Answer to Question #347059 in Statistics and Probability for Yaraaaa

Question

A group of 8 students in your class having the weight of 50, 52, 53, 54, 56, 57 58, and 60.

Answer 1

We have population values 50, 52, 53, 54, 56, 57 58, 60, population size N=8 and sample size n=2.

Mean of population $(\mu)$ = 

$$\dfrac{1}{8}(50+52+53+54+56+57+58+60)=55$$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{8}(25+9+4+1$$ $$+1+4+9+25)=\dfrac{78}{8}=9.75$$

$$\sigma=\sqrt{9.75}\approx3.1225$$

Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{8}C_2=28.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 50,52 & 102/2 \\ \hdashline 2 & 50,53 & 103/2 \\ \hdashline 3 & 50,54 & 104/2 \\ \hdashline 4 & 50,56 & 106/2 \\ \hdashline 5 & 50,57 & 107/2 \\ \hdashline 6 & 50,58 & 108/2 \\ \hdashline 7 & 50,60 & 110/2 \\ \hdashline 8 & 52,53 & 105/2 \\ \hdashline 9 & 52,54 & 106/2 \\ \hdashline 10 & 52,56 & 108/2 \\ \hdashline 11 & 52,57 & 109/2 \\ \hdashline 12 & 52,58 & 110/2 \\ \hdashline 13 & 52,60 & 112/2 \\ \hdashline 14 & 53,54 & 107/2 \\ \hdashline 15 & 53,56 & 109/2 \\ \hdashline 16 & 53,57 & 110/2 \\ \hdashline 17 & 53,58 & 111/2 \\ \hdashline 18 & 53,60 & 113/2 \\ \hdashline 19 & 54,56 & 110/2 \\ \hdashline 20 & 54,57 & 111/2 \\ \hdashline 21 & 54,58 & 112/2 \\ \hdashline 22 & 54,60 & 114/2 \\ \hdashline 23 & 56,57 & 113/2 \\ \hdashline 24 & 56,58 & 114/2 \\ \hdashline 25 & 56,60 & 116/2 \\ \hdashline 26 & 57,58 & 115/2 \\ \hdashline 27 & 57,60 & 117/2 \\ \hdashline 28 & 58,60 & 118/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 102/2 & 1/28 & 102/56 & 10404/112 \\ \hdashline 103/2 & 1/28 & 103/56 & 10609/112 \\ \hdashline 104/2 & 1/28 & 104/56 & 10816/112 \\ \hdashline 105/2 & 1/28 & 105/56 & 11025/112 \\ \hdashline 106/2 & 2/28 & 212/56& 22472/112 \\ \hdashline 107/2 & 2/28 & 214/56 & 22898/112 \\ \hdashline 108/2 & 2/28 & 216/56 & 23328/112 \\ \hdashline 109/2 & 2/28 & 218/56 & 23762/112 \\ \hdashline 110/2 & 4/28 & 440/56 & 48400/112 \\ \hdashline 111/2 & 2/28 & 222/56 & 24642/112 \\ \hdashline 112/2 & 2/28 & 224/56 & 25088/112 \\ \hdashline 113/2 & 2/28 & 226/56 & 25538/112 \\ \hdashline 114/2 & 2/28 & 228/56 & 25992/112 \\ \hdashline 115/2 & 1/28 & 115/56 & 13225/112 \\ \hdashline 116/2 & 1/28 & 116/56 & 13456/112 \\ \hdashline 117/2 & 1/28 & 117/56 & 13689/112 \\ \hdashline 118/2 & 1/28 & 118/56 & 13924/112 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{3080}{56}=55=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{339268}{112}-(55)^2=\dfrac{117}{28}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{\dfrac{117}{28}}\approx2.044$$