Question

If $\bar X$ is the mean deviations of n observations ${x_1},{x_2},...............,{x_n}$, then prove that $\sum\limits_{i = 1}^n {\left( {{x_i} - \bar X} \right)} = 0$. I.e. The algebraic sum of the deviations from mean is zero.

Answer 1

Hint – In this particular type of question use the concept that the mean is the ratio of the sum of all the objects (i.e. sum of all the observations) to the number of objects (i.e. to the total number of observations), later on in the solution consider L.H.S and then expand the summation so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Proof –
Given if, $\bar X$ is the mean of the (n) observations which is given as${x_1},{x_2},...............,{x_n}$
As we all know that the mean is the ratio of the sum of all the objects (i.e. sum of all the observations) to the number of objects (i.e. to the total number of observations).
Therefore, mean of n observations $\left( {\bar X} \right) = \dfrac{{{x_1} + {x_2} + {x_3} + .......... + {x_n}}}{n}$
Therefore, $n\bar X = {x_1} + {x_2} + {x_3} + .......... + {x_n}$.................... (1)
Now take the L.H.S of the given equation we have,
$ \Rightarrow \sum\limits_{i = 1}^n {\left( {{x_i} - \bar X} \right)} $
Now here, i = 1, 2, 3....................n
So expand the summation we have,
$ \Rightarrow \left( {{x_1} - \bar X} \right) + \left( {{x_2} - \bar X} \right) + \left( {{x_3} - \bar X} \right) + .............. + \left( {{x_n} - \bar X} \right)$
Now separate the terms we have,
\[ \Rightarrow \left( {{x_1} + {x_2} + {x_3} + ...... + {x_n} - \left( {\bar X + \bar X + \bar X + \bar X + ................ + \bar X} \right)} \right)\].................... (2)
Now as we see that the mean is added n times so,
\[\left( {\bar X + \bar X + \bar X + \bar X + ................ + \bar X} \right)\] = $n\bar X$
Now from equation (2) we have,
\[ \Rightarrow \left( {{x_1} + {x_2} + {x_3} + ...... + {x_n} - n\bar X} \right)\]
Now from equation (2) substitute the value of $n\bar X$ in the above equation we have,
\[ \Rightarrow {x_1} + {x_2} + {x_3} + ...... + {x_n} - \left( {{x_1} + {x_2} + {x_3} + ...... + {x_n}} \right)\]
Now as we see that all the terms are cancel out so we have,
Therefore, $ \Rightarrow \sum\limits_{i = 1}^n {\left( {{x_i} - \bar X} \right)} $ = 0
= R.H.S
Hence the algebraic sum of the deviations from mean is zero.
Hence proved.

Note – Whenever we face such types of questions the key concept we have to remember is that the formula of the mean which is stated above and, if we add something (say x) n times then the resultant is n multiplied by the x (i.e. nx), so consider the L.H.S of the given expression and expand the summation as above and simplify we will get the required result.