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## Here's the Solution to this Question

We have population values $3,5,7,9$ population size $N=4$ and sample size $n=3.$

Thus, the number of possible samples which can be drawn with replacement is $4^3=64.$

$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 3,3,3 & 3\\ \hdashline 3,3,5 & 11/3\\ \hdashline 3,3,7 & 13/3\\ \hdashline 3,3,9 & 5\\ \hdashline 3,5,3 & 11/3\\ \hdashline 3,5,5 & 13/3\\ \hdashline 3,5,7 & 5\\ \hdashline 3,5,9 & 17/3\\ \hdashline 3,7,3 & 13/3\\ \hdashline 3,7,5 & 5\\ \hdashline 3,7,7 & 17/3\\ \hdashline 3,7,9 & 19/3\\ \hdashline 3,9,3 & 5\\ \hdashline 3,9,5 & 17/3\\ \hdashline 3,9,7 & 19/3\\ \hdashline 3,9,9 & 7\\ \hdashline 5,3,3 & 11/3\\ \hdashline 5,3,5 & 13/3\\ \hdashline 5,3,7 & 5\\ \hdashline 5,3,9 & 17/3\\ \hdashline 5,5,3 & 13/3\\ \hdashline 5,5,5 & 5\\ \hdashline 5,5,7 & 17/3\\ \hdashline 5,5,9 & 19/3\\ \hdashline 5,7,3 & 5\\ \hdashline 5,7,5 & 17/3\\ \hdashline 5,7,7 & 19/3\\ \hdashline 5,7,9 & 7\\ \hdashline 5,9,3 & 17/3\\ \hdashline 5,9,5 & 19/3\\ \hdashline 5,9,7 & 7\\ \hdashline 5,9,9 & 23/3\\ \hdashline 7,3,3 & 13/3\\ \hdashline 7,3,5 & 5\\ \hdashline 7,3,7 & 17/3\\ \hdashline 7,3,9 & 19/3\\ \hdashline 7,5,3 & 5\\ \hdashline 7,5,5 & 17/3\\ \hdashline 7,5,7 & 19/3\\ \hdashline 7,5,9 & 7\\ \hdashline 7,7,3 & 17/3\\ \hdashline 7,7,5 & 19/3\\ \hdashline 7,7,7 & 7\\ \hdashline 7,7,9 & 23/3\\ \hdashline 7,9,3 & 19/3\\ \hdashline 7,9,5 & 7\\ \hdashline 7,9,7 & 23/3\\ \hdashline 7,9,9 & 25/3\\ \hdashline 9,3,3 & 5\\ \hdashline 9,3,5 & 17/3\\ \hdashline 9,3,7 & 19/3\\ \hdashline 9,3,9 & 7\\ \hdashline 9,5,3 & 17/3\\ \hdashline 9,5,5 & 19/3\\ \hdashline 9,5,7 & 7\\ \hdashline 9,5,9 & 23/3\\ \hdashline 9,7,3 & 19/3\\ \hdashline 9,7,5 & 7\\ \hdashline 9,7,7 & 23/3\\ \hdashline 9,7,9 & 25/3\\ \hdashline 9,9,3 & 7\\ \hdashline 9,9,5 & 23/3\\ \hdashline 9,9,7 & 25/3\\ \hdashline 9,9,9 & 9\\ \hdashline \end{array}$

2) The sampling distribution of the sample mean $\bar{X}$ is

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 3 & 1 & 1/64 & 3/64 & 27/192\\ \hdashline & 11/3 & 3 & 3/64 & 11/64 & 121/192 \\ \hdashline & 13/3 & 6 & 6/64 & 26/64 & 338/192\\ \hdashline & 5 & 10 & 10/64 & 50/64 & 750/192 \\ \hdashline & 17/3 & 12 & 12/64 & 68/64 & 1156/192 \\ \hdashline & 19/3 & 12 & 12/64 & 76/64 & 1444/192 \\ \hdashline & 7 & 10 & 10/64 & 70/64 & 1470/192 \\ \hdashline & 23/3 & 6 & 6/64 & 46/64 & 1058/192 \\ \hdashline & 25/3 & 3 & 3/64 & 25/64 & 625/192 \\ \hdashline & 9 & 1 & 1/16 & 9/64 & 243/192 \\ \hdashline Sum= & & 64 & 1 & 6 & 113/3\\ \hdashline \end{array}$

$\mu=\dfrac{3+5+7+9}{4}=6$

$\sigma^2=\dfrac{1}{4}((3-6)^2+(5-6)^2+(7-6)^2+(9-6)^2)$

$=5$

The mean of the sample means is

$\mu_{\bar{X}}=E(\bar{X})=6$

$Var(\bar{X})=\sigma^2_{\bar{X}}=E(\bar{X}^2)-(E(\bar{X}))^2$

$=\dfrac{113}{3}-(6)^2=\dfrac{5}{3}$

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

$\sigma_{\bar{X}}^2=\dfrac{\sigma^2}{n}$