A population consists of the values (1,3,5,7) consider a sample size of two that can be drawn from this population

We have population values $1,3,5,7$ population size $N=4$ and sample size $n=2.$

Thus, the number of possible samples which can be drawn without replacement is $\dbinom{4}{2}=6.$

$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 1,3 & 2\\ \hdashline 1,5 & 3\\ \hdashline 1,7 & 4\\ \hdashline 3,5 & 4\\ \hdashline 3.7 & 5\\ \hdashline 5,7 & 6\\ \hdashline \end{array}$

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

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 2 & 1 & 1/6 & 2/6 & 4/6\\ \hdashline & 3 & 1 & 1/6 & 3/6 & 9/6 \\ \hdashline & 4 & 2 & 2/6 & 8/6 & 32/6\\ \hdashline & 5 & 1 & 1/6 & 5/6 & 25/6 \\ \hdashline & 6 & 1 & 1/16 & 6/6 & 36/6 \\ \hdashline Sum= & & 6 & 1 & 4 & 53/3\\ \hdashline \end{array}$

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

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

$=5$

Check

The mean of the sample means is

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

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

$=\dfrac{53}{3}-(4)^2=\dfrac{5}{3}$

The standard deviation of the sample means is

$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{5/3}$

$\dfrac{\sigma^2}{n}\cdot\dfrac{N-n}{N-1}=\dfrac{5}{3}\cdot \dfrac{4-2}{4-1}=\sigma^2_{\bar{X}}$

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

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