Question

1.   A population consists of the numbers 2, 4, 9, 10, and 5. Let us list all possible samples size of 3 from this population. Find the mean and variance of the sampling distribution of the sample mean.

Answer 1

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

$\begin{pmatrix} N \\ n \end{pmatrix}=\cfrac{N! } {n! \cdot(N-n)! }=\\ =\cfrac{5! } {3! \cdot2! }=\cfrac{4\cdot5}{2}=10.$

All the possible samples of sizes n=3 wich can be drawn without replacement from the population:

$\{ (2,4,9), (2,4,10),(2,4,5),(2,9,10),(2,9,5),\\ (2,10,5),(4,9,10),(4,9,5),(4,10,5),(9,10,5)\}.$

Population mean:

$\mu=\cfrac{2+4+9+10+5}{5}=6.$

Population variance:

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

$X-\mu=\begin{Bmatrix} 2-6,4-6,9-6,10-6,5-6 \end{Bmatrix}=$

$=\begin{Bmatrix} -4, - 2,3,4,-1 \end{Bmatrix},$

$\sigma^2=(-4)^2\cdot \cfrac{1}{5}+(-2)^2\cdot \cfrac{1}{5}+\\ +3^2\cdot \cfrac{1}{5}+4^2\cdot \cfrac{1}{5}+(-1)^2\cdot \cfrac{1}{5}=9.2.$

Mean of the sampling distribution of sample means:

$\mu_{\bar x} =\mu=6.$

Variance of the sampling distribution of sample means:

$\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{9.2}{3}=3.067.$