Answer to Question #268894 in Statistics and Probability for Shah

Question

A population is given by 5 8 10 15 16 20 draw all possible samples of size 2 without replacement. Show that E(x ̅ )=μ

Answer 1

We have population values $5, 8, 10, 15, 16, 20,$ population size $N=6$

$$\mu=\dfrac{5+8+10+15+16+20}{6}=37/3$$

We have population values $5, 8, 10, 15, 16, 20,$ population size $N=6$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{6}{2}=15$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 5,8 & 6.5 \\ \hdashline 2 & 5,10 & 7.5 \\ \hdashline 3 & 5,15 & 10 \\ \hdashline 4 & 5,16 & 10.5 \\ \hdashline 5 & 5,20 & 12.5 \\ \hdashline 6 & 8,10 & 9 \\ \hdashline 7 & 8,15 & 11.5 \\ \hdashline 8 & 8,16 & 12 \\ \hdashline 9 & 8,20 & 14 \\ \hdashline 10 & 10,15 & 12.5 \\ \hdashline 11 & 10,16 & 13 \\ \hdashline 12 & 10,20 & 15 \\ \hline \hdashline 13 & 15,16 & 15.5 \\ \hdashline 14 & 15,20 & 17.5 \\ \hdashline 15 & 16,20 & 18 \\ \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 6.5 & 1 & 1/15 & 6.5/15 & 42.25/15 \\ \hdashline & 7.5 & 1 & 1/15 & 7.5/15 & 56.25/15 \\ \hdashline & 9 &1 & 1/15 & 9/15 & 81/15 \\ \hdashline & 10 & 1 & 1/15 & 10/15 & 100/15 \\ \hdashline & 10.5 & 1 & 1/15 & 10.5/15 & 110.25/15 \\ \hdashline & 11.5 & 1 & 1/15 & 11.5/15 & 132.25/15 \\ \hdashline & 12 & 1 & 1/15 & 12/15 & 144/15 \\ \hdashline & 12.5 & 2 & 2/15 & 25/15 & 312.5/15 \\ \hdashline & 13 & 1 & 1/15 & 13/15 & 169/15 \\ \hdashline & 14 & 1 & 1/15 & 14/15 & 196/15 \\ \hdashline & 15 & 1 & 1/15 & 15/15 & 225/15 \\ \hdashline & 15.5 & 1 & 1/15 & 15.5/15 & 240.25/15 \\ \hdashline & 17.5 & 1 & 1/15 & 17.5/15 & 306.25/15 \\ \hdashline & 18 & 1 & 1/15 & 18/15 & 324/15 \\ \hdashline Total & & 15 & 1 & 37/3 & 814/5 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{37}{3}$$

The mean of the sampling distribution of the sample means is equal to the the mean of the population.

$$E(\bar{X})=\mu_{\bar{X}}=\dfrac{37}{3}=\mu$$