Question

If M is the mean of x1 , x2 , x3 , x4 , x5 and x6, prove that
(x1 − M) + (x2 − M) + (x3 − M) + (x4 − M) + (x5 — M) + (x6 − M) = 0.

Answer 1

Let m be the mean of x₁, x₂ , x₃ , x_4, x₅ and x₆
 

Than M  = `(x_1+ x_2 + x_3 + x_4+ x_5 +x_6)/ 6`

⇒ `x_1+ x_2 + x_3 + x_4+ x_5 +x_6`

To prove :  `( x_1 - M ) + ( x_2 - M ) + ( x_3 - M ) + ( x_4 - M ) + ( x_5 - M ) + ( x_6 - M )`

= `( x_1 + x_2 + x_3 + x_4 + x_5 + x_6 ) - (M + M + M + M + M + M )`

=   6M - 6M
=   0
=   RHS