According to a survey, 50% of Americans were in 2005 satisfied with their job.Assume that the result is true for the current proportion of Americans. A. Find the mean and standard deviation of the proportion for a sample of1000.
Accepted Solution
A:
Answer: [tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]Step-by-step explanation:The probability distribution of sampling distribution [tex]\hat{p}[/tex] is known as it sampling distribution.The mean and standard deviation of the proportion is given by :-[tex]\mu_{\hat{p}}=p\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex], where p =population proportion and n= sample size.Given : According to a survey, 50% of Americans were in 2005 satisfied with their job.i.e. p = 50%=0.50Now, for sample size n= 1000 , the mean and standard deviation of the proportion will be :-[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{0.50(1-0.50)}{1000}}=\sqrt{0.00025}\\\\=0.0158113883008\approx0.0158[/tex]Hence, the mean and standard deviation of the proportion for a sample of 1000:[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]