Answer to Question #339846 in Statistics and Probability for niks

Question

A population consists of 1, 5, 6, 8, 12, 7, and 11. Suppose a sample of size 3. Find the 

mean and variance.

Answer 1

Mean of population 

$$\mu=\dfrac{1+5+6+8+12+7+11}{7}=\dfrac{50}{7}$$ $$\approx7.142857$$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}$$

$$=\dfrac{1}{7}((1-\dfrac{50}{7})^2+(5-\dfrac{50}{7})^2+(6-\dfrac{50}{7})^2$$

$$+(8-\dfrac{50}{7})^2+(12-\dfrac{50}{7})^2+(7-\dfrac{50}{7})^2$$

$$+(11-\dfrac{50}{7})^2)=\dfrac{580}{49}\approx11.836735$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\dfrac{50}{7}=\mu$$

Variance of sampling distribution (without replacement)

$$Var(\bar{X})=\sigma_{\bar{X}}^2=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$=\dfrac{580}{49(7)}(\dfrac{7-2}{7-1})=\dfrac{1450}{1029}$$