91Ó°ÊÓ

$$\begin{gathered} p=59\%=0.59 \\ n=50 \end{gathered}$$

The population proportion $p$is the same as the mean of the sampling distribution of the sample proportions.

${\mathrm{μ}}_{\hat{p}}=p=59\%=0.59$

$$\begin{gathered} p=59\%=0.59 \\ n=50 \end{gathered}$$

We know,

${\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

The population proportion $p$is the same as the mean of the sampling distribution of the sample proportions.

${\mathrm{μ}}_{\hat{p}}=p=59\%=0.59$

The sampling distribution's standard deviation is:

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.59(1-0.59)}{50}} \\ =0.0696 \end{gathered}$$

Therefore, the proportion of couples in which both parents work outside the home among $50$ married couples varies on average by $0.0696$from the mean of $0.59$

$$\begin{gathered} p=59\%=0.59 \\ n=50 \end{gathered}$$

The sampling distribution of the sample proportions $\hat{p}$is about Normal is the large count's condition is satisfied that is when$npâ‰\yen 10$and$n(1-p)â‰\yen 10$

$$\begin{gathered} np=50(0.59)=29.5â‰\yen 10 \\ n(1-p)=50(1-0.59)=20.5â‰\yen 10 \end{gathered}$$

Because the large count's condition is met, the sample proportion's sampling distribution is roughly Normal.