# Obtain the sampling distribution of overline{p}

Obtain the sampling distribution of $\stackrel{―}{p}$
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mhalmantus

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 .
Verify the conditions:
$np=200×0.45$
$=90⇒5$.
$n\left(1—p\right)=200×\left(1—0.45\right)$
And
$=110⇒5$
The conditions are satisfied. Therefore, the sampling distribution of the proportion is normal.
The mean of the $\stackrel{―}{p}$ is $E\left(\stackrel{―}{p}\right)=p$ and standard deviation of $\stackrel{―}{p}$ is ${\sigma }_{\stackrel{―}{p}},=\frac{\sqrt{p\left(1-p\right)}}{n}$
Mean and standard deviation are calculated as below
$E\left(\stackrel{―}{p}\right)=p=0.45$
${\sigma }_{\stackrel{―}{p}}=\sqrt{p\left(1-p\right)}=\frac{\sqrt{0.45×0.55}}{200}=0.0352$

Thus, the sampling distribution of the proportion J of entrepreneurs whose first startup was at 30 years or more is normal with mean $E\left(\stackrel{―}{p}\right)=0.45$ and standard deviation ${\sigma }_{\stackrel{―}{p}},=0.0352$

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