The algebraic sum of the deviations of a set of `n` values from their mean is 0 (b) `n-1` (c) `n` (d) `n+1`

Explanation

`bar(x)=(x_1+x_2+x_3+......x_n)/n`
`n bar(x)=x_1+x_2+x_3+......x_n`....(i)
Algebraic sum of the deviation from the mean is:
Sum`=(x_1-bar(x))+(x_2-bar(x))+(x_3-bar(x))+.....(x_n-n bar(x))`
`=x_1+x_2+x_3+.....+x_n-(bar(x)+bar(x)+bar(x)....+n " times")`
Using (i) we get;
`Sum=n bar(x)-n bar(x)`
`=0`
Hence, algebraic sum of the deviations of a set of n values from mean is 0.