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Question: The average of *n* numbers \(x_{1},x_{2},x_{3},......,x_{n}\) is *M*. If \(x_{n}\) is replaced by \(...

The average of n numbers x1,x2,x3,......,xnx_{1},x_{2},x_{3},......,x_{n} is M. If xnx_{n} is replaced by x′x^{'}, then new average is

A

M−xn+x′M - x_{n} + x^{'}

B

nM−xn+x′n\frac{nM - x_{n} + x^{'}}{n}

C

(n−1)M+x′n\frac{(n - 1)M + x^{'}}{n}

D

M−xn+x′n\frac{M - x_{n} + x^{'}}{n}

Answer

nM−xn+x′n\frac{nM - x_{n} + x^{'}}{n}

Explanation

Solution

M=x1+x2+x3......xnnM = \frac{x_{1} + x_{2} + x_{3}......x_{n}}{n} i.e.

nM=nM−xn=nM−xn+x′=‾nx1+x2+x3+.....xn−1+xnx1+x2+x3+.....xn−1 6mu 6mu 6mu 6mu 6mu 6mux1+x2+x3+.....xn−1+x′‾n\underset{n}{\underline{\underset{nM - x_{n} + x^{'} =}{\underset{nM - x_{n} =}{nM =}}}}\underset{n}{\underline{\underset{x_{1} + x_{2} + x_{3} + .....x_{n - 1} + x^{'}}{\underset{x_{1} + x_{2} + x_{3} + .....x_{n - 1}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}}{x_{1} + x_{2} + x_{3} + .....x_{n - 1} + x_{n}}}}}

∴ New average=nM−xn+x′n= \frac{nM - x_{n} + x^{'}}{n}