Question

random sample of size 4 from a finite population of 3,6,9

Answer 1

a. the mean of the population

$$\mu=\dfrac{2+3+7+8+10}{5}=6$$

the standard deviation of the population

$$\sigma^2=\dfrac{(2-6)^2+(3-6)^2+(7-6)^2+(8-6)^2+(10-6)^2}{5}$$

$=9.2$

$\sigma=\sqrt{9.2}=3.033$

b. There are $\dbinom{5}{4}$ samples of size two which can be drawn without replacement: 

$$\dbinom{5}{4}=\dfrac{5!}{4!(5-4)!}=5$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample\ No. & Sample\ value & Sample\ mean\ \bar{X} \\ \hline 1 & (2,3,7,8) & 5 \\ \hdashline 2 & (2,3,7,10) & 5.5 \\ \hdashline 3 & (2,3,8,10) & 5.75 \\ \hdashline 4 & (2,7,8,10) & 6.75 \\ \hdashline 5 & (3,7,8,10) & 7 \end{array}$$

The sampling distribution of the sample mean $\bar{X}$ and its mean and standard deviation are:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 5 & 1 & 0.2 & 1 & 5 \\ \hdashline 5.5 & 1 & 0.2 & 1.1 & 6.05 \\ \hdashline 5.75 & 1 & 0.2 & 1.15 & 6.6125 \\ \hdashline 6.75 & 1 & 0.2 & 1.35 & 9.1125 \\ \hdashline 7 & 1 & 0.2 & 1.4 & 9.8 \\ \hdashline Total & 5 & 1 & 6 & 36.575 \end{array}$$ $$E(\bar{X})=\sum\bar{X}f(\bar{X})=6$$ $$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(E(\bar{X}))^2=36.575-6^2$$ $$=0.575$$

Verification:

$$\mu=6=E(\bar{X})$$ $$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=$$ $$=\dfrac{9.2}{4}(\dfrac{5-4}{5-1})=0.575$$