Answer to Question #343081 in Statistics and Probability for kiera

Question

A population has mean 75 and standard deviation 12.

a.)Random samples of size 121 are taken. Find the mean and standard deviation of sampling distribution of the means?

b.)How would the answers to part (a) change if the size of the samples were 400 instead of 121?

Answer 1

a)

$$\mu_{\bar{X}}=\mu=75$$

$$\sigma_{\bar{X}}=\sigma/\sqrt{n}=12/\sqrt{121}=\dfrac{12}{11}$$

b)

$$\mu_{\bar{X}}=\mu=75$$

$$\sigma_{\bar{X}}=\sigma/\sqrt{n}=12/\sqrt{400}=\dfrac{12}{20}=\dfrac{3}{5}$$

The mean stays the same with increasing (or decreasing) of $n.$

The standard deviation is decreasing in $\dfrac{\sqrt{n_2}}{\sqrt{n_1}}=\dfrac{\sqrt{400}}{\sqrt{121}}=\dfrac{20}{11}$ times.