Question

Use a normal approximation to find the probability of the indicated number of voters. In this case assume that 119 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.
The probability that fewer than 32 of 19 eligible voters voted is=

Answer 1

I'm sure you meant "fewer than 32 of 119", not "19".

Our "population" proportion is given to be \(0.22=22\%\). We have a sample size of \(119\) voters. This means the distribution of voters has a mean of approximately \(np=119\times0.22=26.18\) and the standard deviation is approximately \(\sqrt{np(1-p)}=\sqrt{119\times0.22\times0.78}\approx4.52\).

To find the probability, we approximate with the normal distribution as follows.
\[P\left(X<32\right)=P\left(\frac{X-26.18}{4.52}<\frac{32-26.18}{4.52}\right)\approx P(Z<1.29)\]
Equivalently, you can compute this probability in terms of the proportions. You just have to adjust the standard deviation, using \(\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22\times0.78}{119}}\approx0.038\) in place of \(\sqrt{np(1-p)}\).
\[P\left(p<\frac{32}{119}\approx0.27\right)=P\left(\frac{p-0.22}{0.038}<\frac{0.27-0.22}{0.038}\right)\approx P(Z<1.32)\]
The discrepancy between \(z=1.29\) and \(z=1.32\) is due to rounding error.