A random sample is to be selected from a population that has a proportion of successes p = 0.61. Determine the mean and standard deviation of the sampling distribution of p? for each of the following sample sizes. (Round your standard deviations to four decimal places.)(a) n = 20mean standard deviation (b) n = 30mean standard deviation (c) n = 40mean standard deviation (d) n = 60mean standard deviation (e) n = 110mean standard deviation (f) n = 210mean standard deviation

Answer with explanation:We know that 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], here p is the proportion of successes ion and n is the sample size.We are given that , Proportion of successes p = 0.61a) n= 20The mean and standard deviation of the sampling distribution of p is given by :-[tex]\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{20}}\\\\=\sqrt{0.011895}\approx0.1091[/tex]So,  mean  = 0.61standard deviation = 0.1091(b) n = 30[tex]\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{30}}\\\\=\sqrt{0.00793}\approx0.0891[/tex]mean=0.61standard deviation  =0.0891(c) n = 40[tex]\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{40}}\\\\\approx0.0771[/tex]mean=0.61standard deviation  =0.0771(d) n = 60  [tex]\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{60}}\\\\\approx0.0630[/tex]mean  =0.61standard deviation  =0.0630(e) n = 110  [tex]\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{110}}\\\\\approx0.0465[/tex]mean  =0.61standard deviation  = 0.0465(f) n = 210  [tex]\mu_{\hat{p}}=0.61\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{0.61(1-0.61)}{210}}\\\\\approx0.0337[/tex]mean  =0.61standard deviation =0.0337