Answer to Question #348515 in Statistics and Probability for bbv

Question

A population consists of the numbers 5, 20, 9, 4, and 2 with sample size of 3

Answer 1

We have population values 2,4,5,9,20, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{2+4+5+9+20}{5}=8$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{36+16+9+1+144}{5}=41.2$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{41.2}\approx6.4187$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_3=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 2,4,5 & 11/3 \\ \hdashline 2 & 2,4,9 & 15/3 \\ \hdashline 3 & 2,4,20 & 26/3\\ \hdashline 4 & 2,5,9 & 16/3 \\ \hdashline 5 & 2,5,20 & 27/3 \\ \hdashline 6 & 2,9,20 & 31/3 \\ \hdashline 7 & 4,5,9 & 18/3 \\ \hdashline 8 & 4,5,20 & 29/3 \\ \hdashline 9 & 4,9,20 & 33/3 \\ \hdashline 10 & 5,9,20 & 34/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 11/3 & 1/10 & 11/30 & 121/90 \\ \hdashline 15/3 & 1/10 & 15/30 & 225/90 \\ \hdashline 16/3 & 1/10 & 16/30 & 256/90 \\ \hdashline 18/3 & 1/10 & 18/30 & 324/90 \\ \hdashline 26/3 & 1/10 & 26/30 & 676/90 \\ \hdashline 27/3 & 1/10 & 27/30 & 729/90 \\ \hdashline 29/3 & 1/10 & 29/30 & 841/90 \\ \hdashline 31/3 & 1/10 & 31/30 & 961/90 \\ \hdashline 33/3 & 1/10 & 33/30 & 1089/90 \\ \hdashline 34/3 & 1/10 & 34/30 & 1156/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=8=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{6378}{90}-(8)^2=\dfrac{103}{15}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{103}{15}}\approx2.6204$$