Consider a sampling distribution with p equals 0.12 and samples of size n each. Using the appropriate​ formulas, find the mean and the standard deviation of the sampling distribution of the sample proportion. (a) For a random sample of size n equals 5000. (b) For a random sample of size n equals 1000. (c) For a random sample of size n equals 500.

Answer with explanation:Consider a sampling distribution with p equals 0.12 and samples of size n each.Then, formula for mean = pFormula for standard deviation= [tex]\sqrt{\dfrac{p(1-p)}{n}}[/tex](a) For a random sample of size n equals 5000.Mean = 0.12Standard deviation= [tex]\sqrt{\dfrac{0.12(1-0.12)}{5000}}[/tex][tex]\\\\0.00459565011723\approx0.0046[/tex](b) For a random sample of size n equals 1000.Mean = 0.12Standard deviation= [tex]\sqrt{\dfrac{0.12(1-0.12)}{1000}}[/tex][tex]\\\\=0.0102761860629\approx0.0103[/tex](c) For a random sample of size n equals 500.Mean = 0.12Standard deviation= [tex]\sqrt{\dfrac{0.12(1-0.12)}{500}}[/tex][tex]\\\\=0.0145327216997\approx0.0145[/tex]