The mean deviation for `n` observations `x_1, x_2, , x_n` from their mean ` X ` is given by

Options

A. ` sum_(i=1)^n(x_i- X )`
B. `1/nsum_(i=1)^n(x_i- X )`
C. `sum_(i=1)^n(x_i- X )^2`
D. `1/nsum_(i=1)^n(x_i- X )^2`

Answer

B

Explanation

We know that algebraic sum of deviations from mean is zero.
`sum_(i=1)^(n)(x_(i)-bar(x)) = (x_(1)-bar(x)) + (x_(2)-bar(x))+(x_(3)-bar(x))+(x_(n)-bar(x))`
`=(x_(1)+ x_(2)+ x_(3)+.....+ x_(n))-nbar(x)`
`sum_(i=1)^(n)x_(i)-nbar(x)=nbar(x)-nbar(x) = 0 " "(therefore sum_(i=1)^(n)x_(i)=nbar(x))`
Here, (b) is the correct answer.