Answer to Question #251371 in Statistics and Probability for Alaric

Question

Random variable of size n=2 are drawn from a population consisting the numbers 8,10, 12,14,and 16. Construct a sampling distribution of the sample mean to answer the questions.

Answer 1

I.

$$mean=\mu=\dfrac{8+10+12+14+16}{5}=12$$

$$variance=\sigma^2=\dfrac{1}{5}((8-12)^2+(10-12)^2$$

$$+(12-12)^2+(14-12)^2+(16-12)^2)=8$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{8}=2\sqrt{2}$$

II. There are $\dbinom{5}{2}=10$ samples of size two which can be drawn without replacement: 

$$\begin{matrix} Sample & Sample\ mean \\ (8,10) & 9 \\ (8,12) & 10 \\ (8,14) & 11 \\ (8,16) & 12 \\ (10,12) & 11 \\ (10,14) & 12 \\ (10,16) & 13\\ (12,14) & 13 \\ (12,16) & 14 \\ (14,16) & 15 \\ \end{matrix}$$

III.

$$\begin{matrix} \bar{X} & P(\bar{X}) \\ 9 & 0.1 \\ 10 & 0.1 \\ 11 & 0.2 \\ 12 & 0.2 \\ 13 & 0.2 \\ 14 & 0.1\\ 15 & 0.1\\ \end{matrix}$$

IV.

$$\mu_{\bar{X}}=9(0.1)+10(0.1)+11(0.2)+12(0.2)$$

$$+13(0.2)+14(0.1)+15(0.1)=12$$

$$\sum_i\bar{X}_i^2P(\bar{X_i})=9^2(0.1)+10^2(0.1)+11^2(0.2)$$

$$+12^2(0.2)+13^2(0.2)+14^2(0.1)+15^2(0.1)=147$$

$$\sigma_{\bar{X}}^2=\sum_i\bar{X}_i^2P(\bar{X_i})-\mu_{\bar{X}}^2=147-12^2=3$$

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{3}$$

$$\mu_{\bar{X}}=12, \sigma_{\bar{X}}=\sqrt{3}$$

V.

The mean $\mu_{\bar{X}}$ and standard deviation $\sigma_{\bar{X}}$ of the sample mean $\bar{X}$ satisfy

$$\mu_{\bar{X}}=12=\mu,$$

$$\sigma_{\bar{X}}=\sqrt{3}=\dfrac{2\sqrt{2}}{\sqrt{2}}\sqrt{\dfrac{5-2}{5-1}}=\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}$$