Question

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.

Answer 1

Answer:

$\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158$

Step-by-step explanation:

The probability distribution of sampling distribution $\hat{p}$ is known as it sampling distribution.

The mean and standard deviation of the proportion is given by :-

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

, 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%=0.50

Now, for sample size n= 1000 , the mean and standard deviation of the proportion will be :-

$\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{0.50(1-0.50)}{1000}}=\sqrt{0.00025}\\\\=0.0158113883008\approx0.0158$

Hence, the mean and standard deviation of the proportion for a sample of 1000:

$\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158$