Question

Consider all possible samples of size 3 that can be drawn without replacement from the population 1, 5, and 8, 6, 10. Compute the following:

1. Population mean

2. Population variance

3. Population standard deviation

4. Illustrate the probability histogram of the sampling distribution of the means

Answer 1

Given population = 1, 5, 8, 6, 10

Population mean $\mu =\dfrac{1+5+8+6+10}{5}=\dfrac{30}{5}=6$

Population variance

$\sigma^2= \sum \dfrac{ (x-\mu)^2}{n}=\dfrac{(1-6)^2+(5-6)^2+(8-6)^2+(6-6)^2+(10-6)^2}{5}=\dfrac{46}{5}=9.2$

Population standard deviation $\sigma=\sqrt{ Variance}=\sqrt{9.2}=3.033$

All possible samples and their means:

1. 1, 5, 8 with mean (1+5+8)/3=4.66
2. 1, 5, 6 with mean 4
3. 1, 5, 10 with mean 5.33
4. 1, 8, 6 with mean 5
5. 1, 8, 10 with mean 6.33
6. 1, 6, 10 with mean 5.66
7. 5, 8, 6 with mean 6.33
8. 5, 8, 10 with mean 7.66
9. 5, 6, 10 with mean 7
10. 8, 6, 10 with mean 8

Sampling distribution of the sample mean:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 4.66 & 1 & 0.1 & 0.466 & 2.17\\ \hdashline & 4 & 1 & 0.1 & 0.4 & 1.6 \\ \hdashline & 5.33 & 1 & 0.1 & 0.533 & 2.84\\ \hdashline & 5 & 1 & 0.1 & 0.5 & 0.25 \\ \hdashline & 6.33& 2 & 0.2 & 1.266 & 8.013\\ \hdashline & 5.66& 1 & 0.1 & 0.566 & 3.2 \\ \hdashline & 7.66 & 1& 0.1& 0.766 & 5.86 \\ \hdashline & 7 & 1 & 0.1& 0.7 & 4.9\\ \hdashline & 8 & 1 & 0.1 & 0.8 & 6.4\\ \hdashline Total= & & 10 & 1 & 6 & 35.23 \\ \hdashline \end{array}$$

Probability histogram of the sampling distribution of the means: