Answer to Question #306874 in Statistics and Probability for tin

Question

1. Given the data 6, 9, 12, 15, 16, 19, 22. Construct a sampling distribution of the sample mean without replacement and repetition by selecting 2 samples at a time.

Answer 1

Let X be the random variable representing the possible values of the sample means

Sample space Sample mean (X)

{6,9} (6+9)/2 = 7.5

{6,12} (6+12)/2 = 9

{6,15} (6+15)/2 = 10.5

{6,16} (6+16)/2 = 11

{6,19} (6+19)/2 = 12.5

{6,22} (6+22)/2 = 14

{9,12} (9+12)/2 = 10.5

{9,15} (9+15)/2 = 12

{9,16} (9+16)/2 = 12.5

{9,19} (9+19)/2 = 14

{9,22} (9+22)/2 = 15.5

{12,15} (12+15)/2 = 13.5

{12,16} (12+16)/2 = 14

{12,19} (12+19) = 15.5

{12,22} (12+22)/2 = 17

{15,16} (15+16)/2 = 15.5

{15,19} (15+19)/2 = 17

{15,22} (15+22)/2 = 18.5

{16,19} (16+19)/2 = 17.5

{16,22} (16+22)/2 = 19

{19,22} (19+22)/2 = 20.5

Therefore, the possible values for the sample means X and the respective probabilities is:

(X) P(X)

7.5 (1/21)

9 (1/21)

10.5 (2/21)

11 (1/21)

12 (1/21)

12.5 (2/21)

13.5 (1/21)

14 (3/21)

15.5 (3/21)

17 (2/21)

17.5 (1/21)

18.5 (1/21)

19 (1/21)

20.5 (1/21)