Answer to Question #324758 in Statistics and Probability for Seika

Question

A population consists of the numbers 3,5,9,10 and 6. Given the sample size of 3 Determine the following:

A. List all the possible samples

B. Compute the mean of each sample.

C. Construct the sampling distribution

D. Consryya histogram

E. Compute the mean of the sampling distribution of the sample means

F. Compute of the variance sampling distribution of the sample means

Answer 1

A. B.

m(3,5,9)=(3+5+9)/3=5.7

m(3,5,10)=)3+5+10)/3=6

m(3,5,6)=(3+5+6)/3=4.7

m(3,9,10)=(3+9+10)/3=7.3

m(3,9,6)=(3+9+6)/3=6

m(3,10,6)=(3+10+6)/3=6.3

m(5,9,10)=(5+9+10)/3=8

m(5,9,6)=(5+9+6)/3=6.7

m(5,10,6)=(5+10+6)/3=7

m(9,10,6)=(9+10+6)/3=8.3

C. Frequency

F(5.7)=F(4.7)=F(7.3)=F(6.3)=F(8)=F(6.7)=F(7)=F(8.3)=1

F(6)=2

Probability

$P(x)=Frequency/\sum Frequency$

P(5.7)=P(4.7)=P(7.3)=P(6.3)=P(8)=P(6.7)=P(7)=P(8.3)=1/10=0.1

P(6)=2/10=0.2

D.

E.

$E(x)=\sum fx=0.1(5.7+4.7+7.3+6.3+8+6.7+7+8.3)+0.2\times 6=6.6$

F.

$\sigma^2=\sum fx^2-(\sum fx)^2=0.1(32.49+22.09+53.29+39.69+64+44.89+48+68.89)+0.2\times36-43.56=0.974$