Question

Refer to Exercise 5.3.

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

Short answer

1. Proved that $\stackrel{\rightharpoonup }{x}$ is an unbiased estimator of
2. 0.805
3. 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:

$$\begin{gathered} {\mu }_x=xp\left(x\right) \\ =\left\{1\left(0.2\right)+2\left(0.3\right)+3\left(0.2\right)+4\left(0.2\right)+5\left(0.1\right)\right\} \\ =\left(0.2+0.6+0.6+0.8+0.5\right) \\ =2.7 \\ E\left(\overline{x}\right)=E\overline{x}p\left(\overline{x}\right) \\ =\left\{\left(1\times 0.04\right)+\left(1.5\times 0.12\right)+\left(2\times 0.17\right)+\left(2.5\times 0.20\right)+\left(3\times 0.20\right)+\left(3.5\times 0.14\right)+\left(4\times 0.08\right)+\left(4.5\times 0.04\right)+\left(5\times 0.01\right)\right\} \\ =\left(0.04+0.18+0.34+0.5+0.6+0.49+0.32+0.18+0.05\right) \\ =2.7 \end{gathered}$$

Therefore as the value of$\mathbf{\mu }$and$\mathrm{E}\left(\stackrel{\rightharpoonup }{\mathrm{x}}\right)$are 2.7,is an unbiased estimator of.

Calculation of the variance σx2

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

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

Calculation of the probability

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

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

Now the range is calculated below:

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

Therefore the probability that will stay within the range (0.906, 4.494) is shown below:

$\mathrm{Probability}=\left\{\mathrm{p}\left(1\right)+\mathrm{p}\left(15\right)+\mathrm{p}\left(2\right)+\mathrm{p}\left(2.5\right)+\mathrm{p}\left(3\right)+\mathrm{p}\left(3.5\right)+\mathrm{p}\left(4\right)+\mathrm{p}\left(4.5\right)\right\}$

localid="1658142468201" $$\begin{gathered} =\left(0.04+0.12+0.17+0.20+0.20+0.14+0.08\right) \\ =0.95 \end{gathered}$$

Therefore the probability thatwill stay within the range (0.906, 4.494) is 0.95