It is given that is the sample proportion of entrepreneurs whose first startup was at 30 years or more is \(p = 0.45\) and the sample size \(n = 200\)

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

Verify the conditions:

\(np = 200 \times 0.45\)

\(=90\Rightarrow 5\).

\(n(1 — p) = 200 \times (1 — 0.45)\)

And

\(=110\Rightarrow 5\)

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 \(\sigma_\overline{p}, = \frac{\sqrt{p(1-p)}}{n}\)

Mean and standard deviation are calculated as below

\(E(\overline{p}) =p = 0.45\)

\(\sigma_{\overline{p}}=\sqrt{p(1-p)}=\frac{\sqrt{0.45\times0.55}}{200}=0.0352\)

\(Py\ 0.45\times 0.55\)

Thus, the sampling distribution of the proportion J of entrepreneurs whose first startup was at 30 years or more is normal with mean \(E(\overline{p}) = 0.45\) and standard deviation \(\sigma_{\overline{p}}, = 0.0352\)

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

Verify the conditions:

\(np = 200 \times 0.45\)

\(=90\Rightarrow 5\).

\(n(1 — p) = 200 \times (1 — 0.45)\)

And

\(=110\Rightarrow 5\)

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 \(\sigma_\overline{p}, = \frac{\sqrt{p(1-p)}}{n}\)

Mean and standard deviation are calculated as below

\(E(\overline{p}) =p = 0.45\)

\(\sigma_{\overline{p}}=\sqrt{p(1-p)}=\frac{\sqrt{0.45\times0.55}}{200}=0.0352\)

\(Py\ 0.45\times 0.55\)

Thus, the sampling distribution of the proportion J of entrepreneurs whose first startup was at 30 years or more is normal with mean \(E(\overline{p}) = 0.45\) and standard deviation \(\sigma_{\overline{p}}, = 0.0352\)