91Ó°ÊÓ

It is given that the population data is 1,2,3,4,5.

We need to determine the mean, ${\mathrm{μ}}_s$, of the variable $\stackrel{¯}{x}$for each of the possible sample sizes.

For the population data: 1,2,3,4,5.

The sample and sample mean for a sample of size $n=1$are shown in the table below.

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{1+2+3+4+5}{5} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{15}{5} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=3 \end{gathered}$$

So when the sample size is 1, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=3$.

For the population data: 1,2,3,4,5.

The sample and sample mean for a sample of size $n=2$are shown in the table below.

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{1.5+2+2.5+3+2.5+3+3.5+3.5+4+4.5}{10} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{30}{10} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=3 \end{gathered}$$

So when the sample size is 2, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=3$.

For the population data: 1,2,3,4,5.

The sample and sample mean for a sample of size $n=3$are shown in the table below.

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{2+2.33+2.67+2.67+3+3.33+3+3.33+3.67+4}{10} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{30}{10} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=3 \end{gathered}$$

So when the sample size is 3, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=3$.

For the population data: 1,2,3,4,5.

The sample and sample mean for a sample of size $n=4$are shown in the table below.

The variable $\stackrel{¯}{x}$has the following mean

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{2.5+2.75+3+3.25+3.5}{5} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{15}{5} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=3 \end{gathered}$$

So when the sample size is 4, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=3$.

So when the sample size is 5, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=3$.

Thus it can be seen that the mean of all potential sample means is the same.

For the given population data: 1,2,3,4,5 the population mean can be given as

$$\begin{gathered} \mathrm{μ}=\frac{1+2+3+4+5}{5} \\ \mathrm{μ}=\frac{15}{5} \\ \mathrm{μ}=3 \end{gathered}$$

So from the results, it can be observed that the population mean is equal to the mean of all potential sample means that is ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$.