Question

Question: Refer to Exercise 5.3.

1. Show that$\overline{\mathrm{x}}$is an unbiased estimator of$\mathrm{\mu }$.
2. Find${\mathrm{\sigma }}_{\mathrm{x}}^2$.
3. Find the x probability that x will fall within$2{\mathrm{\sigma }}_{\mathrm{x}}$of$\mathrm{\mu }$.

Short answer

a) Proved that $\overline{\mathrm{x}}$is an unbiased estimator of $\mathrm{\mu }$

b) 0.805

c) 0.95

Step-by-step solution

List of probabilities

The list of the probabilities found in Exercise 3 corresponding to the respective means 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 the mean μ

The calculation of the meanandis shown below:

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{x}}=\mathrm{xp}\left(\mathrm{x}\right) \\ =\left\{1(0.2)+2(0.3)+3(0.2)+4(0.2)+5(0.1)\right\} \\ =(0.2+0.6+0.6+0.8+0.5) \\ =2.7 \end{gathered}$$

$$\begin{gathered} \mathrm{E}\left(\overline{\mathrm{x}}\right)=\mathrm{E}\overline{\mathrm{x}}\mathrm{p}\left(\overline{\mathrm{x}}\right) \\ =\left\{\left(1\times 0.04\right)+\left(1.5\times 0.12\right)+\left(2\times 0.17\right)+\left(2.5\times 0.20\right)+\left(3\times 0.20\right)+\left(3.5\times 0.14\right) \\ +\left(4\times 0.08\right)+(4.5\times 0.04)+(5\times 0.01)\right\} \\ =\left(0.04+0.18+0.34+0.5+0.6+0.49+0.32+0.18+0.05\right) \\ =2.7 \end{gathered}$$

Therefore as the value of localid="1661429696077" role="math" $\mathbf{\mu }$and localid="1661429661874" $\mathbf{E}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}$are 2.7, localid="1661429676073" $\overline{\mathbf{x}}$is an unbiased estimator of localid="1661429690114" $\mathbf{\mu }$.

Calculation of the variance σx2

b.

The calculation of the variance${\mathrm{\sigma }}_{\mathrm{x}}^2$is shown below:

$$\begin{gathered} {\mathrm{\sigma }}_{\mathrm{x}}^2=\underset{}{\sum {(\mathrm{mean}-\overline{\mathrm{x}})}^2}\mathrm{p}\left(\mathrm{x}\right) \\ =\left[\left\{{\left(1-2.7\right)}^2(0.04)\right\}+\left\{{\left(1.5-2.7\right)}^2\left(0.12\right)\right\}+\left\{{\left(2-2.7\right)}^2\left(0.17\right)\right\} \\ +\left\{{\left(2.5-2.7\right)}^2\left(0.20\right)\right\}+\left\{{\left(3-2.7\right)}^2\left(0.20\right)\right\}+\left\{{\left(3.5-2.7\right)}^2(0.14)\right\} \\ +\left\{{\left(4-2.7\right)}^2\left(0.08\right)\right\}+\left\{{\left(4.5-2.7\right)}^2\left(0.04\right)\right\}+\left\{{\left(5-2.7\right)}^2\left(0.01\right)\right\}\right] \\ =(0.1156+0.1728+0.0833+0.008+0.018+0.0896+0.1352+0.1296+0.0529) \\ =0.805 \end{gathered}$$

Calculation of the probability

In order to find out the probability, the$2{\mathrm{\sigma }}_{\mathrm{x}}$has to be calculated where${\mathrm{\sigma }}_{\mathrm{x}}$is the standard deviation. So,

$$\begin{gathered} {\mathrm{\sigma }}_{\mathrm{x}}=\sqrt{{\mathrm{\sigma }}_{\mathrm{x}}^2} \\ =\sqrt{0.805} \\ =0.897 \\ 2{\mathrm{\sigma }}_{\mathrm{x}}=\left(2\times 0.897\right) \\ =1.794 \end{gathered}$$

Now the range is calculated below:

$$\begin{gathered} 2{\mathrm{\sigma }}_{\mathrm{x}}+\overline{\mathrm{x}}=1.794+2.7 \\ =4.494 \\ \overline{\mathrm{x}}-2{\mathrm{\sigma }}_{\mathrm{x}}=2.7-1.794 \\ =0.906 \end{gathered}$$

Therefore the probability that$\overline{\mathrm{x}}$will stay within the range (0.906, 4.494) is shown below:

$$\begin{gathered} \mathrm{probability}=\left\{\mathrm{p}\left(1\right)+\mathrm{p}(1.5)+\mathrm{p}\left(2\right)+\mathrm{p}(2.5)+\mathrm{p}\left(3\right)+\mathrm{p}(3.5)+\mathrm{p}\left(4\right)+\mathrm{p}(4.5)\right\} \\ =\left(0.04+0.12+0.17+0.20+0.20+0.14+0.08\right) \\ =0.95 \end{gathered}$$

Therefore the probability thatlocalid="1661429738495" $\overline{\mathbf{x}}$will stay within the range (0.906, 4.494) is 0.95