91Ó°ÊÓ

$\text{The following table gives the monthly salaries (in}\$1000\text{s) of the six officers of a company.}$

The population mean monthly salary $\mathrm{μ}$

$$\begin{gathered} =\frac{8+12+16+20+24+28}{6} \\ =\frac{108}{6} \\ =18 \end{gathered}$$

As a result, the average officer pay in the population is $\$18000$

Dot plot showing sample mean sampling distribution

Each dot $(·)$ represents a sample mean

We can observe from the dot plot that there are $7$sample means within $1$unit of $\mathrm{μ}$

The total number of possible sample means (i.e., the total number of samples is $15$)

As a result, the probability that a random sample of size $4\text{'}s$mean wage will be within $1$unit (i.e., $\$1000$) of the population mean $\mathrm{μ}$

$$\begin{gathered} =\frac{\text{Number of sample means within}1\text{unit of}\mathrm{μ}}{\text{Number of all possible of size}4\text{from population of}} \\ =\frac{7}{15} \\ =0.467\text{(approximately)} \end{gathered}$$

The mean of all possible sample means

$$\begin{gathered} ={\mathrm{μ}}_{\stackrel{¯}{x}} \\ =\frac{14+15+16+16+17+18+17+18+19+20+18+19+20+21+22}{15} \\ =\frac{270}{15} \\ =18 \end{gathered}$$

The average of all potential sample means salaries of four officers drawn from a population of six officers is $\$1800$, which is the population mean wage.

Yes, we can get the mean of the variable $\stackrel{¯}{x}$ without doing part(e), because the mean of all alternative means for a certain sample size is always equal to the population mean, i.e., ${\mathrm{μ}}_x=\mathrm{μ}$