Question

Suppose a population consists of the number 97,89,98,93,95,97,97,90,60,94 construct a sampling distribution of all possible samples of size 3 from the population.

Answer 1

We have population values 97,89,98,93,95,97,97,90,60,94, population size N=10 and sample size n=3.

Mean of population 

$$(\mu)=\dfrac{1}{10}(97+89+98+93+95+97$$

 

$$+97+90+60+94)=91$$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{1}{10}(36+4+49+4$$

$$+16+36+36+1+961+9)=115.2$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{115.2}\approx10.7331$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{10}C_3=120.$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\mu=91$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$=\dfrac{115.2}{3}(\dfrac{10-3}{10-1})=\dfrac{89.6}{3}$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{89.6}{3}}\approx5.4650$$