91Ó°ÊÓ

The samples from exercise 7.41 is:

Determine the parts (b) of Exercise $7.41$for samples of size $3$. Calculate the mean height $\left({\mathrm{μ}}_{\stackrel{¯}{x}}\right)$for samples of size $3$ .
As a result, the size 3 samples and their means are obtained as given in the table below:

The number of possible samples $\left(N\right)$of size $3$is $10$. For samples of size $3$ , as illustrated below, the mean of all potential sample means is calculated:
${\mathrm{μ}}_{\stackrel{¯}{x}}=\frac{∑{\stackrel{¯}{x}}_i}{N}$
$=\frac{79.67+77.67+79.67+79+81+89+76.67+78.67+76.67+78}{10}$
$=\frac{786}{10}$
$=78.6$

As a result, the mean height $\left({\mathrm{μ}}_{\stackrel{¯}{x}}\right)$for samples of size $3$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 would be to:
${\mathrm{μ}}_{\overline{x}}=\mathrm{μ}$
$=78.6$
Therefore, the mean height $\left({\mathrm{μ}}_{\stackrel{¯}{x}}\right)$for samples of size $3$is $78.6$.