91Ó°ÊÓ

Given,$$\begin{gathered} n=1540 \\ p=15\%=0.15 \end{gathered}$$

The mean of the sampling distribution is equal to the population proportion:

${\mathrm{μ}}_{\hat{p}}=p=0.15$

Given,$$\begin{gathered} n=1540 \\ p=15\%=0.15 \end{gathered}$$

Formula used:

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

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

$\begin{array}{rcl}{\mathrm{σ}}_{\hat{p}}& =& \sqrt{\frac{p(1-p)}{n}}\\ & =& \sqrt{\frac{0.15(1-0.15)}{1540}}\\ & =& 0.0091\end{array}$

The proportion of adults who jog in a random sample of 1540 adults varies on average by $0.0091$from the mean proportion of $0.15$.

The $10\%$condition is met, because the 1540 adults is less than $10\%$of the population of all adults.

Given,$$\begin{gathered} n=1540 \\ p=15\%=0.15 \end{gathered}$$

The sampling distribution of $\hat{p}$is approximately normal if n p and $n(1-p)$are both at least 10 .

$$\begin{gathered} np=1540×0.15=231 \\ n(1-p)=1540×(1-0.15)=1309 \end{gathered}$$

Since both are greater than the distribution is approximately normal.

Given,$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

The sampling distribution of $\hat{p}$is approximately normal with mean $0.15{\mathrm{σ}}_{\hat{p}}$and standard deviation $0.0091$

The z-score is

$\begin{array}{rcl}z& =& \frac{x-\mathrm{μ}}{\mathrm{σ}}\\ & =& \frac{0.13-0.15}{0.0091}\\ & =& -2.20\\ z& =& \frac{x-\mathrm{μ}}{\mathrm{σ}}\\ & =& \frac{0.17-0.15}{0.0091}\\ & =& 2.20\end{array}$

Finding the corresponding probability using table

$$\begin{gathered} P(0.13≤\hat{p}≤0.17)=P(-2.20<z<2.20)-P(z<-2.20) \\ =0.9861-0.0139 \\ =0.9722 \end{gathered}$$