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We have population values $1,2,3,5,6, 7$ population size $N=6.$

1.

$mean=\mu=\dfrac{1+2+3+5+6+7}{6}=4$

2

$Variance=\sigma^2=\dfrac{1}{6}((1-4)^2+(2-4)^2$

$+(3-4)^2+(5-4)^2+(6-4)^2+(7-4)^2)=\dfrac{14}{3}$

$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{14}{3}}\approx2.160247$

3. The population size $N=6$ and sample size $n=4.$

Thus, the number of possible samples which can be drawn without replacements is $\dbinom{6}{4}=15.$

4.

$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean\ (\bar{x}) \\ \hline 1,2,3,5 & 2.75 \\ \hdashline 1,2,3,6 & 3 \\ \hdashline 1,2,3,7 & 3.25 \\ \hdashline 1,2,5,6 & 3.5 \\ \hdashline 1,2,5,7 & 3.75 \\ \hdashline 1,2,6,7 & 4 \\ \hdashline 1,3,5,6 & 3.75 \\ \hdashline 1,3,5,7 & 4 \\ \hdashline 1,3,6,7 & 4.25 \\ \hdashline 1,5, 6,7 & 4.75 \\ \hdashline 2,3,5,6 & 4 \\ \hdashline 2,3,5,7 & 4.25 \\ \hdashline 2,3,6,7 & 4.5 \\ \hdashline 2,5,6,7 & 5 \\ \hdashline 3,5,6,7 & 5.25 \\ \hdashline \end{array}$

5. The sampling distribution of the sample mean $\bar{x}$ and its mean and standard deviation are:

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{x} & f & f(\bar{x}) & \bar{x}f(\bar{x}) & \bar{x}^2f(\bar{x})\\ \hline & 2.75 & 1 & 1/15 & 11/60 & 121/240 \\ \hdashline & 3 & 1 & 1/15 & 12/60 & 144/240 \\ \hdashline & 3.25 & 1 & 1/15 & 13/60 & 169/240 \\ \hdashline & 3.5 & 1 & 1/15 & 14/60 & 196/240 \\ \hdashline & 3.75 & 2 & 2/15 & 30/60 & 450/240 \\ \hdashline & 4 & 3 & 3/15 & 48/60 & 768/240 \\ \hdashline & 4.25 & 2 & 2/15 & 34/60 & 578/240 \\ \hdashline & 4.5 & 1 & 1/15 & 18/60 & 324/240 \\ \hdashline & 4.75 & 1 & 1/15 & 19/60 & 361/240 \\ \hdashline & 5 & 1 & 1/15 & 20/60 & 400/240 \\ \hdashline & 5.25 & 1 & 1/15 & 21/60 & 441/240 \\ \hdashline Sum= & & 15 & 1 & 4 & 247/15 \\ \hdashline \end{array}$

6.

$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{x}f(\bar{x})=4$

7.

$Var(\bar{X})=\sigma_{\bar{X}}^2=\sum\bar{x}^2f(\bar{x})-(\sum\bar{x}f(\bar{x}))^2$

$=\dfrac{247}{15}-4^2=\dfrac{7}{15}$

$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{7}{15}}\approx0.683130$

Check

$\mu_{\bar{X}}=E(\bar{X})=4=\mu$

$Var(\bar{X})=\dfrac{7}{15}=\dfrac{\sigma^2}{n}\cdot\dfrac{N-n}{N-1}=\dfrac{14}{3(4)}\cdot\dfrac{6-4}{6-1}$