The sum of all deviations taken from Mean =  A) 0  B) -1  C) 1  D) 2

Correct option is: A) 0

Let mean = \(\overline x\) 

\(\because\) Mean = \(\frac {\sum x_i}{n}\) \(\Rightarrow\) \(\sum x_i = n \times Mean = n \overline x\)

Then deviations taken from mean is \(d_i = x_i - \overline x\)

Let total observations be n

\(\therefore\) Sum of deviations taken from mean is

\(\sum \limits _{i=1}^n d_i = \sum \limits _{i=1}^n (x_i- \overline x) = \sum \limits _{i=1}^n\, x_i - \sum \limits _{i=1}^n\, \overline x =\sum \limits _{i=1}^n\, x_i - \overline x \sum \limits _{i=1}^n 1\)

= n \(\overline x\) - n \(\overline x\) = 0