Question

A population consists of the numbers 2, 4, 5, 9, and 10. A random sample of size 2 is

taken from the population.

1. Compute the number of samples using combination.

2. Complete the table on the right and compute the mean µ, variance

σ2 and standard deviation σ of the population.

3. Complete the table on the right and compute

the mean μx̄, variance σx̄2

, and standard

deviation σx̄ of the sampling distribution of

sample means.

Answer 1

We have population values 2,4,5,9,10, population size N=5 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{2+4+5+9+10}{5}=6$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{16+4+1+9+16}{5}=9.2$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{9.2}\approx3.03315$$

A. Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_2=10.$

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 6/2 & 1/10 & 6/20 & 36/40 \\ \hdashline 7/2 & 1/10 & 7/20 & 49/40 \\ \hdashline 9/2 & 1/10 & 9/20 & 81/40 \\ \hdashline 11/2 & 1/10 & 11/20 & 121/40 \\ \hdashline 12/2 & 1/10 & 12/20 & 144/40 \\ \hdashline 13/2 & 1/10 & 13/20 & 169/40 \\ \hdashline 14/2 & 2/10 & 28/20 & 392/40 \\ \hdashline 15/2 & 1/10 & 15/20 & 225/40 \\ \hdashline 19/2 & 1/10 & 19/20 & 361/40 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{120}{20}=6=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{1578}{40}-(6)^2=3.45= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{3.45}\approx1.8574$$