The algebraic sum of the deviations of a frequency distribution from its mean is
Suppose x1, x2, … , xn are n observations with mean as x.
By definition of mean, [i.e. The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations]
We have,
and
nx = x1 + x2 + … + xn …[1]
So, in this case we have assumed mean(a) is equal to mean of the observations(x)
And we know that
di = xi - a
where, di is deviation of a (i.e. assumed mean) from each of xi i.e. observations.
So, In the above case we have
d1 = x1 - x
d2 = x2 - x
.
.
.
dn = xn - x
and sum of deviations
d1 + d2 + … + dn = x1 - x + x2 - x + … + xn - x
= x1 + x2 + … + xn - (x + x + … {upto n times})
= nx - nx [Using 1]
= 0
Hence, sum of deviations is zero.