Chapter 7, Problem 13PT

To determine

Tocalculate:The probability more than 20% of the sample are from poverty households.

Answer to Problem 13PT

The required probability is 0.7995.

Explanation of Solution

Given information:

Proportion of children living in households with less incomes = 22%

Number of children for SRS = 300

Formula used:

The mean of the sampling distribution of $\widehat{p}$ is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Calculation:

Consider the proportion of children living in households with less incomes than the official poverty levelbe p and sample proportion of children be $\widehat{p}$ .

Rewrite it as:

  $p=22\%=0.22$

  $\widehat{p}=20\%=0.20$

For the mean of the sampling distribution of $\widehat{p}$ , use the formula ${\mu }_{\widehat{p}}=p$ .

Substitute 0.22 for p in the formula ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.22$

For the standard deviation of the sampling distribution, use the formula ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.22 for p and 300 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.22\left(1-0.22\right)}{300}}\\ =\sqrt{\frac{0.22\times 0.78}{300}}\\ =\sqrt{\frac{0.1716}{300}}\\ \approx 0.0239\end{array}$

Now, find the z-score by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute 0.20 for $x$ , 0.22 for $\mu$ and 0.0239 for $\sigma$ in the formula $z=\frac{x-\mu }{\sigma }$ and simplify.

  $\begin{array}{c}z=\frac{0.20-0.22}{0.0239}\\ \approx -0.84\end{array}$

Find the probability by using table A as:

  $\begin{array}{c}P\left(\widehat{p}\ge 20\%\right)=P\left(z>-0.84\right)\\ =0.7995\end{array}$

Hence, the required probability is 0.7995.