Consider the numbers 1,2,3,4,5,6,7,8,9 and 10. If 1 is added to each number, the variance of the numbers so obtained is

Given numbers are 1,2,3, 4, 5, 6, 7, 8, 9 and 10.
If 1 is added to each number, then observations will be 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11

$\begin{array}{c}\therefore {\mathrm{\Sigma x}}_{\mathrm{i}}=2+3+4+\dots +11\\ =\frac{10}{2}[2\times 2+9\times 1]=5[4+9]=65\end{array}$

$\begin{array}{c}\mathrm{and} {\mathrm{\Sigma x}}_{\mathrm{i}}^2=2^2+3^2+4^2+5^2+\dots +{11}^2\\ =\left(1^2+2^2+3^2+\dots +{11}^2\right)-\left(1^2\right)\\ =\frac{11\times 12\times 23}{6}-1=505\\ \therefore {\mathrm{s}}^2=\frac{{\mathrm{\Sigma x}}_{\mathrm{i}}^2}{\mathrm{n}}-{\left(\frac{{\mathrm{\Sigma x}}_{\mathrm{i}}}{\mathrm{n}}\right)}^2=\frac{505}{10}-{\left(\frac{65}{10}\right)}^2\\ =50.5-(6.5{)}^2\\ =50.5-42.25\\ =8.25\end{array}$