Answer to Question #337111 in Statistics and Probability for Xyka Sarte

Question

Given the population 3 5,7,9,11,13

How many samples can be made from the population with sample size of 3?

Calculate the mean of the sampling distribution

Compute the variance o the sampling distribution

Answer 1

The number of possible samples which can be selected without replacement is

$\begin{pmatrix} N \\ n \end{pmatrix}=\cfrac{N! } {n! \cdot(N-n)! }=\\ =\cfrac{6! } {3! \cdot3! }=\cfrac{4\cdot5\cdot6}{2\cdot3}=20.$

Population mean:

$\mu=\cfrac{3+5+7+9+11+13}{6}=8.$

Population variance:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\\ =\begin{Bmatrix} 3-8,5-8,7-8,9-8,11-8,13-8 \end{Bmatrix}=$

$=\begin{Bmatrix} -5, - 3,-1,1,3,5 \end{Bmatrix},$

$\sigma^2=(-5)^2\cdot \cfrac{1}{6}+(-3)^2\cdot \cfrac{1}{6}+\\ +(-1)^2\cdot \cfrac{1}{6}+1^2\cdot \cfrac{1}{6}+3^2\cdot \cfrac{1}{6}+5^2\cdot \cfrac{1}{6}=11.667.$

Mean of the sampling distribution of sample means:

$\mu_{\bar x} =\mu=8.$

Variance of the sampling distribution of sample means:

$\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{11.667}{3}=3.889.$