Consider the following probability distribution:

1. Findand${\sigma }^2$.
2. Find the sampling distribution of the sample mean x for a random sample of n = 2 measurements from this distribution
3. Show that $x$is an unbiased estimator of $\mu$. [Hint: Show that.]$\sum _{}^{}\left(\overline{x}\right)=\sum _{\overline{}}^{}\overline{x}p\left(\overline{x}\right)=\mu$.]
4. Find the sampling distribution of the sample variance$s^2$for a random sample of n = 2 measurements from this distribution.

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.

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

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,$\underset{}{\overset{}{\sum \left(\overline{x}\right)=\sum _{}^{}}}\overline{x}p\left(\overline{x}\right)=\mu$is proved. So,$x$is an unbiased estimator of$\mu$

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

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$\text{3}\frac{\text{1}}{\text{3}}$