Answer to Question #189340 in Statistics and Probability for Joy

Question

Problem: Random samples of size N 2 are drawn from a finite

population consisting of the number 5, 6, 7, 8, and 9. Compute for the

Mean, Variance and Standard Deviation of the Population, and the Mean,

Variance and Standard Deviation of the Sample Means.

Answer 1

We have population values 5,6,7,8 and 9. ,population size N=5 and sample size n=2.

Thus, the number of possible samples which can be drawn without replacement is

$\binom{N}{n}=\binom{5}{2}= 10$

Sample : (5,6) , (5,7), (5,8) , (5,9) , (6,7) , (6,8) , (6,9) , (7,8) , (7,9) , (8,9)

Mean of each sample : 5.5 , 6 , 6.5 , 7 , 6.5 , 7 , 7.5 , 7.5 , 8 , 8.5

Sample Mean =$\dfrac{5.5+6+6.5+7+6.5+7+7.5+7.5+8+8.5}{10}=\dfrac{70}{10}=7$

Population Mean : $\bar x = \dfrac{\sum x}{n}=\dfrac{5+6+7+8+9}{5}=7$

To find population and sample variance and standard deviation

Following data are to be calculated:

Population Variance : $\sigma^2=\dfrac{\sum x^2-\frac{(\sum x)^2}{n}}{n}$

$=\dfrac{255-\frac{(35)^2}{5}}{5}\\=\dfrac{255-245}{5}\\=2$

Sample Variance: $S^2=\dfrac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} ​$

$=\dfrac{255-\frac{(35)^2}{5}}{4}\\\dfrac{255-245}{4}\\=2.5$

Population Standard Deviation : $\sigma=\sqrt{\dfrac{\sum x^2-\frac{(\sum x)^2}{n}}{n}} ​$

$\sigma=\sqrt{2}=1.414$

Sample Standard Deviation : $S=\sqrt{\dfrac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1}} ​ ​ ​ ​$

$= \sqrt{2.5}=1.5811$