According to data from the 2010 United States Census, 43.1% of Americans over the age of 65 were male. Suppose Maria, a researcher, takes a random sample of 60 Americans over the age of 65 and finds that 18 are male. Let p represent the sample proportion of Americans over the age of 65 that were male. What are the mean and standard deviation of the sampling distribution of p?

Answer: [tex]\mu_{\hat{p}}=0.431\\\\ \sigma_{\hat{p}}=0.064[/tex]Step-by-step explanation:The mean and standard deviation of the sampling distribution of p is given by :-[tex]\mu_{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex], 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%=0.431sample size : n= 60Then, the mean and standard deviation of the sampling distribution of p will be :-[tex]\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[/tex]Hence, the mean and standard deviation of the sampling distribution of p :[tex]\mu_{\hat{p}}=0.431\\\\ \sigma_{\hat{p}}=0.064[/tex]