Answer to Question #334523 in Statistics and Probability for JAYJAY

Question

A population consist of numbers 2 5 6 8 and 9, construct the sampling distribution of sample mean with sample size of 3. make a histogram of the distribution.

Answer 1

We have population values 2,5,6,8,9, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{2+5+6+8+9}{5}=6$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{16+1+0+4+9}{5}=6$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{6}$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,5,6 & 13/3 \\ \hdashline 2 & 2,5,8 & 15/3 \\ \hdashline 3 & 2,5,9 & 16/3 \\ \hdashline 4 & 2,6,8 & 16/3 \\ \hdashline 5 & 2,6,9 & 17/3 \\ \hdashline 6 & 2,8,9 & 19/3 \\ \hdashline 7 & 5,6,8 & 19/3 \\ \hdashline 8 & 5,6,9 & 20/3 \\ \hdashline 9 & 5,8,9 & 22/3 \\ \hdashline 10 & 6,8,9 & 23/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 13/3 & 1/10 & 13/30 & 169/90 \\ \hdashline 15/3 & 1/10 & 15/30 & 225/90 \\ \hdashline 16/3 & 2/10 & 32/30 & 512/90 \\ \hdashline 17/3 & 1/10 & 17/30 & 289/90 \\ \hdashline 19/3 & 2/10 & 38/30 & 722/90 \\ \hdashline 20/3 & 1/10 & 20/30 & 400/90 \\ \hdashline 22/3 & 1/10 & 22/30 & 484/90 \\ \hdashline 23/3 & 1/10 & 23/30 & 529/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=37-(6)^2=1= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{1}=1$$