Question

Using the number 5, 10, 15, 20, 25, and 30 as the element of the population, construct the sampling distribution of the sample means with a sample size of 2. Find the population and sample mean, population and sample variance, and population and sample standard deviation.

Answer 1

We have population values 5,10,15,20,25,30, population size N=6 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{5+10+15+20+25+30}{6}=17.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{6}(156.25+56.25+6.25$$ $$+6.25+56.25+156.25)=\dfrac{437.5}{6}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{437.5}{6}}\approx8.5391$$

Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_2=15.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 5,10 & 15/2 \\ \hdashline 2 & 5,15 & 20/2 \\ \hdashline 3 & 5,20 & 25/2 \\ \hdashline 4 & 5,25 & 30/2 \\ \hdashline 5 & 5,30 & 35/2 \\ \hdashline 6 & 10,15 & 25/2 \\ \hdashline 7 & 10,20 & 30/2 \\ \hdashline 8 & 10,25 & 35/2 \\ \hdashline 9 & 10,30 & 40/2 \\ \hdashline 10 & 15,20 & 35/2 \\ \hdashline 11 & 15,25 & 40/2 \\ \hdashline 12 & 15,30 & 45/2 \\ \hdashline 13 & 20,25 & 45/2 \\ \hdashline 14 & 20,30 & 50/2 \\ \hdashline 15 & 25,30 & 55/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 15/2 & 1/15 & 3/6 & 45/12 \\ \hdashline 20/2 & 1/15 & 4/6 & 80/12 \\ \hdashline 25/2 & 2/15 & 10/6 & 250/12 \\ \hdashline 30/2 & 2/15 & 12/6 & 360/12 \\ \hdashline 35/2 & 3/15 & 21/6 & 735/12 \\ \hdashline 40/2 & 2/15 & 16/6 & 640/12 \\ \hdashline 45/2 & 2/15 & 18/6 & 810/12 \\ \hdashline 50/2 & 1/15 & 10/6 & 500/12 \\ \hdashline 55/2 & 1/15 & 11/6 & 605/12 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{105}{6}=17.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{4025}{12}-(17.5)^2=\dfrac{175}{6}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{175}{6}}\approx5.4006$$