91Ó°ÊÓ

Total number of athletes$=316$

Number of$SRS=550$

Mean of the sampling distribution of $\hat{p}$is, ${\mathrm{μ}}_{\hat{p}}=p$

Standard deviation of the sampling distribution of $\hat{p}$is, ${\mathrm{σ}}_{\widehat{p}}=\sqrt{\frac{p(1−p)}{n}}$

We can suppose $x$and $p$in terms of the number of athletes who have been barred from studying at least once by their coaches.

Hence, $p=0.40$and $n=50$

Determine the product of $n$and $p$:

localid="1650010231214" $$\begin{gathered} \begin{array}{r}np=50×0.40\end{array} \\ =20>10 \end{gathered}$$

Due to the values of $np,n(1−p)â‰\yen 10$, the sampling distribution of $\hat{p}$is approximately normal.

Here, sampling distribution of $\hat{p}$is normally distributed and the mean is ${\mathrm{μ}}_{\hat{p}}$and standard deviation is ${\mathrm{σ}}_{\hat{p}}$.

Substitute the values of $p=0.40$in ${\mathrm{μ}}_{\hat{p}}=p$.

${\mathrm{μ}}_{\hat{p}}=0.40$

Substitute the values of $p=0.40$and $n=50$by using the given formula:

localid="1650010234438" $$\begin{gathered} {\mathrm{σ}}_{\widehat{p}}=\sqrt{\frac{p(1−p)}{n}}. \\ {\mathrm{σ}}_p^{\mathrm{λ}}=\sqrt{\frac{0.40(1−0.40)}{50}} \\ =\sqrt{0.0048} \\ =0.0693 \end{gathered}$$

Calculate the probability that at least $\frac{15}{50}=0.30\left(30\mathrm{\%}\right)$participants have been told by the coaches to neglect their studies at least once, as follows:

localid="1650010288971" $$\begin{gathered} \begin{array}{c}P(\widehat{p}â‰\yen 15)=P\left(\frac{\widehat{p}−{\mathrm{μ}}_{\widehat{p}}}{{\mathrm{σ}}_{\widehat{p}}}â‰\yen \frac{0.30−0.40}{0.0693}\right)\end{array} \\ =P(zâ‰\yen 1.44) \\ =1−P(z<−1.44) \\ =1−0.0749 \\ =0.9251 \end{gathered}$$