91Ó°ÊÓ

The calculation of the mean in the case of the three values of x is shown below.

Therefore, the mean is$\text{= 1}\frac{\text{2}}{\text{3}}$

The calculation of the varianceof the three values of x is shown below.

$\begin{array}{rcl}\mathrm{μ}& =& {\left(\text{0 -}\frac{\text{5}}{\text{3}}\right)}^{\text{2}}×\frac{\text{1}}{\text{3}}+{\left(\text{1 -}\frac{\text{5}}{\text{3}}\right)}^{\text{2}}×\frac{\text{1}}{\text{3}}+{\left(\text{4 -}\frac{\text{5}}{\text{3}}\right)}^{\text{2}}×\frac{\text{1}}{\text{3}}\\ & & \text{=}\frac{\text{25}}{\text{27}}\text{+}\frac{\text{4}}{\text{27}}\text{+}\frac{\text{49}}{\text{27}}\\ & & \text{=}\frac{\text{78}}{\text{27}}\\ & & \text{= 2}\frac{\text{8}}{\text{9}}\\ & & \\ & & \end{array}$

Therefore, the median is$\text{= 2}\frac{\text{8}}{\text{9}}$

1. The calculation of the mean of the two values (which are samples) of x is shown below.

The probabilities of the nine sample means are calculated below.

Therefore, for all the means, the probability is.$\frac{\text{1}}{\text{9}}$

The calculation ofis shown below.

Therefore, the equation,$\underset{}{\overset{}{∑\left(\overline{x}\right)=\underset{}{\overset{}{∑}}}}\overline{x}p\left(\overline{x}\right)=\mathrm{μ}$is proved. So,$x$is an unbiased estimator of$\mathrm{μ}$

$\begin{array}{rcl}{s}^2& =& E\left[{\left\{\overline{x}-E\left(\overline{x}\right)\right\}}^2\right]\\ & =& \underset{}{\overset{}{∑{\left(\overline{x}-\mathrm{μ}\right)}^2}}p\left(\overline{x}\right)\end{array}$

The calculation of is shown below.

$\begin{array}{rcl}& & {\text{s}}^{\text{2}}\text{=}{\left(\text{0 -}\frac{\text{5}}{\text{3}}\right)}^{\text{2}}\text{×}\left(\frac{\text{1}}{\text{3}}\right)\text{×}{\left(\text{1 -}\frac{\text{5}}{\text{3}}\right)}^{\text{2}}\text{×}\left(\frac{\text{1}}{\text{3}}\right)\text{+}{\left(\text{4 -}\frac{\text{5}}{\text{3}}\right)}^{\text{2}}\text{×}\left(\frac{\text{1}}{\text{3}}\right)\\ & & \text{=}\frac{\text{25}}{\text{9}}\text{×}\left(\frac{\text{1}}{\text{3}}\right)\text{+}\frac{\text{16}}{\text{9}}\text{×}\left(\frac{\text{1}}{\text{3}}\right)\text{+}\frac{\text{49}}{\text{9}}\text{×}\left(\frac{\text{1}}{\text{3}}\right)\\ & & \text{=}\frac{\text{25}}{\text{27}}\text{+}\frac{\text{16}}{\text{27}}\text{+}\frac{\text{49}}{\text{27}}\\ & & \text{=}\frac{\text{90}}{\text{27}}\\ & & \text{=}\frac{\text{10}}{\text{3}}\\ & & \text{= 3}\frac{\text{1}}{\text{3}}\\ & & \end{array}$

The value of$s^2$is$\text{3}\frac{\text{1}}{\text{3}}$