A group of students got the following scores in a test:9,12,15,18,21 and 24. Consider samples of size 3thag can be drawn from this population. List all the possible samples and the corresponding determine and list all possible samples and the corresponding sample means

The number of possible samples of size 3 without replacement can be calculated as $\dbinom{6}{3}=\dfrac{6!}{3!(6-3)!}=20.$

$\mu=\dfrac{9+12+15+18+21+24}{6}=16.5$

$\sigma^2=\dfrac{1}{6}((9-16.5)^2+(12-16.5)^2+(12-16.5)^2$

$+(18-16.5)^2+(21-16.5)^2+(24-16.5)^2)=\dfrac{157.5}{6}$

$\def\arraystretch{1.5} \begin{array}{c:c:c} & Sample & Mean \\ \hline & 9,12,15 & 12\\ \hdashline & 9,12,18 & 13\\ \hdashline & 9,12,21 & 14\\ \hdashline & 9,12,24 & 15\\ \hdashline & 9,15,18 & 14\\ \hdashline & 9,15,21 & 15\\ \hdashline & 9,15,24 & 16\\ \hdashline & 9,18, 21 & 16\\ \hdashline & 9,18,24 & 17\\ \hdashline & 9,21,24 & 18\\ \hdashline & 12,15,18 & 15\\ \hdashline & 12,15,21 & 16\\ \hdashline & 12,15,24 & 17\\ \hdashline & 12,18, 21 & 17\\ \hdashline & 12,18, 24 & 18\\ \hdashline & 12,21, 24 & 19\\ \hdashline & 15,18, 21 & 18\\ \hdashline & 15,18, 24 & 19\\ \hdashline & 15,21, 24 & 20\\ \hdashline & 18,21, 24 & 21\\ \hdashline \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Mean, \bar{x_i} & f_i & p(\bar{x}_i) \\ \hline & 12 & 1 & 1/20 \\ \hdashline & 13 & 1 & 1/20 \\ \hdashline & 14 & 2 & 2/20 \\ \hdashline & 15 & 3 & 3/20 \\ \hdashline & 16 & 3 & 3/20 \\ \hdashline & 17 & 3 & 3/20 \\ \hdashline & 18 & 3 & 3/20 \\ \hdashline & 19 & 2 & 2/20 \\ \hdashline & 20 & 1 & 1/20 \\ \hdashline & 21 & 1 & 1/20 \\ \hdashline \end{array}$

Check

$\mu_{\bar{X}}=12(1/20)+13(1/20)+14(2/20)+15(3/20)$

$+16(3/20)+17(3/20)+18(3/20)+19(2/20)$

$+20(1/20)+21(1/20)=330/20=16.5=\mu$