Question

The algebraic sum of deviation of a set of n observations from their mean is
A. 0
B. $\dfrac{{n(n + 1)}}{2}$
C. $\dfrac{{n(n - 1)}}{2}$
D. $\dfrac{{n + 1}}{2}$

Answer 1

Hint: First we have to know what deviation from mean is, given the set of observations we found out the mean which is given by the ratio of sum of all observations to the total number of observations, and now the value of each observation may or may not be equal to the mean of the total observations, the value of each observation may be greater than or less than the value of the mean, the difference between each value of observation and the mean is called as the deviation from mean.

Complete step-by-step solution:
Given there are n observations and we have to find out the sum of deviations of all the observations.
Let the set of n observations be ${x_1},{x_2},{x_3},....{x_n}$.
The mean of these observations be $\overline x $,
Then the deviation from each observation would be :
Deviation of first observation is $({x_1} - \overline x )$
Deviation of second observation is $({x_2} - \overline x )$
Deviation of third observation is $({x_3} - \overline x )$ and so on…
Deviation of last observation is $({x_n} - \overline x )$
Now the sum of the deviations of all observations is given by:
$ \Rightarrow \sum\limits_i {({x_i} - \overline x )} $
$ \Rightarrow \sum\limits_i {({x_i} - \overline x )} = ({x_1} - \overline x ) + ({x_2} - \overline x ) + ({x_3} - \overline x ) \cdot \cdot \cdot \cdot + ({x_n} - \overline x )$
Now let us take an example of a real set of observations which are given below:
1, 2, 6, 8, 11, 15, 20.
Let these be a set of 7 observations.
The mean is given by the ratio of the sum of observations to the total no. of observations :
$ \Rightarrow \overline x = \dfrac{{1 + 2 + 6 + 8 + 11 + 15 + 20}}{7}$
$ \Rightarrow \overline x = \dfrac{{63}}{7}$
$ \Rightarrow \overline x = 9$
$\therefore $The mean of the observations is 9.
Now computing the sum of the deviations the mean of each observation is given by:
$ \Rightarrow \sum\limits_i {({x_i} - \overline x )} = (1 - 9) + (2 - 9) + (6 - 9) + (8 - 9) + (11 - 9) + (15 - 9) + (20 - 9)$
$ \Rightarrow \sum\limits_i {({x_i} - \overline x )} = - 8 - 7 - 3 - 1 + 2 + 6 + 11$
$ \Rightarrow \sum\limits_i {({x_i} - \overline x )} = 0$
$\therefore \sum\limits_i {({x_i} - \overline x )} = 0$, the sum of deviations from the mean is zero.

The algebraic sum of deviation of a set of n observations from their mean is zero.

Note: The most important fact is that the algebraic sum of deviation of a set of any number of observations from their mean is always zero.