Officer Salaries. The following table gives the monthly salaries (in $\backslash (1000$) of the six officers of a company.

a. Calculate the population mean monthly salary,$\mu$

There are $15$possible samples of size $4$from the population of six officers. They are listed in the first column of the following table.

b. Complete the second and third columns of the table.

c. Complete the dot plot for the sampling distribution of the sample mean for samples of size $4$Locate the population means on the graph.

d. Obtain the probability that the mean salary of a random sample of four officers will be within 1 (i.e., $\backslash )1000$) of the population mean.

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

The population mean monthly salary $\mu$

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

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 $\mu$

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 $\mu$

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

The mean of all possible sample means

$$\begin{gathered} ={\mu }_{\overline{x}} \\ =\frac{14+15+16+16+17+18+17+18+19+20+18+19+20+21+22}{15} \\ =\frac{270}{15} \\ =18 \end{gathered}$$

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 $\overline{x}$ 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., ${\mu }_x=\mu$