Chapter 7, Problem 4CRE

a.

To determine

To find: The mean of the sampling distribution of $\widehat{p}$

Answer to Problem 4CRE

The mean is 0.15.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population proportion}\left(p\right)=0.15\\ \text{Sample size}\left(n\right)=1540\end{array}$

The mean of the sampling distribution can be calculated as:

  $\begin{array}{c}{\mu }_{\widehat{p}}=p\\ =0.15\end{array}$

Here, mean is 0.15

b.

To determine

To find: The standard deviation of the sample proportion.

Answer to Problem 4CRE

The standard deviation is 0.0091.

Explanation of Solution

The standard deviation of the sample proportion is computed as:

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.15\left(1-0.15\right)}{1540}}\\ =0.0091\end{array}$

Thus, the standard deviation is 0.0091.

c.

To determine

To find: Whether the normality assumptions are satisfied or not.

Explanation of Solution

Here,

  $\begin{array}{l}np=1540\left(0.15\right)=231>10\\ n\left(1-p\right)=1540\left(1-0.15\right)=1309>10\end{array}$

Thus, normal approximation could be used here.

d.

To determine

To find: The probability that between 13% and 17% of the sample of 1540 adults are joggers.

Answer to Problem 4CRE

The probability is 0.9722.

Explanation of Solution

The probability that between 13% and 17% of the sample of 1540 adults are joggersis computed as:

  $\begin{array}{c}P\left(0.13\le \widehat{p}\le 0.17\right)=P\left(\frac{0.13-0.15}{0.0091}\le Z\le \frac{0.17-0.15}{0.0091}\right)\\ =P\left(-2.20\le Z\le 2.20\right)\\ =0.9722\end{array}$

Thus, the probability is 0.9722.