In a group of data, there are n observations ${\mathrm{x}}_1, {\mathrm{x}}_2...{\mathrm{x}}_{\mathrm{n}}.$ If $\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}+1\right)}^2=9\mathrm{n}$ and $\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}+1\right)}^2=5\mathrm{n}$ the standard deviation of the data is

$$\begin{gathered} \sum {\left({\mathrm{x}}_{\mathrm{i}}+1\right)}^2=9\mathrm{n} -\left(1\right) \\ \sum {\left({\mathrm{x}}_{\mathrm{i}}-1\right)}^2=5\mathrm{n} -\left(2\right) \\ \left(1\right) + \left(2\right) \Rightarrow \\ 2\sum \left({{\mathrm{x}}_{\mathrm{i}}}^2+1\right)=14\mathrm{n} \\ \sum \left({{\mathrm{x}}_{\mathrm{i}}}^2+1\right)=7\mathrm{n} \Rightarrow \sum {{\mathrm{x}}_{\mathrm{i}}}^2+\mathrm{n}=7\mathrm{n} \\ \Rightarrow \sum {{\mathrm{x}}_{\mathrm{i}}}^2=6\mathrm{n} \\ \left(1\right) - \left(2\right) \Rightarrow \\ 4\sum {\mathrm{x}}_{\mathrm{i}}=4\mathrm{n} \Rightarrow \frac{\sum {\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}=1 \end{gathered}$$

 $\begin{array}{rcl}variance& =& \frac{\sum {{\mathrm{x}}_{\mathrm{i}}}^2}{\mathrm{n}}-{\left(\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}\right)}^2\\ & =& 6-1\\ & =& 5\\ \mathrm{S}.\mathrm{D}& =& \sqrt{5}\end{array}$