Mathematics - Statistics Question with Solution | TestHub

We have $\frac{1}{n}{\sum }_{i=1}^n{\left(x_i+2\right)}^2=18$ and $$\begin{array}{l}\frac{1}{n}{\sum }_{i=1}^n{\left(x_i-2\right)}^2=10\\ \Rightarrow {\sum }_{i=1}^n{\left(x_i+2\right)}^2=18n\text{ and }\end{array}$$ $$\begin{array}{l}{\sum }_{i=1}^n{\left(x_i-2\right)}^2=10n\\ \Rightarrow {\sum }_{i=1}^n{\left(x_i+2\right)}^2+{\sum }_{i=1}^n{\left(x_i-2\right)}^2=28n\end{array}$$ and ${\sum }_{i=1}^n{\left(x_i+2\right)}^2-{\sum }_{i=1}^n{\left(x_i-2\right)}^2=8n$ $$\begin{array}{l}\Rightarrow 2{\sum }_{i=1}^n{\left(x_i+4\right)}^2=28n\text{ and }2{\sum }_{i=1}^{n}4x_i=8n\\ \Rightarrow {\sum }_{i=1}^n{x}_1^2+4n=14n\text{ and }{\sum }_{i=1}^n{x}_i=n\\ \Rightarrow {\sum }_{i=1}^n{x}_i^2=10n\text{ and }{\sum }_{i=1}^n{x}_i=n\\ \therefore \sigma =\sqrt{\frac{1}{n}{\sum }_{i=1}^n{x}_i^2-{\left(\frac{1}{n}{\sum }_{i=1^{\prime }}^n{x}_i\right)}^2}=\sqrt{\frac{10n}{n}-{\left(\frac{n}{n}\right)}^2}=3\end{array}$$