Question

Consider a population consisting of 1,2,3,4 and 5. Supposed samples of size 3 are drawn from this population.Find the sampling distribution of the sample means.

Answer 1

We have population values 1,2,3,4,5, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{1+2+3+4+5}{5}=3$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{4+1+0+1+4}{5}=2$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{2}\approx1.4142$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_3=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,2,3 & 6/3 \\ \hdashline 2 & 1,2,4 & 7/3 \\ \hdashline 3 & 1,2,5 & 8/3\\ \hdashline 4 & 1,3,4 & 8/3 \\ \hdashline 5 & 1,3,5 & 9/3 \\ \hdashline 6 & 1,4,5 & 10/3 \\ \hdashline 7 & 2,3,4 & 9/3 \\ \hdashline 8 & 2,3,5 & 10/3 \\ \hdashline 9 & 2,4,5 & 11/3 \\ \hdashline 10 & 3,4,5 & 12/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 6/3 & 1/10 & 6/30 & 36/90 \\ \hdashline 7/3 & 1/10 & 7/30 & 49/90 \\ \hdashline 8/3 & 2/10 & 16/30 & 128/90 \\ \hdashline 9/3 & 2/10 & 18/30 & 162/90 \\ \hdashline 10/3 & 2/10 & 20/30 & 200/90 \\ \hdashline 11/3 & 1/10 & 11/30 & 121/90 \\ \hdashline 12/3 & 1/10 & 12/30 & 144/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=3=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{840}{90}-(3)^2=\dfrac{1}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{1}{3}}\approx0.57735$$