Chapter 6: Q. 63 (page 392)

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 $X_{2}$ are the insurer's income from the two insurance policies, the insurer's average income $W$ on the two policies is
$W = X 1 + X 222^{W} = \frac{X_{1} + X_{2}}{2}$
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

Expert verified

The required value for Mean of W,

$μ_{W} = $ 303.35$

The required value for Standard deviation of W ,

$σ_{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,

$μ_{X} = $ 303.35$

The standard deviation (SD)

$σ_{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:

μ_{a X + b Y} = a_{μ_{X}} + b_{μ_{Y}}

Property variance:

σ_{a X + b Y}^{2} = a^{2} μ_{X}^{2} + b^{2} μ_{Y}^{2}

Since W represents insurer's average income.

Then

W = \frac{X_{1} + X_{2}}{2} = 0.5 X_{1} + 0.5 X_{2}

Thus,

We have

Mean of W,

\begin{array}{rcl} μ_{W} & = & μ_{0.5 x_{1} + 0.5 x_{2}} \\ = & 0.5 μ_{X} + 0.5 μ_{X} \\ = & 0.5 (303.35) + 0.5 (303.35) \\ = & $ 303.35 \end{array}

Standard deviation of W,

\begin{array}{rcl} σ_{W} & = & σ_{0.5 x_{1} + 0.5 x_{2}} \\ = & \sqrt{0 . 5^{2} σ_{x}^{2} + 0 . 5^{2} σ_{x}^{2}} \\ = & \sqrt{0.25 (9707.57)^{2} + 0.25 (9707.57)^{2}} \\ \approx & $ 6864.29 \end{array}

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Short Answer

Step by step solution

Given Information

Find the mean and standard deviation of W

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