The winner of the 2012-2013 National Basketball Association (NBA) championship was the Miami Heat. One possible starting lineup for that team is as follows.

a. Determine the population mean height, $\mu$, of the five players:

b. Consider samples of size $2$without replacement. Use your answer to Exercise $7.11\left(b\right)$on page $295$and Definition $3.11$on page $140$to find the mean, ${\mu }_{\mathrm{r}}$, of the variable $\hat{x}$.

c. Find ${\mu }_{x^{*}}$using only the result of part (a).

Given in the question that a table

Here we need to find the population mean height of the players

The equation will be

$$\begin{gathered} \mu =\frac{\sum _{i=1}^n x_i}{n} \\ =\frac{83+76+80+74+80}{5} \\ =\frac{393}{5} \\ =78.6 \end{gathered}$$

Therefore the population mean height of the players will be$78.6$

Given in the question that a table

Here we need to obtain the mean height ${\mu }_{\overline{x}}$for sample size $2$

The below table shows the corresponding mean obtained

$\begin{array}{|lll|}\hline \text{Sample size}& \text{Height}& \text{Mean}\left(\overline{x}\right)\\ \mathrm{B},\mathrm{W}& 83,76& \frac{83+76}{2}=79.5\\ \mathrm{B},\mathrm{J}& 83,80& \frac{83+80}{2}=81.5\\ \mathrm{B},\mathrm{C}& 83,74& \frac{83+74}{2}=78.5\\ \mathrm{B},\mathrm{H}& 83,80& \frac{83+80}{2}=81.5\\ \text{W,J}& 76,80& \frac{76+80}{2}=78.0\\ \mathrm{W},\mathrm{C}& 76,74& \frac{76+74}{2}=75.0\\ \mathrm{W},\mathrm{H}& 76,80& \frac{76+80}{2}=78.0\\ \mathrm{J},\mathrm{C}& 80,74& \frac{80+74}{2}=77.0\\ \mathrm{J},\mathrm{H}& 80,80& \frac{80+80}{2}=80.0\\ \mathrm{C},\mathrm{H}& 74,80& \frac{74+80}{2}=77.0\\ \hline\end{array}$

The mean of all possible sample mean is get in the table is

$$\begin{gathered} {\mu }_{\overline{x}}=\frac{\sum \overline{x_i}}{N} \\ =\frac{\left(\begin{array}{l}79.5+81.5+78.5+81.5+78.0+\\ 75.0+78.0+77.0+80.0+77.0\end{array}\right)}{10} \\ =\frac{786}{10} \\ =78.6 \end{gathered}$$

Given in the question that a table

Here we need to compare the population mean and mean height that has sample size $2$.

So, the mean of the sample mean is equal to the population mean

$$\begin{gathered} \left({\mu }_x\right)=\mu \\ =78.6 \end{gathered}$$

So, here both are same.