Question

A population consists of the values (1, 3, 4). If the samples of size 2 will be drawn from the population, find the variance of the sampling distribution of the means.

Answer 1

The population mean:

$\mu=\cfrac{1+3+4}{3}=\cfrac{8}{3}.$

The population variance:

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

$X-\mu=\begin{Bmatrix} 1-\cfrac{8}{3},3-\cfrac{8}{3},4-\cfrac{8}{3} \end{Bmatrix}=$

$=\begin{Bmatrix} -\cfrac{5}{3}, \cfrac{1}{3},\cfrac{4}{3} \end{Bmatrix},$

$\sigma^2=\bigg(\cfrac{-5}{3}\bigg)^2\cdot \cfrac{1}{3}+\bigg(\cfrac{1}{3}\bigg)^2\cdot \cfrac{1}{3}+\bigg(\cfrac{4}{3}\bigg)^2\cdot \cfrac{1}{3}=\\ =\cfrac{42}{27}=\cfrac{14}{9}=1.556.$

The variance of the sampling distribution of sample means:

$\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{1.556}{2}=0.778.$