91Ó°ÊÓ

Let the average salary be $\$2,000$per year with a standard deviation of $\$8,000$. We randomly survey $1,000$residents of that country.

Definition for random variable $X:$

$X=$the yearly income of someone in a third world country

Definition for mean random variable $\overline{X}$:

$\overline{X}=$the average salary from samples of 1,000 residents of a third world country.

The mean random variable is defined as, $\stackrel{¯}{X}~N\left(\mathrm{μ},\frac{\mathrm{σ}}{\sqrt{n}}\right)$

On substituting the values of given information, we get$\stackrel{¯}{X}~N\left(2000,\frac{8000}{\sqrt{1000}}\right)$

Possibility for the standard deviation to be greater than the average.

Very wide differences in data values can have averages smaller than standard deviations.

The distribution of the sample mean will have higher probabilities closer to the population mean.

$$\begin{gathered} P(2000<\stackrel{¯}{x}<2100)=\text{normalcdf}\left(2000,2100,2000,\frac{8000}{\sqrt{1000}}\right)=0.1537 \\ P(2100<\stackrel{¯}{x}<2200)=\text{normalcdf}\left(2100,2200,2000,\frac{{\displaystyle 8000}}{{\displaystyle \sqrt{1000}}}\right)=0.1317 \end{gathered}$$