Question

If $\overline{\mathrm{X}}$ is the mean of n values x₁, x₂, ......,x_n of variable X, then $\sum _{\mathrm{i}=1}^{\mathrm{n}}({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{X}})$ is always equal to _____________.

Answer 1

$$\begin{gathered} \mathrm{Let}\overline{X}\mathrm{is}\mathrm{the}\mathrm{mean}\mathrm{of}{x}_1,x_{2,}.....x_n\mathrm{for}\mathrm{variable}X. \\ \mathrm{then}\sum _{\mathrm{i}=1}^n\left(x_i-\overline{X}\right) \\ =\sum _{i=1}^n{x}_i-\sum _{i=1}^n\overline{X} \\ =\sum _{i=1}^n{x}_i-\overline{X}\sum _{i=1}^{n}1 \\ =\sum _{i=1}^n{x}_i-n\overline{X} \\ \begin{array}{l}\mathrm{Since}\overline{X}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ \mathrm{i}.\mathrm{e}\sum _{i=1}^n{x}_i=n\overline{X}\end{array} \\ =n\overline{X}-n\overline{X} \\ =0 \\ \therefore \sum _{i=1}^n\left(x_i-\overline{X}\right)=\mathrm{zero} \end{gathered}$$