Question

Find the mean of the set 7 10 14 17 20 of data and construct a sampling distribution by selecting 3 samples at a time

Answer 1

We have population values 7, 10, 14, 17, 20, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{7+10+14+17+20}{5}=13.6$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{109.2}{5}=21.84$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{21.84}\approx4.6733$$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 7,10,14 & 31/3 \\ \hdashline 2 & 7, 10, 17 & 34/3 \\ \hdashline 3 & 7,10,20 & 37/3\\ \hdashline 4 & 7,14,17 & 38/3 \\ \hdashline 5 & 7,14,20 & 41/3 \\ \hdashline 6 & 7,17,20 & 44/3 \\ \hdashline 7 & 10,14,17 & 41/3 \\ \hdashline 8 & 10,14,20 & 44/3 \\ \hdashline 9 & 10,17,20 & 47/3 \\ \hdashline 10 & 14,17,20 & 51/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 31/3 & 1/10 & 31/30 & 961/90 \\ \hdashline 34/3 & 1/10 & 34/30 & 1156/90 \\ \hdashline 37/3 & 1/10 & 37/30 & 1369/90 \\ \hdashline 38/3 & 1/10 & 38/30 & 1444/90 \\ \hdashline 41/3 & 2/10 & 82/30 & 3362/90 \\ \hdashline 44/3 & 2/10 & 88/30 & 3872/90 \\ \hdashline 47/3 & 1/10 & 47/30 & 2209/90 \\ \hdashline 51/3 & 1/10 & 51/30 & 2601/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=13.6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{16974}{90}-(13.6)^2=\dfrac{327.6}{90}=3.64= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{3.64}\approx1.9079$$