I.

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

III.

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

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