ_{0.25 X_{1} + 0.25 X_{2} + 0.25 X_{s} + 0.25 X_{4}} \\ = & 0.25 μ_{X_{1}} + 0.25 μ_{X_{2}} + 0.25 μ_{X_{3}} + 0.25 μ_{X_{4}} \\ = & μ_{X} \\ = & $ 303.35 \end{array}

Standard deviation of V,

$\begin{array}{rcl} σ_{V} & = & σ_{0.25 x_{1} + 0.25 x_{2} + 0.25 x_{3} + 0.25 x_{4}} \\ = & \sqrt{0 . 25^{2} σ_{X_{1}}^{2} + 0 . 25^{2} σ_{X_{2}}^{2} + 0 . 25^{2} σ_{X_{1}}^{2} + 0 . 25^{2} 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}} \\ \approx & $ 4853.79 \end{array}$