The AM of the nnumbers `x_(1),x_(2)......x_(n)` is M. If `x_(1)` is replaced by x then the new AM is

Options

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

Answer

B

Explanation

`M=(x_(1)+x_(2)+x_(3)..x_(n))/(n)`
`implies nM=x_(1)+x_(2)+x_(3)+..+x_(n-1)+x_(n)`
`implies nM-x_(n)=x_(1)+x_(2)+x_(3)+..+x_(n-1)`
`implies nM-x_(n)+x'=x_(1)+x_(2)+x_(3)+..+x_(n-1)+x'`
`implies (nM-x_(n)+x')/(n)=(x_(1)+x_(2)+x_(3)+..+x_(n-1)+x')/(n)`
`therefore " New average"=(nM-x_(n)+x')/(n)`