91Ó°ÊÓ

The calculation of the mean$\mathbf{μ}$in the case of the three values of x is shown below.localid="1658059082670" $$\begin{gathered} \mathbf{μ}\mathbf{=}\mathbf{0}\mathbf{×}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{×}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\mathbf{4}\mathbf{×}\frac{\mathbf{1}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\frac{\mathbf{2}}{\mathbf{3}} \end{gathered}$$.

Therefore, the mean is $\mathbf{1}\frac{\mathbf{2}}{\mathbf{3}}$.

The calculation of the variance${\mathbf{σ}}^{\mathbf{2}}$of the three values of x is shown below.$$\begin{gathered} \mathbf{μ}\mathbf{=}{\mathbf{(}\mathbf{0}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{×}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}{\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{×}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{×}\frac{\mathbf{1}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{49}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{78}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\frac{\mathbf{8}}{\mathbf{9}} \end{gathered}$$

Therefore, the median is $\mathbf{2}\frac{\mathbf{8}}{\mathbf{9}}$.

b.

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 $\frac{\mathbf{1}}{\mathbf{9}}$.

c.

The calculation of$\underset{\mathbf{}}{\mathbf{∑}}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}$is shown below.

The calculation of$\underset{\mathbf{}}{\mathbf{∑}}\overline{\mathbf{x}}\mathbf{p}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}$is shown below

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

localid="1658060043017" $$\begin{gathered} {\mathbf{s}}^{\mathbf{2}}\mathbf{=}\mathbf{E}\mathbf{\left[}{\left\{\overline{x}-E\left(\overline{x}\right)\right\}}^{\mathbf{2}}\mathbf{\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\underset{\mathbf{}}{\mathbf{∑}}{\mathbf{(}\overline{\mathbf{x}}\mathbf{-}\mathbf{μ}\mathbf{)}}^{\mathbf{2}}\mathbf{p}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)} \end{gathered}$$

.

The calculation of${\mathbf{s}}^{\mathbf{2}}$is shown below.

localid="1658060107622" $$\begin{gathered} {\mathbf{s}}^{\mathbf{2}}\mathbf{=}{\mathbf{(}\mathbf{0}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{×}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}{\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{×}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{×}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{9}}\mathbf{×}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{16}}{\mathbf{9}}\mathbf{×}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{49}}{\mathbf{9}}\mathbf{×}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{16}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{49}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{90}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{10}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\frac{\mathbf{1}}{\mathbf{3}} \end{gathered}$$.

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