91Ó°ÊÓ

To determine the sample for exercise $7.41$for samples of size$4$. The given sample in exercise 7.41 as:

For samples of size $4$, calculate the mean height using $\left({\mathrm{μ}}_{\overline{x}}\right)$.
As a result, as indicated in the table below, the samples of size $4$ and their respective means are obtained:

As a result, the number of feasible size $4$samples$\left(N\right)$is $5$. The mean of all potential sample means is calculated as follows for samples of size $4$:
${\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{∑{\stackrel{¯}{x}}_i}{N}$
$=\frac{78.25+79.75+78.25+79.25+77.5}{5}$
$=\frac{393}{5}$
$=78.6$
As a result, the mean height $\left({\mathrm{μ}}_{\stackrel{¯}{x}}\right)$for samples of size 4 is$78.6$.

Calculate the mean height $\left({\mathrm{μ}}_{\stackrel{¯}{x}}\right)$.
The average height of five players in the population is $78.6$ inches.
The population mean is equal to the mean of the sample mean.
That is to state,
${\mathrm{μ}}_{\overline{x}}=\mathrm{μ}$

$=78.6$

As a result, the mean height $\left({\mathrm{μ}}_{\stackrel{¯}{x}}\right)$ for size $4$ samples is$78.6$.