91Ó°ÊÓ

Data on the population:$2,5,8$.

For the variable $x$, we have population data, which is $2,5,8$.

The mean ${\mathrm{μ}}_{\stackrel{¯}{x}}$of the variable $\overline{x}$for each of the samples is calculated as follows:

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

The variable $\overline{x}$has the following mean ${\mathrm{μ}}_{\stackrel{¯}{x}}$:

${\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{2+5+8}{3}$

$=\frac{15}{3}$

$=5$

The variable $\overline{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}$ of $5$.

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

The variable $\overline{x}$has the following mean ${\mathrm{μ}}_{\stackrel{¯}{x}}$:

role="math" localid="1650970365589" ${\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{3.5+2+6.5}{3}$

role="math" localid="1650970431300" $=\frac{15}{3}$

$=5$

The variable $\overline{x}$ has a mean value of $5$ (${\mathrm{μ}}_{\stackrel{¯}{x}}$).

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

Interpretation: We can see from the foregoing that the mean of all potential sample means is the same.

Data on the population:$2,5,8$.

The following is a definition of the population mean:

$\mathrm{μ}=\frac{∑x_i}{N}$

$=\frac{2+5+8}{3}$

$=5$

$5$is the population average.

We can see that $\mathrm{μ}=5$ based on the results of parts (a) and (b).

The population mean is equal to the mean of all feasible sample means.