Answer to Question #283154 in Statistics and Probability for Zaryab Khan

Question

Draw all possible sample of size 2 from the set A={2,3,4,6,8} with replacement and find their mean values. Now construct a frequency distribution of these mean values. then find the mean of this frequency distribution. then finally verify that this mean is equal to the mean of the set A.

Answer 1

When drawn with replacement, the total number of combinations that can be obtained is $n^r,$ where $r$ is the number of elements chosen and $n$ the number of elements to Choose from.

We have $5^2=25$ combinations.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample& Sample\ mean, \bar{x} \\ \hline 2,2 & 2 \\ 2,3 & 2.5 \\ 2,4 & 3 \\ 2,6 & 4 \\ 2,8 & 5 \\ 3,2 & 2.5 \\ 3,3 & 3 \\ 3,4 & 3.5 \\ 3,6 & 4.5 \\ 3,8 & 5.5 \\ 4,2 & 3 \\ 4,3 & 3.5 \\ 4,4 & 4 \\ 4,6 & 5 \\ 4,8 & 6 \\ 6,2 & 4 \\ 6,3 & 4.5 \\ 6,4 & 5 \\ 6,6 & 6 \\ 6,8 & 7 \\ 8,2 & 5 \\ 8,3 & 5.5 \\ 8,4 & 6 \\ 8,6 & 7 \\ 8,8 & 8 \\ \end{array}$$

Make a probability distribution of the sample means

$$\def\arraystretch{1.5} \begin{array}{c:c:c} \bar{x} & f & probability \\ 2 & 1 & 1/25=0.04 \\ 2.5 & 2 & 2/25=0.08 \\ 3 & 3 & 3/25=0.12 \\ 3.5 & 2 & 2/25=0.08 \\ 4 & 3 & 3/25=0.12 \\ 4.5 & 2 & 2/25=0.08 \\ 5 & 4 & 4/25=0.16 \\ 5.5 & 2 & 2/25=0.08 \\ 6 & 3 & 3/25=0.12 \\ 7 & 2 & 2/25=0.08 \\ 8 & 1 & 1/25=0.04 \\ \end{array}$$

$$\mu_{\bar{x}}=2(0.04)+2.5(0.08)+3(0.12)+3.5(0.08)$$

$$+4(0.12)+4.5(0.08)+5(0.16)+5.5(0.08)$$

$$+6(0.12)+7(0.08)+8(0.04)=4.6$$

$$\mu=\dfrac{2+3+4+6+8}{5}=4.6$$

The mean of the frequency distribution of mean values is equal to the mean of the set A

$$\mu_{\bar{x}}=4.6=\mu$$