Compute xbar, s², and sfor each of the following data sets. If appropriate, specify the units in which your answer is expressed.

a.3, 1, 10, 10, 4

b.8 feet, 10 feet, 32 feet, 5 feet

c.-1, -4, -3, 1, -4, -4

d.1/5 ounce, 1/5 ounce, 1/5 ounce, 2/5 ounce, 1/5 ounce, 4/5 ounce

x = 3, 1, 10, 10, 4
$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\displaystyle \sum \mathrm{x}}}{\mathrm{n}} \\ =\frac{3+1+10+10+4}{5} \\ =\frac{28}{5} \\ =5.6 \end{gathered}$$

$$\begin{gathered} {\text{s}}^{\text{2}}\text{=}\frac{{\displaystyle \sum {(\text{χ}-\stackrel{-}{\chi })}^{\text{2}}}}{\text{n}-\text{1}} \\ =\frac{69.2}{5-1} \\ =\frac{69.2}{4} \\ =17.3 \end{gathered}$$

$\begin{array}{c}\text{s=}\sqrt{{\text{s}}^{\text{2}}}\\ \text{=}\sqrt{\text{17.3}}\\ \text{=4.159}\end{array}$

Therefore, localid="1651308857080" $\overline{\mathbf{x}}$= 5.6, s² = 17.3, s = 4.159

x = 8 feet, 10 feet, 32 feet, 5 feet

$$\begin{gathered} \stackrel{-}{\chi }=\frac{{\displaystyle \sum \text{χ}}}{\text{n}} \\ \text{=}\frac{\text{8+10+32+5}}{\text{4}} \\ \text{=}\frac{\text{55}}{\text{4}} \\ \text{=13.75} \end{gathered}$$

$$\begin{gathered} {\text{s}}^{\text{2}}\text{=}\frac{{\displaystyle \sum {(\text{χ}-\stackrel{-}{\chi })}^{\text{2}}}}{\text{n}-\text{1}} \\ \text{=}\frac{\text{456.75}}{\text{4}-\text{1}} \\ \begin{array}{c}\text{=}\frac{\text{456.75}}{\text{3}}\\ \text{=152.25}\end{array} \end{gathered}$$

$\begin{array}{c}\text{s=}\sqrt{{\text{s}}^{\text{2}}}\\ =\sqrt{152.25}\\ =12.338\end{array}$

Therefore,$\overline{\mathbf{x}}$ = 13.75 feet, s² = 152.25, s = 12.338 feet

x = -1, -4, -3, 1, -4, -4

$$\begin{gathered} \stackrel{-}{\chi }=\frac{{\displaystyle \sum \text{χ}}}{\text{n}} \\ =\frac{(-1)+(-4)+(-3)+1+(-4)+(-4)}{6} \\ =\frac{-15}{6} \\ =-2.5 \end{gathered}$$

$$\begin{gathered} {\text{s}}^{\text{2}}\text{=}\frac{{\displaystyle \sum {(\text{χ}-\stackrel{-}{\chi })}^{\text{2}}}}{\text{n}-\text{1}} \\ \text{=}\frac{\text{21.75}}{\text{6}-\text{1}} \\ \text{=}\frac{\text{21.75}}{\text{5}} \\ \text{=4.35} \end{gathered}$$

$$\begin{gathered} \text{s=}\sqrt{{\text{s}}^{\text{2}}} \\ \text{=}\sqrt{\text{4.35}} \\ \text{=2.085} \end{gathered}$$

Therefore, $\overline{\mathbf{x}}$= -2.5, s² = 4.35, s = 2.085

x = 1/5 (0.2) ounce, 1/5 (0.2) ounce, 1/5 (0.2) ounce, 2/5 (0.4) ounce, 1/5 (0.2) ounce, 4/5 (0.8) ounce

$$\begin{gathered} \stackrel{-}{\chi }=\frac{{\displaystyle \sum \text{χ}}}{\text{n}} \\ \text{=}\frac{\text{0.2+0.2+0.2+0.4+0.2+0.8}}{\text{6}} \\ \text{=}\frac{\text{2}}{\text{6}} \\ \text{=0.33} \end{gathered}$$

$$\begin{gathered} {\text{s}}^{\text{2}}\text{=}\frac{{\displaystyle \sum {(\text{χ}-\stackrel{-}{\chi })}^{\text{2}}}}{\text{n}-\text{1}} \\ \text{=}\frac{\text{0.2934}}{\text{6}-\text{1}} \\ \text{=}\frac{\text{0.2934}}{\text{5}} \\ \text{=0.05868} \end{gathered}$$

$\begin{array}{c}\text{s=}\sqrt{{\text{s}}^{\text{2}}}\\ \text{=}\sqrt{\text{0.05868}}\\ \text{=0.242}\end{array}$

Therefore, $\overline{\mathbf{x}}$= 0.33 ounce, s² = 0.05868, s = 0.242 ounce.