Answer to Question #342485 in Statistics and Probability for Fema Shien

Question

Solve the given problems

1. A population consists of the four numbers 1. 2. 4 and 5. List all the possible

samples of size n = 3 which can be drawn with replacement from the population

Find the following

a Population mean

b. Population variance

c. Population standard deviation

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f Standard deviation of the sampling distribution of sample means

Answer 1

We have population values $1,2,4,5$ population size $N=4$ and sample size $n=3.$

Thus, the number of possible samples which can be drawn with replacement is $4^3=64.$

a. Population mean:

$$\mu=\dfrac{1+2+4+5}{4}=3$$

b. Population variance:

$$\sigma^2=\dfrac{1}{4}((1-3)^2+(2-3)^2+(4-3)^2+(5-3)^2)$$ $$=2.5$$

c. Population standard deviation:

$$\sigma=\sqrt{\sigma^2}=\sqrt{2.5}\approx1.58114$$

d.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean(\bar{X}) \\ \hline 1,1,1 & 1\\ \hdashline 1,1,2 & 4/3\\ \hdashline 1,1,4 & 2\\ \hdashline 1,1,5 & 7/3\\ \hdashline 1,2,1 & 4/3\\ \hdashline 1,2,2 & 5/3\\ \hdashline 1,2,4 & 7/3\\ \hdashline 1,2,5 & 8/3\\ \hdashline 1,4,1 & 2\\ \hdashline 1,4,2 & 7/3\\ \hdashline 1,4,4 & 3\\ \hdashline 1,4,5 & 10/3\\ \hdashline 1,5,1 & 7/3\\ \hdashline 1,5,2 & 8/3\\ \hdashline 1,5,4 & 10/3\\ \hdashline 1,5,5 & 11/3\\ \hdashline 2,1,1 & 4/3\\ \hdashline 2,1,2 & 5/3\\ \hdashline 2,1,4 & 7/3\\ \hdashline 2,1,5 & 8/3\\ \hdashline 2,2,1 & 5/3\\ \hdashline 2,2,2 & 2\\ \hdashline 2,2,4& 8/3\\ \hdashline 2,2,5 & 3\\ \hdashline 2,4,1 & 7/3\\ \hdashline 2,4,2 & 8/3\\ \hdashline 2,4,4 & 10/3\\ \hdashline 2,4,5 & 11/3\\ \hdashline 2,5,1 & 8/3\\ \hdashline 2,5,2 & 3\\ \hdashline 2,5,4 & 11/3\\ \hdashline 2,5,5 & 4\\ \hdashline 4,1,1 & 2\\ \hdashline 4,1,2 & 7/3\\ \hdashline 4,1,4 & 3\\ \hdashline 4,1,5 & 10/3\\ \hdashline 4,2,1 & 7/3\\ \hdashline 4,2,2 & 8/3\\ \hdashline 4,2,4 & 10/3\\ \hdashline 4,2,5 & 11/3\\ \hdashline 4,4,1 & 3\\ \hdashline 4,4,2 & 10/3\\ \hdashline 4,4,4 & 4\\ \hdashline 4,4,5 & 13/3\\ \hdashline 4,5,1 & 10/3\\ \hdashline 4,5,2 & 11/3\\ \hdashline 4,5,4 & 13/3\\ \hdashline 4,5,5 & 14/3\\ \hdashline 5,1,1 & 7/3\\ \hdashline 5,1,2 & 8/3\\ \hdashline 5,1,4 & 10/3\\ \hdashline 5,1,5 & 11/3\\ \hdashline 5,2,1 & 8/3\\ \hdashline 5,2,2 & 3\\ \hdashline 5,2,4 & 11/3\\ \hdashline 5,2,5 &4\\ \hdashline 5,4,1 & 10/3\\ \hdashline 5,4,2 & 11/3\\ \hdashline 5,4,4 & 13/3\\ \hdashline 5,4,5 & 14/3\\ \hdashline 5,5,1 &11/3\\ \hdashline 5,5,2 & 4\\ \hdashline 5,5,4 & 14/3\\ \hdashline 5,5,5 & 5\\ \hdashline \end{array}$$

The sampling distribution of the sample mean $\bar{X}$ is

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline & 1 & 1 & 1/64 & 1/64 & 3/192\\ \hdashline & 4/3 & 3 & 3/64 & 4/64 & 16/192 \\ \hdashline & 5/3 & 3 & 3/64 & 5/64 & 25/192\\ \hdashline & 2 & 4 & 4/64 & 8/64 & 48/192 \\ \hdashline & 7/3 & 9 & 9/64 & 21/64 & 147/192 \\ \hdashline & 8/3 & 9 & 9/64 & 24/64 & 192/192 \\ \hdashline & 3 & 6 & 6/64 & 18/64 & 162/192 \\ \hdashline & 10/3 & 9 & 9/64 & 30/64 & 300/192 \\ \hdashline & 11/3 & 9 & 9/64 & 33/64 & 363/192 \\ \hdashline & 4 & 4 & 4/64 & 16/64 & 192/192 \\ \hdashline & 13/3 & 3 & 3/64 & 13/64 & 169/192 \\ \hdashline & 14/3 & 3 & 3/64 & 14/64 & 196/192 \\ \hdashline & 5 & 1 & 1/64 & 5/64 & 75/192 \\ \hdashline Sum= & & 64 & 1 & 3 & 59/6 \\ \hdashline \end{array}$$

e. The mean of the sample means is

$$\mu_{ \bar{X}}=E(\bar{X})=3$$

$$Var(\bar{X})=\sigma^2_{\bar{X}}=E(\bar{X}^2)-(E(\bar{X}))^2$$

$$=\dfrac{59}{6}-(3)^2=\dfrac{5}{6}$$ $$\mu_{\bar{X}}=\mu$$ $$\sigma_{\bar{X}}^2=\dfrac{\sigma^2}{n}$$

f. Standard deviation of the sampling distribution of sample means:

$$\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{5}{6}}\approx0.91287$$