Answer :
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 = 0.61
a) n= 20
The 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 = 0.61
standard 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
=0.61
standard 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
=0.61
standard 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 =0.61
standard 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 =0.61
standard 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 =0.61
standard deviation =0.0337