91Ó°ÊÓ

We are given a data in the table as

And we need to find the population mean of the data.

From the table, the population data is given as $83,76,80,74,80$.

So the population mean height for the five players is given as

$$\begin{gathered} \mathrm{μ}=\frac{83+76+80+74+80}{5} \\ \mathrm{μ}=\frac{393}{5} \\ \mathrm{μ}=78.6 \end{gathered}$$

For the population data: $83,76,80,74,80$.

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

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{79.5+81.5+78.5+81.5+78+75+78+77+80+77}{10} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{786}{10} \\ {\mathrm{μ}}_{\stackrel{¯}{x}}=78.6 \end{gathered}$$

So when the sample size is 2, the variable $\stackrel{¯}{x}$has a mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=78.6$.

We know that mean of the sample mean is equal to the population mean irrespective of the sample size.

Here, the population mean is $\mathrm{μ}=78.6$.

So the mean of the sample mean of sample size 2 is

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ} \\ =78.6 \end{gathered}$$