Studious athletes A university is concerned about the academic standing of its intercollegiate athletes. A study committee chooses an SRS of 50 of the 316 athletes to interview in detail. Suppose that $40\%$ of the athletes have been told by coaches to neglect their studies on at least one occasion. What is the probability that at least 15 in the sample are among this group?

Total number of athletes$=316$

Number of$SRS=550$

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

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

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

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

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

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

${\mu }_{\hat{p}}=0.40$

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

localid="1650010234438" $$\begin{gathered} {\sigma }_{\widehat{p}}=\sqrt{\frac{p(1-p)}{n}}. \\ {\sigma }_p^{\lambda }=\sqrt{\frac{0.40(1-0.40)}{50}} \\ =\sqrt{0.0048} \\ =0.0693 \end{gathered}$$

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" $$\begin{gathered} \begin{array}{c}P(\widehat{p}\ge 15)=P\left(\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\ge \frac{0.30-0.40}{0.0693}\right)\end{array} \\ =P(z\ge 1.44) \\ =1-P(z<-1.44) \\ =1-0.0749 \\ =0.9251 \end{gathered}$$