According to government data, $22\%$American children under the age of six live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of $300$children. Find the probability that more than $20\%$of the sample is from poverty households. Be sure to check that you can use the Normal approximation.

Households with incomes less than the official poverty level is$22\%$

Choose SRS of $300$children

Find the probability that more than $20\%$of the sample is from poverty households.

The given data is written as

$p=22\%=0.22$

$\hat{p}=20\%=0.20$

$n=300$

The population percentage is equal to the mean of the sampling distribution of the sample proportion $\hat{p}$

${\mu }_{\hat{p}}=p=0.22$

The square root of the products of the probabilities$pand1-p$ divided by the sample size $n$equals the standard deviation of the sampling distribution of the sample proportion $\hat{p}$:

localid="1650250733252" $$\begin{gathered} {\sigma }_{\hat{p}}=\frac{p(1-p)}{n} \\ =\sqrt{\frac{0.22(1-0.22)}{300}} \\ \approx 0.0239\left(\right) \end{gathered}$$

The $\hat{p}$is approximately normal because $np=300(0.22)=66andn(1-p)=300(1-0.22)=23$ are both at least $10$

The z-score is the difference between the mean and the standard deviation:

localid="1650250746686" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.20-0.22}{0.0239} \\ \approx -0.84 \end{gathered}$$

The probability is determined using table A as

$P(\hat{p}>0.20)=P(z>-0.84)=P(z<-0.84)=0.7995$

$P(\hat{p}>0.20)=0.7995$.