Chapter 7.2, Problem 39E

To determine

Tocalculate:The probability that 75% or more of the women in the sample have been on a diet.

Answer to Problem 39E

The required probability is 0.0375.

Explanation of Solution

Given information:

Number of SRS = 267

Proportion of women which have been on a diet within the past 12 months = 70%

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 the proportion of women which have been on a diet within the past 12 months be p and proportion of women in the sample which have been on a diet be $\widehat{p}$ .

So, we can write:

  $p=70\%=0.70$

  $\widehat{p}=75\%=0.75$

For the mean of the sampling distribution of $\widehat{p}$ , we can use the formula ${\mu }_{\widehat{p}}=p$ .

Substitute 0.70 for p in the formula ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.70$

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

Substitute 0.70 for p and 267 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.70\left(1-0.70\right)}{267}}\\ =\sqrt{\frac{0.70\times 0.3}{267}}\\ =\sqrt{\frac{0.21}{267}}\\ \approx 0.0280449\end{array}$

Now, find the z-score by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute 0.75 for $x$ , 0.70 for $\mu$and 0.0280449 for $\sigma$in the formula $z=\frac{x-\mu }{\sigma }$and simplify.

  $\begin{array}{c}z=\frac{0.75-0.70}{0.0280449}\\ \approx 1.78\end{array}$

Find the probability by using table A.

  $\begin{array}{c}P\left(\widehat{p}\ge 75\%\right)=P\left(z>1.78\right)\\ =P\left(Z<-1.78\right)\\ =0.0375\end{array}$

Hence, the required probability is 0.0375.