The average of n numbers `x_(1),x_(2), …,x_(n)` is M. If `x_(n)` is replaced by x' , then the new average is

Options

A. `(M-x_(n)+x')/(n)`
B. `((n-1)M+x')/(n)`
C. `(nM-x_(n)+x')/(n)`
D. `M-x_(n)+x'`

Answer

C

Explanation

We have,
`M=(x_(1)+x_(2)+...+x_(n))/(n)`
`rArr x_(1)+x_(2)+ ...+x_(n-1) +x_(n) = nM`
`rArr x_(1)+x_(2)+ ...+x_(n-1)=nM-x_(n) " " `...(i)
Let M' be the average of `x_(1)+x_(2)+ ...+x_(n-1)+x' `. Then,
`M'=(x_(1)+x_(2)+ ...+x_(n-1)+x')/(n)`
`rArr M'=(nM-x_(n)+x')/(n) " " `[Using (i)]