Problem:
Compute the sample standard deviation of ${-4, -10, -4, -6, -12, -2}$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Find the (sample) standard deviation of the list}: \\
 (-4,-10,-4,-6,-12,-2)\text{} \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{The }\text{standard }\text{deviation }\text{is }\text{given }\text{by}: \\
 \sqrt{(\text{variance})} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{(sample) }\text{variance }\text{of }\text{a }\text{list }\text{of }\text{numbers }X=\left\{X_1,X_2,\ldots ,X_n\right\} \text{with }\text{mean }\mu =\frac{X_1+X_2+\ldots +X_n}{n} \text{is }\text{given }\text{by}: \\
 \frac{\left| X_1-\mu \right| {}^2+\left| X_2-\mu \right| {}^2+\ldots +\left| X_n-\mu \right| {}^2}{n-1} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{There }\text{are }n=6 \text{elements }\text{in }\text{the }\text{list }X=\{-4,-10,-4,-6,-12,-2\}: \\
 \frac{\left| X_1-\mu \right| {}^2+\left| X_2-\mu \right| {}^2+\left| X_3-\mu \right| {}^2+\left| X_4-\mu \right| {}^2+\left| X_5-\mu \right| {}^2+\left| X_6-\mu \right| {}^2}{6-1} \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{The }\text{elements }X_i \text{of }\text{the }\text{list }X=\{-4,-10,-4,-6,-12,-2\} \text{are:} \\
 
\begin{array}{l}
 X_1=-4 \\
 X_2=-10 \\
 X_3=-4 \\
 X_4=-6 \\
 X_5=-12 \\
 X_6=-2 \\
\end{array}
 \\
\end{array}
 \\
 \frac{| -\mu -4| ^2+| -\mu -10| ^2+| -\mu -4| ^2+| -\mu -6| ^2+| -\mu -12| ^2+| -\mu -2| ^2}{6-1} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{mean }(\mu ) \text{is }\text{given }\text{by$\backslash $n$\backslash $n}\mu =\frac{X_1+X_2+X_3+X_4+X_5+X_6}{6}=\frac{-4-10-4-6-12-2}{6}=-\frac{19}{3}: \\
 \frac{\left| -\frac{-19}{3}-4\right| ^2+\left| -\frac{-19}{3}-10\right| ^2+\left| -\frac{-19}{3}-4\right| ^2+\left| -\frac{-19}{3}-6\right| ^2+\left| -\frac{-19}{3}-12\right| ^2+\left| -\frac{-19}{3}-2\right| ^2}{6-1} \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{The values of the differences are:} \\
 
\begin{array}{l}
 -\frac{-19}{3}-4=\frac{7}{3} \\
 -\frac{-19}{3}-10=-\frac{11}{3} \\
 -\frac{-19}{3}-4=\frac{7}{3} \\
 -\frac{-19}{3}-6=\frac{1}{3} \\
 -\frac{-19}{3}-12=-\frac{17}{3} \\
 -\frac{-19}{3}-2=\frac{13}{3} \\
 6-1=5 \\
\end{array}
 \\
\end{array}
 \\
 \frac{\left| \frac{7}{3}\right| ^2+\left| -\frac{11}{3}\right| ^2+\left| \frac{7}{3}\right| ^2+\left| \frac{1}{3}\right| ^2+\left| -\frac{17}{3}\right| ^2+\left| \frac{13}{3}\right| ^2}{5} \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{The }\text{values }\text{of }\text{the }\text{terms }\text{in }\text{the }\text{numerator }\text{are}: \\
 
\begin{array}{l}
 \left| \frac{7}{3}\right| ^2 \text{= }\frac{49}{9} \\
 \left| -\frac{11}{3}\right| ^2 \text{= }\frac{121}{9} \\
 \left| \frac{7}{3}\right| ^2 \text{= }\frac{49}{9} \\
 \left| \frac{1}{3}\right| ^2 \text{= }\frac{1}{9} \\
 \left| -\frac{17}{3}\right| ^2 \text{= }\frac{289}{9} \\
 \left| \frac{13}{3}\right| ^2 \text{= }\frac{169}{9} \\
\end{array}
 \\
\end{array}
 \\
 \frac{\frac{49}{9}+\frac{121}{9}+\frac{49}{9}+\frac{1}{9}+\frac{289}{9}+\frac{169}{9}}{5} \\
\end{array}
 \\
 
\begin{array}{l}
 \frac{49}{9}+\frac{121}{9}+\frac{49}{9}+\frac{1}{9}+\frac{289}{9}+\frac{169}{9}=\frac{226}{3}: \\
 \frac{\frac{226}{3}}{5} \\
\end{array}
 \\
 
\begin{array}{l}
 \frac{\frac{226}{3}}{5}=\frac{226}{15}: \\
 \frac{226}{15} \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{The standard deviation is given by} \\
 \sqrt{(\text{variance})} \text{= }\sqrt{\frac{226}{15}}: \\
\end{array}
 \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \sqrt{\frac{226}{15}} \\
\end{array}
$} \\
\end{array}
 \\
\end{array}
{'source': 'AMPS', 'file': 'amps/mathematica/counting_and_statistics/variance_and_std_w_steps/9658.txt'}