91Ó°ÊÓ

From the given problem, the distribution of X follows normal with mean$\mathrm{μ}=1.5$ and a standard deviation$\mathrm{σ}=0.2$

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

The mean of the sampling distribution of $\stackrel{¯}{x}$ is the population mean $\mathrm{μ}$. That is,

${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$

$E\left(\stackrel{¯}{x}\right)=\mathrm{μ}$

Thus,

$E\left(\stackrel{¯}{x}\right)=1.5$

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

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

Since,

$Var\left(\stackrel{¯}{x}\right)=\frac{{\mathrm{σ}}^2}{n}$

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

Thus,

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

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 $\stackrel{¯}{x}$ is normal.

Consider $\stackrel{¯}{x}=1.52$

$\mathrm{μ}=1.5$ and $Var\left(\stackrel{¯}{x}\right)=0.0004$

The z-score is,

$z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{\sqrt{Var\left(\stackrel{¯}{x}\right)}}$

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

Thus, the required z-score is 1.

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

Therefore, Probability of $\stackrel{¯}{x}$ 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.