Refer to Exercise $7.10$on page $295$.

a. Use your answers from Exercise $7.10\left(b\right)$to determine the mean, ${\mu }_i$, of the variable $\hat{x}$for each of the possible sample sizes.

b. For each of the possible sample sizes, determine the mean, ${\mu }_{i+}$of the variable $\hat{x}$, using only your answer from Exercise $7.10\left(a\right)$

Given in the question that

Population data: $2,3,5,5,7,8$.

The population data for the variable $x$is $2,3,5,5,7,8$

Here determine the role="math" localid="1651232434578" ${\mu }_{\overline{x}}$of the variable $\overline{x}$for each sample in the question

The possible sample and sample means for a sample of size $n=1$is given in the table

$\begin{array}{|ll|}\hline \text{Sample}& \overline{x}\\ 2& 2\\ 3& 3\\ 5& 5\\ 5& 5\\ 7& 7\\ 8& 8\\ \hline\end{array}$

Therefore the mean ${\mu }_{\overline{x}}$of the variable $\overline{x}$will be,

role="math" localid="1651232793762" $$\begin{gathered} {\mu }_{\overline{x}}=\frac{2+3+5+5+7+8}{6} \\ =\frac{30}{6} \\ =5^6 \end{gathered}$$

The mean${\mu }_{\overline{x}}$of the variable$\overline{x}$will be$5$

Now we need to calculate the possible sample and sample means for sample of size $n=2$

The table will be

$\begin{array}{|llll|}\hline \text{Sampie}& \overline{x}& \text{Sample}& \overline{x}\\ 2.3& 2.5& 3.7& 5\\ 2.5& 3.5& 3.8& 5.5\\ 2.5& 3.5& 5.5& 5\\ 2,7& 4.5& 5,7& 6\\ 2,8& 5& 5,8& 6.6\\ 3,5& 4& 5,7& 6\\ 3.5& 4& 5.8& 6.5\\ & & 7.8& 7.5\\ \hline\end{array}$

The mean ${\mu }_{\overline{x}}$for the variable $\overline{x}$will be

role="math" localid="1651234078644" $$\begin{gathered} {\mu }_x=\frac{\left\{\begin{array}{l}3.3+3.3+4+4.3+4+4.7+5+4.7+5+5.7+4.3+5+5.3+5+\\ 5.3+6+5.7+6+6.7+6.7\end{array}\right\}}{20} \\ =\frac{100}{20} \\ =5 \end{gathered}$$

The mean ${\mu }_{\overline{x}}$of the variable $\overline{x}$when sample size $n=2$is $5$

Given in the question that

Population data: $2,3,5,5,7,8$.

Here we need to find the population mean

The calculation for the population mean will be,

$$\begin{gathered} \mu =\frac{\sum x_i}{N} \\ =\frac{2+3+5+5+7+8}{6} \\ =5 \end{gathered}$$

So, The population mean will be$5$.