Answer to Question #204182 in Statistics and Probability for ellea

Question

Consider a population of Senior High School consisting of the values 1, 2, 3,4, 5, and 6.

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

1.

We have population values $1,2,3,4,5,6$ population size $N=6.$ 

$$\mu=\dfrac{1+2+3+4+5+6}{6}=3.5$$

2.

$$\sigma^2=\dfrac{1}{N}\sum_i(x_i-\mu)^2$$

$$\sigma^2=\dfrac{1}{6}\big((1-3.5)^2+(2-2.5)^2+(3-3.5)^2+(4-3.5)^2$$

$$(5-3.5)^2+(6-3.5)^2\big)=\dfrac{17.5}{6}\approx2.916667$$

3.

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{17.5}{6}}\approx1.707825$$

4.

Let sample size $n=2.$Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{6}{2}=15$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,2 & 3/2 \\ \hdashline 2 & 1,3 & 4/2 \\ \hdashline 3 & 1,4 & 5/2 \\ \hdashline 4 & 1,5 & 6/2\\ \hdashline 5 & 1,6 & 7/2 \\ \hdashline 6 & 2,3 & 5/2 \\ \hline 7 & 2,4 & 6/2 \\ \hline 8 & 2,5 & 7/2 \\ \hline 9 & 2,6 & 8/2 \\ \hline 10 & 3,4 & 7/2 \\ \hline 11 & 3,5 & 8/2 \\ \hline 12 & 3,6 & 9/2 \\ \hline 13 & 4, 5 & 9/2 \\ \hline 14 & 4,6 & 10/2 \\ \hline 15 & 5,6 & 11/2 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 3/2 & 1& 1/15 & 3/30 & 9/60 \\ \hdashline 4/2 & 1 & 1/15 & 4/30 & 16/60 \\ \hdashline 5/2 & 2 & 2/15 & 10/30 & 50/90 \\ \hdashline 6/2 & 2 & 2/15 & 12/30 & 72/60 \\ \hdashline 7/2 & 3 & 3/10 & 21/30 & 147/60 \\ \hdashline 8/2 & 2 & 2/15 & 16/30 & 128/60 \\ \hdashline 9/2 & 2& 2/15 & 18/30 & 162/60 \\ \hdashline 10/2 & 1 & 1/15 & 10/30 & 100/60 \\ \hdashline 11/2 & 1 & 1/15 & 11/30 & 121/60 \\ \hdashline Total & 15 & 1 & 105/30 & 805/60 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{105}{30}=3.5$$

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

the mean of the population.

$$E(\bar{X})=3.5=\mu$$

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

$$=\dfrac{805}{60}-(\dfrac{105}{30})^2=\dfrac{7}{6}\approx1.666667$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{7}{6}}\approx1.080123$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{17.5}{6(2)}(\dfrac{6-2}{6-1})$$ $$=\dfrac{3.5}{3}\approx1.666667, True$$