Answer:
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=0.0785[/tex]
Step-by-step explanation:
We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-
[tex]\mu_{\hat{p}}=p[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]
, where p=population proportion and n= sample size.
Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.
i.e. p= 19%=0.19
The for sample size n= 25
The mean and the standard error of the sampling distribution of the sample proportions will be :-
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785[/tex]
Hence , the mean and the standard error of the sampling distribution of the sample proportions :
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=0.0785[/tex]
