Question

samples of three cards are drawn at random from a population of six cards numbered from 1 to 6. Construct the sampling Distribution of Sample means.

Answer 1

We have population values 1,2,3,4,5 and 6 population size N=6 and sample size n=3.

Mean of population $(\mu)$ = 

$$\dfrac{1+2+3+4+5+6}{6}=3.5$$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{17.5}{6}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{17.5}{6}}\approx1.7078$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_3=20.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,2,3 & 6/3 \\ \hdashline 2 & 1,2,4 & 7/3 \\ \hdashline 3 & 1,2,5 & 8/3 \\ \hdashline 4 & 1,2,6 & 9/3 \\ \hdashline 5 & 1,3,4 & 8/3 \\ \hdashline 6 & 1,3,5 & 9/3 \\ \hdashline 7 & 1,3,6 & 10/3 \\ \hdashline 8 & 1,4,5 & 10/3 \\ \hdashline 9 & 1,4,6 & 11/3 \\ \hdashline 10 & 1,5,6 & 12/3 \\ \hdashline 11 & 2,3,4 & 9/3 \\ \hdashline 12 & 2,3,5 & 10/3 \\ \hdashline 13 & 2,3,6 & 11/3 \\ \hdashline 14 & 2,4,5 & 11/3 \\ \hdashline 15 & 2,4,6 & 12/3 \\ \hdashline 16 & 2,5,6 & 13/3 \\ \hdashline 17 & 3,4,5 & 12/3 \\ \hdashline 18 & 3,4,6 & 13/3 \\ \hdashline 19 & 3,5,6 & 14/3 \\ \hdashline 20 & 4,5,6 & 15/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 6/3 & 1/20 & 6/60 & 36/180\\ \hdashline 7/3 & 1/20 & 7/60 & 49/180\\ \hdashline 8/3 & 2/20 & 16/60 & 128/180\\ \hdashline 9/3 & 3/20 & 27/60 & 243/180 \\ \hdashline 10/3 & 3/20 & 30/60 & 300/180\\ \hdashline 11/3 & 3/20 & 33/60 & 363/180 \\ \hdashline 12/3 & 3/20 & 36/60 & 432/180 \\ \hdashline 13/3 & 2/20 & 26/60 & 338/180 \\ \hdashline 14/3 & 1/20 & 14/60 & 196/180 \\ \hdashline 15/3 & 1/20 & 15/60 & 225/180 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=3.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{2310}{180}-(3.5)^2=\dfrac{7}{12}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{7}{12}}\approx0.7638$$