Question

A population N = 5 consists of the following values: 11, 13, 15, 17, and 19. Estimate the population mean (u) using a random variable of size n = 3

Answer 1

a. We have population values 11, 13, 15, 17, and 19, population size N=4 and sample size n=3.

Mean of population $(\mu)$ = 

$$\dfrac{11+13+15+17+19}{5}=15$$

The number of possible samples which can be drawn without replacement is $\dbinom{5}{3}=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 &11,13,15 & 39/3 \\ \hdashline 2 & 11,13,17 & 41/3 \\ \hdashline 3 & 11,13,19 & 43/3 \\ \hdashline 4 & 11,15,17 & 43/3 \\ \hdashline 5 & 11,15,19 & 45/3 \\ \hdashline 6 & 11,17,19 & 47/3 \\ \hdashline 7 & 13,15,17 & 45/3 \\ \hdashline 8 & 13,15,19 & 47/3 \\ \hdashline 9 & 13,17,19 & 49/3 \\ \hdashline 10 & 15,17,19 & 51/3 \\ \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})\\ \hline 39/3 & 1/10 & 39/30\\ \hdashline 41/3 & 1/10 & 41/30\\ \hdashline 43/3 & 2/10 & 86/30\\ \hdashline 45/3 & 2/10 & 90/30\\ \hdashline 47/3 & 2/10 & 94/30\\ \hdashline 49/3 & 1/10 & 49/30\\ \hdashline 51/3 & 1/10 & 51/30\\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{450}{30}=15=\mu$$