Answer to Question #300634 in Statistics and Probability for Jeyem

Question

Given the data 6, 9, 12, 15, 16, 19, 22. Construct a sampling distribution of the sample mean without replacement and repetition by selecting 2 samples at a time.

Answer 1

We have population values $6,9,12,15,16,19,22$ population size $N=7$  and sample size $n=2.$

Thus, the number of possible samples which can be drawn without replacements is $\dbinom{7}{2}=21.$

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean\ (\bar{x}) \\ \hline 6,9 & 7.5 \\ \hdashline 6,12 & 9 \\ \hdashline 6,15 & 10.5 \\ \hdashline 6,16 & 11 \\ \hdashline 6,19 & 12.5 \\ \hdashline 6,22 & 14 \\ \hdashline 9,12 & 10.5 \\ \hdashline 9,15 & 12 \\ \hdashline 9,16 & 12.5 \\ \hdashline 9,19 & 14 \\ \hdashline 9,22 & 15.5 \\ \hdashline 12,15 & 13.5 \\ \hdashline 12,16 & 14 \\ \hdashline 12,19 & 15.5 \\ \hdashline 12,22 & 17 \\ \hdashline 15,16 & 15.5 \\ \hdashline 15,19 & 17 \\ \hdashline 15,22 & 18.5 \\ \hdashline 16,19 & 17.5 \\ \hdashline 16,22 & 19 \\ \hdashline 19,22 & 20.5 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} & \bar{x} & f & f(\bar{x}) & \bar{x}f(\bar{x})\\ \hline & 7.5 & 1 & 1/21 & 5/14 \\ \hdashline & 9 & 1 & 1/21 & 3/7 \\ \hdashline & 10.5 & 2 & 2/21 & 1 \\ \hdashline & 11 & 1 & 1/21 & 11/21 \\ \hdashline & 12 & 1 & 1/21 & 4/7 \\ \hdashline & 12.5 & 2 & 2/21 & 25/21 \\ \hdashline & 13.5 & 1 & 1/21 & 9/14 \\ \hdashline & 14 & 3 & 1/7 & 2 \\ \hdashline & 15.5 & 3 & 1/7 & 31/14 \\ \hdashline & 17 & 2 & 2/21 & 34/21 \\ \hdashline & 17.5 & 1 & 1/21 & 5/6 \\ \hdashline & 18.5 & 1 & 1/21 & 37/42 \\ \hdashline & 19 & 1 & 1/21 & 19/21 \\ \hdashline & 20.5 & 1 & 1/21 & 41/42 \\ \hdashline Sum= & & 21 & 1 & 99/7 \\ \hdashline \end{array}$$

$$\mu_{\bar{x}}=E(\bar{X})=99/7$$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & p(x)\\ \hline 7.5 & 1/21\\ \hdashline 9 & 1/21 \\ \hdashline 10.5 & 2/21 \\ \hdashline 11 & 1/21 \\ \hdashline 12 & 1/21 \\ \hdashline 12.5 & 2/21 \\ \hdashline 13.5 & 1/21\\ \hdashline 14 & 1/7\\ \hdashline 15.5 & 1/7 \\ \hdashline 17& 2/21 \\ \hdashline 17.5 & 1/21 \\ \hdashline 18.5& 1/21 \\ \hdashline 19& 1/21 \\ \hdashline 20.5& 1/21 \\ \end{array}$$