The sampling distribution of the proportion is approximately normal if \(np \Rightarrow 5\ and\ n(1 — p) \Rightarrow 5\).

Verify the conditions:

\(np = 200 \times 0.55\)

\(= 110\Rightarrow 5\)

\(n(1 — p) = 200 \times (1 — 0.55)\)

And

\(=90 \Rightarrow 5\)

The conditions are satisfied. Therefore, the sampling distribution of the proportion is normal.

The mean of the \(\overline{p}\ is\ E(\overline{p})) = p\) and standard deviation of \(\overline{p}\ is\ \sigma_{\overline{p}} = \sqrt{p(1-p)}/n\)

In this context, bar p is the sample proportion of entrepreneurs whose first startup was at 29 years or less

The mean of \(\overline{p}\) is

\(E \overline{p})=p= 0.55\)

The standard deviation of \(\overline{p}\) is

\(\sigma_{p} = \frac{\sqrt{p(1-p)}}{n}=\frac{\sqrt{0.55\times 0.45}}{200}= 0.0352\)

Thus, the sampling distribution of the proportion \(\overline{p}\) proportion of entrepreneurs whose first startup was at 29 years or less is normal with mean \(E(\overline{p}) = 0.55\) and standard deviation \(\sigma_{p} = 0.0352\)