Question

If M is the mean of X1,X2,X3, X4, X5 and X6 then prove that (X1-M)+(X2-M)+(X3-M)+(X4-M)+(X5-M)+(X6-M)=0

Answer 1

Answer:

mean of X1,X2,..IS M

BY SUBTRACTING M FROM THE GIVEN ABSVs SUBTRACTS M FROM THE REAL MEAN

Answer 2

(X1 - M) + (X2 - M) + (X3 - M) + (X4 - M) + (X5 - M) +(X6 - M) = 0 (Proved)

Step-by-step explanation:

Given that M is the mean of X1, X2, X3, X4, X5, and X6.

So, $M = \frac{X1 + X2 + X3 + X4 + X5 + X6}{6}$

Now, we have to prove that, (X1 - M) + (X2 - M) + (X3 - M) + (X4 - M) + (X5 - M) +(X6 - M) = 0

Now, we have, $M = \frac{X1 + X2 + X3 + X4 + X5 + X6}{6}$

⇒ 6M = X1 + X2 + X3 + X4 + X5 + X6

⇒ X1 + X2 + X3 + X4 + X5 + X6 - 6M = 0

⇒ (X1 - M) + (X2 - M) + (X3 - M) + (X4 - M) + (X5 - M) +(X6 - M) = 0 (Proved)