Answer to Question #334010 in Statistics and Probability for Lhay

Question

Random sample of n=2 are drawn from a finite population consisting of the numbers 5,6,7,8,and 9

A.Find the mean population

B.find the standard deviation of the population

C. Find the mean of the sampling distribution of the sample means.

D.FIND The standard deviation of the sampling distribution of the sample means

E.Verify the central limit theorem

Answer 1

A. We have population values 5,6,7,8,9, population size N=5 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{5+6+7+8+9}{5}=7$

B. Variance of population 

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

$$=2$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{2}\approx1.4142$$

C. The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_2=10.$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 11/2 & 1/10 & 11/20 & 121/40 \\ \hdashline 12/2 & 1/10& 12/20 & 144/40 \\ \hdashline 13/2 & 2/10 & 26/20 & 338/40 \\ \hdashline 14/2 & 2/10 & 28/20 & 392/40 \\ \hdashline 15/2 & 2/10 & 30/20 & 450/40 \\ \hdashline 16/2 & 1/10 & 16/20 & 256/40 \\ \hdashline 17/2 & 1/10 & 17/20 & 89/40 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=7=\mu$$

D. The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1990}{40}-(7)^2=\dfrac{3}{4}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}\approx0.8660$$

E.

$$\mu_{\bar{X}}=E(\bar{X})=7=\mu$$

$$\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})= \dfrac{2}{2}(\dfrac{5-2}{5-1})=\dfrac{3}{4}=\sigma^2_{\bar{X}}$$