Answer to Question #341324 in Statistics and Probability for Killua

Question

1. A population consists of six values (6, 9, 12, 15, 18, and 21). (with illustration)

e. Determine the mean, variance and standard deviation of the sample mean.

2. The scores of individual students on a national test have a normal distribution with mean of 18.6 and a standard deviation of 5.9. At Federico Ramos Rural High School, 76 students took the test. If the scores at this school have the same distribution as national scores, solve for the following: (with illustration)

a. determine the mean and standard deviation of the sampling distribution of the sample mean.

Answer 1

1. We have population values 6,9,12,15,18,21, population size N=6 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{6+9+12+15+18+21}{6}=13.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{6}(56.25+20.25+2.25$$

$$+2.25+20.25+56.25)=26.25$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{26.25}\approx5.1235$$

Select a random sample of size 3 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_3=20.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 6,9,12 & 9 \\ \hdashline 2 & 6,9,15 & 10 \\ \hdashline 3 & 6,9,18 & 11 \\ \hdashline 4 & 6,9,21 & 12 \\ \hdashline 5 & 6,12,15 & 11 \\ \hdashline 6 & 6,12,18 & 12 \\ \hdashline 7 & 6,12, 21 & 13 \\ \hdashline 8 & 6,15,18 & 13 \\ \hdashline 9 & 6,15,21 & 14 \\ \hdashline 10 & 6, 18,21 & 15 \\ \hdashline 11 & 9,12,15 & 12 \\ \hdashline 12 & 9, 12,18 & 13 \\ \hdashline 13 & 9, 12, 21 & 14 \\ \hdashline 14 & 9,15,18 & 14 \\ \hdashline 15 & 9,15,21 & 15 \\ \hdashline 16 & 9, 18,21 & 16 \\ \hdashline 17 & 12, 15,18 & 15 \\ \hdashline 18 & 12, 15,21 & 16 \\ \hdashline 19 & 12, 18,21 & 17 \\ \hdashline 20 & 15, 18,21 & 18 \\ \hdashline \end{array}$$

B.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 9 & 1/20 & 9/20 & 81/20 \\ \hdashline 10 & 1/20 & 10/20 & 100/20 \\ \hdashline 11 & 2/20 & 22/20 & 242/20 \\ \hdashline 12 & 3/20 & 36/20 & 432/20 \\ \hdashline 13 & 3/20 & 39/20 & 507/20 \\ \hdashline 14 & 3/20 & 42/20 & 588/20 \\ \hdashline 15 & 3/20 & 45/20 & 675/20 \\ \hdashline 16 & 2/20 & 32/20 & 512/20 \\ \hdashline 17 & 1/20 & 17/20 & 289/20 \\ \hdashline 18 & 1/20 & 18/20 & 324/20 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{270}{20}=13.5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{3750}{20}-(13.5)^2=5.25= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{5.25}\approx2.2913$$

2.

$$\mu_{\bar{X}}=\mu=18.6$$

$$\sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{5.9}{\sqrt{76}}=0.6768$$