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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.
Solution
Let m be the mean of x1, x2 , x3 , x4, 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
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