Answer to Question #256790 in Statistics and Probability for jay

Question

C. Solve. Given a population of the scores of BSHS honor students in Mathematics as 1, 3, 4, 6, and 8. Suppose a sample size of 3 have to be drawn from it for the first, second and third rank;

Answer 1

a.

We have population values $1,3,4,6,8,$ population size $N=5$

$$\mu=\dfrac{1+3+4+6+8}{5}=4.4$$

b.

$$\sigma^2=\dfrac{1}{5}((1-4.4)^2+(3-4.4)^2+(4-4.4)^2$$ $$+(6-4.4)^2)+(8-4.4)^2)=5.84$$

c.

$$\sigma=\sqrt{\sigma^2}=\sqrt{5.84}\approx2.4166$$

d. We have population values $1,3,4,6,8$ population size $N=5$ and sample size $n=3.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{3}=10$$

The sampling distribution of the sample means.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=4.4$$

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

$$E(\bar{X})=\mu_{\bar{X}}=4.4=\mu$$

e.

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

$$=\dfrac{61}{3}-(4.4)^2=\dfrac{2.92}{3}$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{2.92}{3}}\approx0.9866$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5.84}{3}(\dfrac{5-3}{5-1})$$

$$=\dfrac{2.92}{3}, True$$