Answer to Question #331774 in Statistics and Probability for Aina

Question

A group of students got the following scores in a test: 6, 11, 22, 17, 18, and 21. Consider samples of size 3 that can be drawn from this population.

List all the possible samples and the corresponding mean.

Construct the sampling distribution of the sample means.

Answer 1

m(6,11,22)=(6+11+22)/3=13

m(6,11,17)=(6+11+17)/3=11.3

m(6,11,18)=(6+11+18)/3=11.6

m(6,11,21)=(6+11+21)/3=12.7

m(6,22,17)=(6+22+17)/3=15

m(6,22,18)=(6+22+18)/3=15.3

m(6,22,21)=(6+22+21)/3=16.3

m(6,17,18)=(6+17+18)/3=13.7

m(6,17,21)=(6+17+21)/3=14 7

m(6,18,21)=(6+18+21)/3=15

m(11,22,17)=(11+22+17)/3=16.7

m(11,22,18)=(11+22+18)/3=17

m(11,22,21)=(11+22+21)/3=18

m(11,17,18)=(11+17+18)/3=15.3

m(11,17,21)=(11+17+21)/3=16.3

m(11,18,21)=(11+18+21)/3=16.7

m(22,17,18)=(22+17+18)/3=19

m(22,17,21)=(22+17+21)/3=20

m(22,18,21)=(22+18+21)/3=20.3

m(17,18,22)=(17+18+21)/3=18.7

Frequency

F(13)=F(11.3)=F(11.6)=F(12.7)=F(13.7)=F(14.7)=F(19)=F(20)=F(20.3)=F(18.7)=F(17)=F(18)=1

F(15)=F(15.3)=F(16.7)=F(16.3)=2

Probabilitys $P(x)=F(x)/\sum F(x)$

P(13)=P(11.3)=P(11.6)=P(12.7)=P(13.7)=P(14.7)=P(19)=P(20)=P(20.3)=P(18.7)=P(17)=FlP(18)=1/20

P(15)=P(15.3)=P(16.7)=P(16.3)=2/20=1/10