Question

According to data from the 2010 United States Census, 43.1% of Americans over the age of 65 were male.

Answer 1

Answer: $\mu_{\hat{p}}=0.431\\\\ \sigma_{\hat{p}}=0.064$

Step-by-step explanation:

The mean and standard deviation of the sampling distribution of p is given by :-

$\mu_{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}$

, where p= population proportion.

n= sample size.

Let p represent the sample proportion of Americans over the age of 65 that were male.

Given : The proportion of Americans over the age of 65 were male.

p= 43.1%=0.431

sample size : n= 60

Then, the mean and standard deviation of the sampling distribution of p will be :-

$\mu_{\hat{p}}=p =0.431\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.431(1-0.431)}{60}}\\\\=\sqrt{0.0041}\approx0.064$

Hence, the mean and standard deviation of the sampling distribution of p :

$\mu_{\hat{p}}=0.431\\\\ \sigma_{\hat{p}}=0.064$