Question

Refer to Exercise 5.3 and find $\mathbf{E}\mathbf{=}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\mu }$. Then use the sampling distribution of$\overline{\mathbf{x}}$found in Exercise 5.3 to find the expected value of$\overline{\mathbf{x}}$. Note that$\mathbf{E}\mathbf{=}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathbf{\mu }$.

Short answer

$\mathbf{E}\mathbf{=}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{7}$

Step-by-step solution

Determination of the probabilities of the means

The list of the probabilities found in Exercise 5.3 is shown below.

Mean	Probability
1	0.04
1.5	0.12
2	0.17
2.5	0.20
3	0.20
3.5	0.14
4	0.08
4.5	0.04
5	0.01

Calculation of Ex

The calculation of $\mathit{E}\left(\overline{x}\right)$is shown below.

$$\begin{gathered} \mathit{E}\left(\overline{x}\right)\mathbf{=}\mathbf{\sum }_{\mathbf{}}\left[xp\left(x\right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=1\times 0.04+1.5\times 0.12+2\times 0.17+2.5\times 0.20+3\times 0.20+3.5\times 0.14+4\times 0.08+4.5\times 0.04+5\times 0.01 \\ =0.04+0.18+0.34+0.50+0.60+0.49+0.32+0.18+0.05 \\ =2.7 \end{gathered}$$

Here, x= variables, and p(x)=probabilities.

Therefore, the final answer is 2.7.