Chapter 7.2, Problem 35E

(a)

To determine

To Explain: the mean of the sampling distribution of $\text{<mover accent="true"> <mo>p</mo> <mo>^</mo></mover>}$

Answer to Problem 35E

0.20

Explanation of Solution

Given:

  $\begin{array}{l}p\text{=20\% =0}\text{.20}\\ n\text{ =30}\end{array}$

Calculation:

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

  ${\mu }_{\widehat{p}}=p=20\%=0.20$

(b)

To determine

To Calculate: and interpret the standard deviation of the sampling distribution of $\text{<mover accent="true"><mo> p</mo> <mo>^</mo></mover>}$ .

Answer to Problem 35E

0.0730

Explanation of Solution

Given:

  $\begin{array}{l}p\text{=20\% =0}\text{.20}\\ n\text{ =30}\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

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

  ${\mu }_{\widehat{p}}=p=20\%=0.20$

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is the square root of the product of $p$ and $\left(1-p\right)$ divided by the sample size $n$ .

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.20\left(1-0.20\right)}{30}}\\ =0.0730\end{array}$

The proportion of orange candies among 30 randomly selected Skittles varies on average by $0.0730$ from the mean of $\text{0}\text{.20}$ .

(c)

To determine

To Explain: the shape of the sampling distribution of $\text{<mover accent="true"> <mo>p</mo> <mo>^</mo></mover>}$ .

Answer to Problem 35E

Approximate to the normal

Skewed to right

Explanation of Solution

Given:

  $\begin{array}{c}p\text{=20\% =0}\text{.20}\\ n\text{ =30}\end{array}$

Formula used:

Calculation:

The sampling distribution of the sample proportions $\widehat{p}$ is about Normal is the large counts condition is that is when $np\ge 10$ and $n\left(1-p\right)\ge 10$ .

  $\begin{array}{c}np=30\left(0.20\right)=6<10\\ n\left(1-n\right)=30\left(1-0.20\right)=24\ge 10\end{array}$

Since the population proportion of $0.20$ is closer to 0 than 1, the sample proportion is more likely to be close to 0 than 1 and therefore the sampling distribution of the sample proportions will be skewed to the right.