Given the problem, let's determine the standard deviation of the sampling distribution of the sample proportion [tex]$\hat{p}$[/tex], using the following formula:
[tex]\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
\][/tex]
where:
- [tex]\( p \)[/tex] is the true proportion of the tennis player's serve-return rate, which is 0.71.
- [tex]\( n \)[/tex] is the sample size, which is 55.
Substituting these values into the formula, we get:
[tex]\[
\sigma_{\hat{p}} = \sqrt{\frac{0.71 \times (1 - 0.71)}{55}}
\][/tex]
Simplifying inside the square root:
[tex]\[
\sigma_{\hat{p}} = \sqrt{\frac{0.71 \times 0.29}{55}}
\][/tex]
Calculating the numerator:
[tex]\[
0.71 \times 0.29 = 0.2059
\][/tex]
Now, dividing by the sample size:
[tex]\[
\frac{0.2059}{55} = 0.003743636
\][/tex]
Taking the square root of this quotient yields:
[tex]\[
\sigma_{\hat{p}} = \sqrt{0.003743636} \approx 0.061
\][/tex]
Thus, we find that the standard deviation of the sampling distribution of [tex]$\hat{p}$[/tex] is approximately 0.061.
Therefore, the most appropriate choice from the given options is:
[tex]\[
\sigma_{\hat{p}} = 0.061
\][/tex]
This means that in Simple Random Samples (SRSs) of size 55, the sample proportion of the tennis player's serve-return rate typically varies by about 0.061 from the true proportion [tex]\( p = 0.71 \)[/tex]. Thus, the correct answer is:
[tex]\[
\sigma_{\hat{p}} = 0.061. \text{In SRSs of size 55, the sample proportion of this tennis player's serve-return rate typically varies 0.061 from the true proportion p=0.71.}
\][/tex]
