Calculate the range, variance, and standard deviation for the following samples:

a.4, 2, 1, 0, 1

b.1, 6, 2, 2, 3, 0, 3

c.8, -2, 1, 3, 5, 4, 4, 1, 3, 3

d.0, 2, 0, 0, -1, 1, -2, 1, 0, -1, 1, -1, 0, -3,-2, -1, 0, 1

Range = 4 – 0 = 4

$\begin{array}{c}\text{Mean}\left(\stackrel{-}{\chi }\right)\text{=}\frac{{\displaystyle \sum \text{χ}}}{\text{N}}\\ \text{=}\frac{\text{4+2+1+0+1}}{\text{5}}\\ \text{=}\frac{\text{8}}{\text{5}}\\ \text{=1.6}\\ \chi \text{=1.6}\end{array}$

$\begin{array}{c}\text{Mean}\left(\stackrel{-}{\chi }\right)\text{=}\frac{{\displaystyle \sum \text{χ}}}{\text{N}}\\ \text{=}\frac{\text{4+2+1+0+1}}{\text{5}}\\ \text{=}\frac{\text{8}}{\text{5}}\\ \text{=1.6}\end{array}$

$\stackrel{-}{\chi }\text{=1.6}$

$\begin{array}{c}\text{Variance=}\frac{{\displaystyle \sum \text{(χ}-\stackrel{-}{\chi }{\text{)}}^{\text{2}}}}{\text{N}-\text{1}}\\ \text{=}\frac{\text{9.2}}{\text{5}-\text{1}}\\ \text{=}\frac{\text{9.2}}{\text{4}}\\ \text{=2.3}\end{array}$

Therefore, the variance is 2.3

$\begin{array}{c}\text{Standarddeviation=}\sqrt{\text{Variance}}\\ \text{=}\sqrt{\text{2.3}}\\ \text{=1.52}\end{array}$

Therefore, the standard deviation is 1.52.

Range = 6 – 0 = 6

localid="1651305337400" $\begin{array}{c}\text{Mean}\left(\stackrel{-}{\chi }\right)\text{=}\frac{{\displaystyle \sum \text{χ}}}{\text{N}}\\ =\frac{1+6+2+2+3+0+3}{7}\\ =\frac{17}{7}\\ =2.43\end{array}$

localid="1651305447844" $\stackrel{-}{\chi }\text{=2.43}$

localid="1651305675002" $\begin{array}{c}\text{Variance=}\frac{{\displaystyle \sum \text{(χ}-\stackrel{-}{\chi }{\text{)}}^{\text{2}}}}{\text{N}-\text{1}}\\ \text{=}\frac{\text{21.7143}}{\text{7}-\text{1}}\\ \text{=}\frac{\text{21.7143}}{\text{6}}\\ \text{=3.619}\end{array}$

Therefore, the variance is 3.619.

$\begin{array}{c}\text{Standarddeviation=}\sqrt{\text{Variance}}\\ \text{=}\sqrt{\text{3.619}}\\ \text{=1.90}\end{array}$

Therefore, the standard deviation is 1.90.

Range = 8 – (-2) = 10

$\begin{array}{c}\text{Mean}\left(\stackrel{-}{\chi }\right)\text{=}\frac{{\displaystyle \sum \text{χ}}}{\text{N}}\\ \text{=}\frac{\text{8+}(-\text{2})\text{+1+3+5+4+4+1+3+3}}{\text{10}}\\ \text{=}\frac{\text{30}}{\text{10}}\\ \text{=3}\end{array}$

$\stackrel{-}{\chi }=3$

$\begin{array}{c}\text{Variance=}\frac{{\displaystyle \sum {\text{(χ}-\stackrel{-}{\chi })}^{\text{2}}}}{\text{N}-\text{1}}\\ \text{=}\frac{\text{64}}{\text{10}-\text{1}}\\ \text{=}\frac{\text{64}}{\text{9}}\\ \text{=7.11}\end{array}$

Therefore, the variance is 7.11.

$\begin{array}{c}\text{Standarddeviation=}\sqrt{\text{Variance}}\\ \text{=}\sqrt{\text{7.11}}\\ \text{=2.66}\end{array}$

Therefore, the standard deviation is 2.66.

Range = 2 – (-3) = 5

$$\begin{gathered} \text{Mean}\left(\stackrel{-}{\chi }\right)\text{=}\frac{{\displaystyle \sum \text{χ}}}{\text{N}} \\ \text{=}\frac{\text{0+2+0+0+}(-\text{1})\text{+1+}(-\text{2})\text{+1+0+}(-\text{1})\text{+1+}(-\text{1})\text{+0+}(-\text{3})\text{+}(-\text{2})\text{+}(-\text{1})\text{+0+1}}{\text{18}} \\ \text{=}\frac{-\text{5}}{\text{18}} \\ \text{=}-\text{0.27} \end{gathered}$$

$\stackrel{-}{\chi }=-\text{0.27}$

$$\begin{gathered} \text{Variance=}\frac{{\displaystyle \sum \text{(χ}-\stackrel{-}{\chi }{\text{)}}^{\text{2}}}}{\text{N}-\text{1}} \\ \text{=}\frac{\text{8.4432}}{\text{18}-\text{1}} \\ \begin{array}{c}=\frac{\text{8.4432}}{\text{17}}\\ \text{=0.4966}\end{array} \end{gathered}$$

Therefore, the variance is 0.4966.

$\begin{array}{c}\text{Standarddeviation=}\sqrt{\text{Variance}}\\ \text{=}\sqrt{\text{0.4966}}\\ \text{=0.70}\end{array}$

Therefore, the standard deviation is 0.70.