Chapter 7.2, Problem 34E

To determine

Tocalculate:The probability that at least 15 in the sample are among the group.

Answer to Problem 34E

The required probability is 0.9251.

Explanation of Solution

Given information:

Total number of athletes = 316

Number of SRS = 550

Formula used:

The mean of the sampling distribution of $\widehat{p}$is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of the sampling distribution of $\widehat{p}$is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Consider that number of athletes which have been told by coaches to neglect their studies on at least one occasion and the corresponding proportion be $x$and $\widehat{p}$respectively.

So, we can write $p=0.40$and $n=50$

Now, find the product of n and p as per the requirement.

  $\begin{array}{c}np=50\times 0.40\\ =20\\ >10\end{array}$

Also, find the product of nand $1-p$ .

  $\begin{array}{c}n\left(1-p\right)=50\times \left(1-0.40\right)\\ =50\times 0.60\\ =30\\ >10\end{array}$

Here, the sampling distribution of $\widehat{p}$is approximately because the values of $np,n\left(1-p\right)\ge 10$ .

This means, the sampling distribution of $\widehat{p}$is normally distributed whose mean is ${\mu }_{\widehat{p}}$and standard deviation is ${\sigma }_{\widehat{p}}$ .

Substitute 0.40 for p in the formula of sampling distribution ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.40$

Substitute 0.40 for $p$and 50 for n in the standard deviation formula ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.40\left(1-0.40\right)}{50}}\\ =\sqrt{0.0048}\\ =0.0693\end{array}$

Write the probability at least $\frac{15}{50}=0.30\left(30\%\right)$in the sample have been told by the coaches to neglect their studies at least one occasion as,

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

Hence, the required probability is 0.9251.