Respuesta :
Answer:
[tex]\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997[/tex]
Step-by-step explanation:
We know that the mean and standard deviation of the sampling distribution of the sample proportion([tex]\hat{p}[/tex]) is given by :-
[tex]\mu _{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]
, where p= Population proportion and n = sample size.
Let p be the proportion of blue chips.
As per given , we have
p= 0.275
n= 5
Now , the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be :
[tex]\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{ 0.275(1- 0.275)}{5}}\\\\=0.19968725547\approx0.1997[/tex]
Hence, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 are :
[tex]\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997[/tex]
The standard deviation and mean is
[tex]\\\mu = 0.275\\\sigma = 0.1997[/tex]
Given that ;
No. of blue chips contain in a bag = P = 27.5 % = 0.27
No. of chips random selected at a time with replacement = n = 5
We know that the mean and standard deviation of the sampling distribution of the sample proportion (p) is given by :-
[tex]\mu\-_p[/tex] = p
= 0.275
[tex]\sigma_p[/tex] = [tex]\sqrt{\frac{p (1-p)}{n} }[/tex]
Where p= Population proportion and n = sample size.
Let p be the proportion of blue chips.
As per given , we have
p= 0.275
n= 5
= [tex]\sqrt{\frac{0.275 ( 1 - 0.275)}{5} }[/tex]
= 0.1997
Now , the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be :
Hence, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 are :
[tex]\mu = 0.275\\\sigma = 0.1997[/tex]
For more details about standard deviation and mean click the link given below;
https://brainly.com/question/8170300
