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.

Respuesta :

Answer with explanation:

Consider a sampling distribution with p equals 0.12 and samples of size n each.

Then, formula for mean = p

Formula for standard deviation= [tex]\sqrt{\dfrac{p(1-p)}{n}}[/tex]

(a) For a random sample of size n equals 5000.

Mean = 0.12

Standard 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 = 0.12

Standard 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 = 0.12

Standard deviation= [tex]\sqrt{\dfrac{0.12(1-0.12)}{500}}[/tex]

[tex]\\\\=0.0145327216997\approx0.0145[/tex]