Question

Consider a population with values (2,5,7,8)

a. Find the population mean.

b. Find the population variance.

c. Find the population standard deviation.

d. Find the possible samples of size 2 that can be drawn with replacement from this population.

Answer 1

a. Mean

$$\mu=\dfrac{2+5+7+8}{4}=5.5$$

b. Variance

$$\sigma^2=\dfrac{1}{4}\big((2-5.5)^2+(5-5.5)^2+(7-5.5)^2$$

$$+(8-5.5)^2\big)=\dfrac{21}{4}=5.25$$

c. Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{21}{4}}=\dfrac{\sqrt{21}}{2}\approx2.2913$$

d. We have population values $2,5,7,8,$ population size $N=4$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{4}{2}=6$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,5 & 3.5 \\ \hdashline 2 & 2,7 & 4.5 \\ \hdashline 3 & 2,8 & 5 \\ \hdashline 4 & 5,7 & 6 \\ \hdashline 5 & 5,8 & 6.5 \\ \hdashline 6 & 7,8 & 7.5 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 7/2 & 1& 1/6 & 7/12 & 49/24 \\ \hdashline 9/2 & 1 & 1/6 & 9/12 & 81/24 \\ \hdashline 5 & 1 & 1/6 & 10/12 & 100/24 \\ \hdashline 6 & 1 & 1/6 & 12/12 & 144/24 \\ \hdashline 13/2 & 1& 1/6 & 13/12 & 169/24 \\ \hdashline 15/2 & 1 & 1/6 & 15/12 & 225/24 \\ \hdashline Total & 6 & 1 & 66/12 & 768/24 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{66}{12}=5.5$$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$$E(\bar{X})=5.5=\mu$$

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$

$$=\dfrac{768}{24}-(\dfrac{66}{12})^2=\dfrac{42}{24}=\dfrac{7}{4}=1.75$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{7}{4}}=\dfrac{\sqrt{7}}{2}\approx1.3229$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{\dfrac{21}{4}}{2}(\dfrac{4-2}{4-1})$$

$$=\dfrac{7}{4}=1.75, True$$