Question

A sample survey interviews an SRS of $267$college women. Suppose (as is roughly true) that $70\%$of college women have been on a diet within the past$12$months. What is the probability that $75\%$or more of the women in the sample have been on a diet? Follow the four-step process.

Short answer

The probability that $75\%$or more of the women in the sample have been on a diet$P(\hat{p}\ge 75\%)=0.0375$.

Step-by-step solution

Given Information

Given

$p=70\%=0.70$

$\hat{p}=75\%=0.75$

$n=267$

Explanation

The mean of the sampling distribution of $\hat{p}$is equal to the population proportion $p$:

$$\begin{gathered} {\mu }_{\hat{p}}=p \\ =0.70 \end{gathered}$$

The standard deviation of the sampling distribution of $\hat{p}$is:

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.70(1-0.70)}{267}} \\ \approx 0.0280449 \end{gathered}$$

The z-score is the value decreased by the mean, divided by the standard deviation:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.75-0.70}{0.0280449} \\ \approx 1.78 \end{gathered}$$

Determine the corresponding probability using table A:

$$\begin{gathered} P(\hat{p}\ge 75\%)=P(z>1.78) \\ =P(Z<-1.78) \\ =0.0375 \end{gathered}$$