Question

Given population 15, 12, 6, 9, and 17. Suppose samples of size 3 are drawn from this population. Complete the table and draw the sketch of histogram. Describe the distribution

Answer 1

1. We have population values 6,9,12,15,17, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{6+9+12+15+17}{5}=11.8$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{5}(33.64+7.84+0.04$$

$$+10.24+27.04)=15.76$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{15.76}\approx3.9699$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 6,9,12 & 27/3 \\ \hdashline 2 & 6,9,15 & 30/3 \\ \hdashline 3 & 6,9,17 & 32/3 \\ \hdashline 4 & 6,12,15 & 33/3 \\ \hdashline 5 & 6,12,17 & 35/3 \\ \hdashline 6 & 6,15,17 & 38/3 \\ \hdashline 7 & 9,12,15 & 36/3 \\ \hdashline 8 & 9,12,17& 38/3 \\ \hdashline 9 & 9,15,17 & 41/3 \\ \hdashline 10 & 12,15,17 & 44/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 27/3 & 1/10 & 27/30 & 729/90 \\ \hdashline 30/3 & 1/10& 30/30 & 900/90 \\ \hdashline 32/3 & 1/10 & 32/30 & 1024/90 \\ \hdashline 33/3 & 1/10 & 33/30 & 1089/90 \\ \hdashline 35/3 & 1/10 & 35/30 & 1225/90 \\ \hdashline 36/3 & 1/10 & 36/30 & 1296/90 \\ \hdashline 38/3 & 2/10 & 76/30 & 2888/90 \\ \hdashline 41/3 & 1/10 & 41/30 & 1681/90 \\ \hdashline 44/3 & 1/10 & 44/30 & 1936/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=11.8=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{12768}{90}-(11.8)^2=\dfrac{7.88}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{7.88}{3}}\approx1.6207$$