Variable life insurance return rates. Refer to the International Journal of Statistical Distributions (Vol. 1, 2015) study of a variable life insurance policy, Exercise 4.97 (p. 262). Recall that a ratio (x) of the rates of return on the investment for two consecutive years was shown to have a normal distribution, with $\mu =1.5$, $\sigma =0.2$. Consider a random sample of 100 variable life insurance policies and let$\overline{x}$represent the mean ratio for the sample.

a. Find E(x) and interpret its value.

b. Find Var(x).

c. Describe the shape of the sampling distribution of$\overline{x}$.

d. Find the z-score for the value $\overline{x}=1.52$.

e. Find $P\left(\overline{x}>1.52\right)$

f. Would your answers to parts a–e change if the rates (x) of return on the investment for two consecutive years was not normally distributed? Explain.

From the given problem, the distribution of X follows normal with mean$\mu =1.5$ and a standard deviation$\sigma =0.2$

From the central theorem, as the sample size is large the mean of the sample follows normal distribution with mean ${\mu }_{\overline{x}}\left(=\mu \right)$ and variance $\frac{{\sigma }^2}{n}$.

The mean of the sampling distribution of $\overline{x}$ is the population mean $\mu$. That is,

${\mu }_{\overline{x}}=\mu$

$E\left(\overline{x}\right)=\mu$

Thus,

$E\left(\overline{x}\right)=1.5$

The mean ratio of the rates of return on the investment for two consecutive years for the sampling distribution of $\overline{x}$ is 1.5.

b. From the given problem the sample size is $n=100$

Since,

$Var\left(\overline{x}\right)=\frac{{\sigma }^2}{n}$

$$\begin{gathered} =\frac{{\left(0.2\right)}^2}{100} \\ =\frac{0.04}{100} \\ =0.0004 \end{gathered}$$

Thus,

$Var\left(\overline{x}\right)=0.0004$

From the central limit theorem as the sample size is large the mean of the sample follows normal distribution. The shape of the sampling distribution of $\overline{x}$ is normal.

Consider $\overline{x}=1.52$

$\mu =1.5$ and $Var\left(\overline{x}\right)=0.0004$

The z-score is,

$z=\frac{\overline{x}-\mu }{\sqrt{Var\left(\overline{x}\right)}}$

$$\begin{gathered} =\frac{1.52-1.5}{\sqrt{0.0004}} \\ =\frac{0.02}{0.02} \\ =1 \end{gathered}$$

Thus, the required z-score is 1.

$\begin{array}{rcl}P\left(\overline{x}>1.52\right)& =& P\left(\frac{\overline{x}-\mu }{\sqrt{Var\left(\overline{x}\right)}}>\frac{1.52-1.5}{\sqrt{0.0004}}\right)\\ & =& P\left(z>1\right)\\ & =& 1-P\left(z<1\right)\\ & =& 1-\left\{0.5+P\left(0<z<1\right)\right\}\\ & =& 1-0.5-0.3413\\ & =& 0.1587\\ & & \end{array}$

Therefore, Probability of $\overline{x}$ greater than 1.52 is 0.1587.

Thus, the answers in the parts a-e would not change if the rates were not normally distributed.