Question

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.

Short answer

a.

$\mu$=$\text{1}\frac{\text{2}}{\text{3}}$

and

${\sigma }^2=\text{2}\frac{\text{8}}{\text{9}}$=.

b.

Samples	Means
0	$\frac{\text{1}}{\text{9}}$
1	$\frac{\text{1}}{\text{9}}$
4	$\frac{\text{1}}{\text{9}}$
0.5	$\frac{\text{1}}{\text{9}}$
0.5	$\frac{\text{1}}{\text{9}}$
2	$\frac{\text{1}}{\text{9}}$
2	$\frac{\text{1}}{\text{9}}$
2.5	$\frac{\text{1}}{\text{9}}$
2.5	$\frac{\text{1}}{\text{9}}$

c. It is proved.

d. The required answer is $3\frac{1}{3}$

Step-by-step solution

Calculation of the mean μ 

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

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

Calculation of the variance σ2

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$\text{= 2}\frac{\text{8}}{\text{9}}$

Calculation of the value of the sample mean 

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

Samples	Means
0,0	0
1,1	1
4,4	4
0,1	0.5
1,0	0.5
0,4	2
4,0	2
1,4	2.5
4,1	2.5

Computation of the sample distribution

The probabilities of the nine sample means are calculated below.

Samples	Means
0	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
1	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
4	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
0.5	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
0.5	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
2	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
2	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
2.5	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$
2.5	$\frac{\text{1}}{\text{3}}\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{9}}$

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

Calculation of ∑x

c .The calculation of $\sum _{}^{}\left(\overline{x}\right)$is shown below.

Samples	Means
0	$0\times \frac{\text{1}}{\text{3}}=0$
1	$1\times \frac{\text{1}}{\text{3}}=\frac{\text{1}}{\text{3}}$
4	$4\times \frac{\text{1}}{\text{3}}=\frac{\text{4}}{\text{3}}$
Total	$\frac{\text{5}}{\text{3}}$

Computation of∑xpx in Part (c)

The calculation ofis shown below.

Samples	Means
0	$0\times \frac{\text{1}}{\text{9}}=0$
1	$1\times \frac{\text{1}}{\text{9}}=\frac{\text{1}}{\text{9}}$
4	$4\times \frac{\text{1}}{\text{9}}\text{=}\frac{\text{4}}{\text{9}}$
0.5	$0.5\times \frac{1}{9}=\frac{\text{0.5}}{\text{9}}$
0.5	$0.5\times \frac{\text{1}}{\text{9}}=\frac{\text{0.5}}{\text{9}}$
2	$2\times \frac{1}{9}=\frac{2}{9}$
2	$2\times \frac{\text{1}}{\text{9}}=\frac{2}{9}$
2.5	$2.5\times \frac{1}{9}=\frac{\text{2.5}}{\text{9}}$
2.5	$\text{2.5×}\frac{\text{1}}{\text{9}}\text{=}\frac{\text{2.5}}{\text{9}}$
Total	$\frac{\text{5}}{\text{3}}$

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$\mu$

Formula to calculate s2

d.From Part (a), the value of $\sum _{}^{}\left(\overline{x}\right)$found is$\frac{5}{3}$, and this value can be used in the formula of$s^2$, as shown below.

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

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