Answer 1

If `bar( X) `be the mean of the n observations q `X_i,...,X_n`, then we have

`bar(X) = 1/nsum_(i=1)^n X_i`

`⇒ sum_(i=1)^n X_i = nbar(X)`

Let ` bar(X)` be the mean of n values `X_i,...,X_n` . So, we have

`bar(X) = 1/nsum_(i=1)^n X_i`

`⇒ sum_(i=1)^n X_i = nbar(X)`

The sum of the deviations of n values `X_i,...,X_n` from their mean`bar(X)` is

`(x_1 - bar(X) ) +(x_2 - bar(X) )+......+(x_n-bar(X))`

`= sum_(i=1)^n(x_i- bar(X))`

`= sum_(i=1)^n x_i- sum_(i=1)^n  bar(X)`

`= nbar(X) - nbar(X)`

=0