Question

Consider a sampling distribution with p equals 0.15p=0.15 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 5000n=5000. b. For a random sample of size n equals 1000n=1000. c. For a random sample of size n equals 500n=500.

Answer 1

Answer:

$a.\ \mu_p=750\ \ , \sigma_p=0.005\\\\b.\ \mu_p=150\ \ , \sigma_p=0.0113\\\\c.\ \mu_p=75\ \ , \sigma_p=0.0160$

Step-by-step explanation:

a. Given p=0.15.

-The mean of a sampling proportion  of n=5000 is calculated as:

$\mu_p=np\\\\=0.15\times 5000\\\\=750$

-The standard deviation is calculated using the formula:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\=\sqrt{\frac{0.15(1-0.15)}{5000}}\\\\=0.0050$

Hence, the sample mean is μ=750 and standard deviation is σ=0.0050

b. Given that p=0.15 and n=1000

#The mean of a sampling proportion  of n=1000 is calculated as:

$\mu_p=np\\\\=1000\times 0.15\\\\\\=150$

#-The standard deviation is calculated as follows:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{1000}}\\\\\\=0.0113$

Hence, the sample mean is μ=150 and standard deviation is σ=0.0113

c. For p=0.15 and n=500

#The mean is calculated as follows:

$\mu_p=np\\\\\\=0.15\times 500\\\\=75$

#The standard deviation of the sample proportion is calculated as:

$\sigma_p=\sqrt{\frac{p(1-p)}{n}}\\\\\\=\sqrt{\frac{0.15\times 0.85}{500}}\\\\\\=0.0160$

Hence, the sample mean is μ=75 and standard deviation is σ=0.0160