Do you jog? The Gallup Poll once asked a random sample of$1540$adults, "Do you happen to jog?" Suppose that in fact $15\%$of all adults jog.

(a) What is the mean of the sampling distribution of $\hat{p}$? Justify your answer.

(b) Find the standard deviation of the sampling distribution of $\hat{p}$. Check that the $10\%$condition is met.

(c) Is the sampling distribution of $\hat{p}$approximately Normal? Justify your answer.

(d) Find the probability that between $13\%$and $17\%$of a random sample of $1540$adults are joggers. Show your work.

Population proportion $\left(p\right)=0.15$,

Sample size $\left(n\right)=1540$.

The sample distribution's mean can be computed as:

${\mu }_{\hat{p}}=p$

$=0.15$

The standard deviation of $\hat{p}$'s sampling distribution must be determined. Make sure that the $10\%$ criteria is met.

The sample proportion's standard deviation is calculated as:

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

$=\sqrt{\frac{0.15(1-0.15)}{1540}}$

$=0.0091$

Given in the question that, The Gallup Poll once asked a random sample of $1540$adults.

We have to find is the sampling distribution of p close to Normal? Justify your response.

Here,

$np=1540(0.15)$

$=231>10$

$n(1-p)=1540(1-0.15)$

$=1309>10$

As a result, normal approximation might be appropriate.

We need to calculate the likelihood that between $13$ and $17$ percent of a random sample of$1540$ people are joggers. Display your work.

The probability that between $13\%$and $17\%$of the sample of $1540$adults are joggers is calculated as:

$P(0.13\le \hat{p}\le 0.17)=P\left(\frac{0.13-0.15}{0.0091}\le Z\le \frac{0.17-0.15}{0.0091}\right)$