Answer to Question #247692 in Statistics and Probability for Waleed

Question

Suppose a small finite population consists of only N = 5 numbers:

54 55 59 63 64

I. Find population mean and standard deviation
II. Draw all possible samples of size 2 with replacement
III. Construct sampling distribution
IV. Find sample mean and standard deviation
V. Verify/justify your results.

Answer 1

I.

$$mean=\mu=\dfrac{54 +55 +59+ 63 +64}{5}=59$$

$$variance=\sigma^2=\dfrac{1}{5}((54-59)^2+(55-59)^2$$

$$+(59-59)^2+(63-59)^2+(64-59)^2)=16.4$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{16.4}=2\sqrt{4.1}$$

II. There are $5^2=25$ samples of size two which can be drawn with replacement: 

$$\begin{matrix} Sample & Sample\ mean \\ (54,54) & 54 \\ (54,55) & 54.5 \\ (54,59) & 56.5 \\ (54,63) & 58.5 \\ (54,64) & 59 \\ (55,54) & 54.5 \\ (55,55) & 55\\ (55,59) & 57 \\ (55,63) & 59 \\ (55,64) & 59.5 \\ (59,54) & 56.5 \\ (59,55) & 57 \\ (59,59) & 59 \\ (59,63) & 61 \\ (59,64) & 61.5 \\ (63,54) & 58.5 \\ (63,55) & 59\\ (63,59) & 61 \\ (63,63) & 63 \\ (63,64) & 63.5 \\ (64,54) & 59 \\ (64,55) & 59.5 \\ (64,59) & 61.5 \\ (64,63) & 63.5 \\ (64,64) & 64 \\ \end{matrix}$$

III.

$$\begin{matrix} \bar{X} & P(\bar{X}) \\ 54 & 1/25 \\ 54.5 & 2/25 \\ 55 & 1/25 \\ 56.5 & 2/25 \\ 57 & 2/25 \\ 58.5 & 2/25 \\ 59 & 5/25\\ 59.5 & 2/25 \\ 61 & 2/25 \\ 61.5 & 2/25 \\ 63 & 1/25 \\ 63.5 & 2/25 \\ 64 & 1/25 \\ \end{matrix}$$

IV.

$$\mu_{\bar{X}}=54(\dfrac{1}{25})+54.5(\dfrac{2}{25})+55(\dfrac{1}{25})+56.5(\dfrac{2}{25})$$

$$+57(\dfrac{2}{25})+58.5(\dfrac{2}{25})+59(\dfrac{5}{25})+59.5(\dfrac{2}{25})$$

$$+61(\dfrac{2}{25})+61.5(\dfrac{2}{25})+63(\dfrac{1}{25})+63.5(\dfrac{2}{25})$$

$$+64(\dfrac{1}{25})=59$$

$$\sum_i\bar{X}_i^2P(\bar{X_i})=54^2(\dfrac{1}{25})+54.5^2(\dfrac{2}{25})+55^2(\dfrac{1}{25})$$

$$+56.5^2(\dfrac{2}{25})+57^2(\dfrac{2}{25})+58.5^2(\dfrac{2}{25})+59^2(\dfrac{5}{25})$$

$$+59.5^2(\dfrac{2}{25})+61^2(\dfrac{2}{25})+61.5^2(\dfrac{2}{25})+63^2(\dfrac{1}{25})$$

$$+63.5^2(\dfrac{2}{25})+64^2(\dfrac{2}{25})=3489.2$$

$$\sigma_{\bar{X}}^2=\sum_i\bar{X}_i^2P(\bar{X_i})-\mu_{\bar{X}}^2=3489.2-59^2=8.2$$

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{8.2}$$

$$\mu_{\bar{X}}=59, \sigma_{\bar{X}}=\sqrt{8.2}$$

V.

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

$$\mu_{\bar{X}}=59=\mu, \sigma_{\bar{X}}=\sqrt{8.2}=\dfrac{2\sqrt{4.1}}{\sqrt{2}}=\dfrac{\sigma}{\sqrt{n}}$$