Answer to Question #128719 in Statistics and Probability for maham

Question

Given the six-element population 8,12,15,19,21, 25, 32, and 36. How many samples of size 2 can be
drawn, without replacement from this population? Compute the sampling distribution of the mean for
samples of size 2. Compute the mean and standard deviation of this distribution. Also verify the results.

Answer 1

N - number of samples that can be drawn, N=C_8,2=28

(8,12), mean=10

(8,15), mean=11.5

(8,19), mean=13.5

(8,21), mean=14.5

(8,25), mean=16.5

(8,32), mean=20

(8,36), mean=22

(12,15), mean=13.5

(12,19), mean=15.5

(12,21), mean=16.5

(12,25), mean=18.5

(12,32), mean=22

(12,36), mean=24

(15,19), mean=17

(15,21), mean=18

(15,25), mean=20

(15,32), mean=23.5

(15,36), mean=25.5

(19,21), mean=20

(19,25), mean=22

(19,32), mean=25.5

(19,36), mean=27.5

(21,25), mean=23

(21,32), mean=26.5

(21,36), mean=28.5

(25,32), mean=28.5

(25,36), mean=30.5

(32,36), mean=34

mean of distribution=(10+11.5+...+34)/28=588/28=21

variance=((10-21)²+...+(34-21)²)/(28-1)=978/27=36.2

standard deviation=$\sqrt{variance}=6.018$