The average of n numbers x1,x2,...,xn is M. If xn is replaced by x′, then the new average is
A
M−xn+x′n
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B
(n−1)M+x′n
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C
nM−xn+x′n
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D
M−xn+x′
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Solution
The correct option is AnM−xn+x′n We have, M=x1+x2+...+xnn ⇒x1+x2+...+xn−1+xn=nM ⇒x1+x2+...+xn−1=nM−xn Let M′ be the average of x1,x2,...,xn−1,x′. Then M′=x1+x2+...+xn−1+x′n⇒M′=nM−xn+x′n