We have binomial experiment with \(n = 100\) and \(p = 0.23\)

\(np = 100(0.23)\)

\(np = 23\)

\(nq = 100(1 — 0.23)\)

\(nq = 77\)

Since both the values np and ng are greater than 5, hence, we can approximate the \(\hat{p}\) distribution by a normal distribution.

The formula for the mean of the \(\hat{p}\) distribution is \(\mu_{hat{p}} = \hat{p}\).

\(\mu_{hat{p}} = 0.23\)

The formula for the standard error of the normal approximation to the \(\hat{p}\) distribution is

\(\sigma_{\hat{p}}=\sqrt{\frac{pq}{n}}\)

\(\sigma_{\hat{p}}=\sqrt{0.23\frac{1-0.23}{100}}\)

\(\sigma_{hat{p}}=0.042\)

\(np = 100(0.23)\)

\(np = 23\)

\(nq = 100(1 — 0.23)\)

\(nq = 77\)

Since both the values np and ng are greater than 5, hence, we can approximate the \(\hat{p}\) distribution by a normal distribution.

The formula for the mean of the \(\hat{p}\) distribution is \(\mu_{hat{p}} = \hat{p}\).

\(\mu_{hat{p}} = 0.23\)

The formula for the standard error of the normal approximation to the \(\hat{p}\) distribution is

\(\sigma_{\hat{p}}=\sqrt{\frac{pq}{n}}\)

\(\sigma_{\hat{p}}=\sqrt{0.23\frac{1-0.23}{100}}\)

\(\sigma_{hat{p}}=0.042\)