The average of n numbers x1,x2,x3,…,xn is M . If xn is replaced by x′ , then new average is

M=x1+x2+x3……xnn⇒nM=x1+x2+x3+…+xn−1+xn⇒nM−xn=x1+x2+x3+…+xn−1⇒nM−xn+x′=x1+x2+x3+…+xn−1+x′⇒nM−xn+x′n=x1+x2+x3+…+xn−1+x′n∴ New average =nM−xn+x′n

The average of n numbers x1,x2,x3,…,xn is M . If xn is replaced by x′ , then new average is

M=x1+x2+x3……xnn⇒nM=x1+x2+x3+…+xn−1+xn⇒nM−xn=x1+x2+x3+…+xn−1⇒nM−xn+x′=x1+x2+x3+…+xn−1+x′⇒nM−xn+x′n=x1+x2+x3+…+xn−1+x′n∴ New average =nM−xn+x′n

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$The average of n numbers x_{1}, x_{2}, x_{3}, \dots, x_{n} is M . If x_{n} is replaced by x^{'}, then new average is$

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$M - x_{n} + x^{'}$

$\frac{n M - x_{n} + x^{'}}{n}$

$\frac{(n - 1) M + x^{'}}{n}$

$\frac{M - x_{n} + x^{'}}{n}$

answer is B.

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Detailed Solution

$\begin{array}{l} M = \frac{x_{1} + x_{2} + x_{3} \dots \dots x_{n}}{n} \\ \Rightarrow n M = x_{1} + x_{2} + x_{3} + \dots + x_{n - 1} + x_{n} \\ \Rightarrow n M - x_{n} = x_{1} + x_{2} + x_{3} + \dots + x_{n - 1} \\ \Rightarrow n M - x_{n} + x^{'} = x_{1} + x_{2} + x_{3} + \dots + x_{n - 1} + x^{'} \\ \Rightarrow \frac{n M - x_{n} + x^{'}}{n} = \frac{x_{1} + x_{2} + x_{3} + \dots + x_{n - 1} + x^{'}}{n} \\ ∴ New average = \frac{n M - x_{n} + x^{'}}{n} \end{array}$