91影视

Given data,

${\text{s}}^{\text{2}}=\frac{危{(\text{蠂}鈭�{\displaystyle \stackrel{-}{蠂}})}^2}{\text{n}鈭�1}$

The calculation is given below:

The sample (1, 1) $\stackrel{-}{蠂}=1$

$\begin{array}{c}{\text{s}}^{\text{2}}=[{(1鈭�1)}^2+{(1鈭�1)}^2]/1\\ =0\end{array}$

The sample (1, 3) $\stackrel{-}{蠂}=1$

$$\begin{gathered} {\text{s}}^{\text{2}}=[{(1鈭�1)}^2+{(1鈭�1)}^2]/1 \\ =0 \end{gathered}$$

The sample (1, 2) $\stackrel{-}{蠂}=1.5$

$$\begin{gathered} {\text{s}}^{\text{2}}=[{(1鈭�15)}^2+{(2鈭�1.5)}^2]/1 \\ =0 \end{gathered}$$

To calculate the ${\mathbf{\text{s}}}^{\mathbf{\text{2}}}$values for every sample remaining

Sample Distribution of ${\text{s}}^{\text{2}}$is

The calculation is given below:

$$\begin{gathered} \text{Populationvariance}{蟽}^{\text{2}} \\ ={(1鈭�2.7)}^2\left(0.2\right)+{(2鈭�2.7)}^2\left(0.3\right)+{(3鈭�2.7)}^2\left(0.2\right)+{(4鈭�2.7)}^2\left(0.2\right)+{(5鈭�2.7)}^2\left(0.1\right) \\ =0.578+.147+.018+.338+.529 \\ =1.61 \end{gathered}$$

The calculation is given below:

To demonstrate the ${\text{s}}^{\text{2}}$is an unbiased estimator ${蟽}^{\text{2}}$

Using the sampling distribution of ${\text{s}}^{\text{2}}$

$$\begin{gathered} =\left(0\right)\left(0.22\right)+\left(0.5\right)\left(0.36\right)+\left(2\right)\left(0.24\right)+\left(4.5\right)+\left(.14\right)+\left(8\right)\left(.04\right) \\ =0+.18+.48+.63+.32 \\ =1.61 \end{gathered}$$

The calculation is given below:

The calculation is given below:

S from the sampling distribution

$$\begin{gathered} =\left(0\right)\left(0.22\right)+\left(0.707\right)\left(0.36\right)+\left(1.414\right)\left(0.24\right)+\left(2.121\right)+\left(0.14\right)+\left(2.828\right)\left(.04\right) \\ =0+.255+.339+.297+.113 \\ =1.004 \end{gathered}$$

But,

$$\begin{gathered} 蟽=\sqrt{{蟽}^2} \\ =\sqrt{1.61} \\ =1.269 \\ 鈮�1.004 \end{gathered}$$

So, s is a biased estimator of localid="1662357666676" $蟽$.