Answer to Question #313081 in Statistics and Probability for Donniee

Question

A population consists of the four numbers {2, 3, 6, 9}. Consider all possible samples of size 3 can be drawn without replacement from this population. Find the following:

a. mean of the population

b. population variance 

c. standard deviation of the population

d. mean of the sampling distribution of means

e. the standard deviation of the sampling distribution of means

f. List down all the possible samples of size 3.

g. construct a sampling distribution

Answer 1

$a:\\\mu =\frac{2+3+6+9}{4}=5\\b:\\\sigma ^2=\frac{\sum{\left( x_i-\mu \right) ^2}}{n}=\frac{\left( 2-5 \right) ^2+\left( 3-5 \right) ^2+\left( 6-5 \right) ^2+\left( 9-5 \right) ^2}{4}=7.5\\c:\\\sigma =\sqrt{\sigma ^2}=\sqrt{7.5}=2.73861\\d:\\\bar{x}=\mu =55\\e:\\s=\frac{\sigma}{\sqrt{3}}=1.58114\\f:\\\left( 2,3,6 \right) ,\bar{x}=\frac{11}{3}\\\left( 2,3,9 \right) ,\bar{x}=\frac{14}{3}\\\left( 2,6,9 \right) ,\bar{x}=\frac{17}{3}\\\left( 3,6,9 \right) ,\bar{x}=6\\g:\\P\left( \bar{x}=\frac{11}{3} \right) =P\left( \bar{x}=\frac{14}{3} \right) =P\left( \bar{x}=\frac{17}{3} \right) =P\left( \bar{x}=6 \right) =\frac{1}{4}\\$