The mean of the series `x_1, x_2,...x_n` is `barX`. If `x_2` is replaced by `lambda` then the new mean is

Options

A. `barX-x_(2)+lambda`
B. `(barX-x_(2)-lambda)/(n)`
C. `(barX-x_(2)+lambda)/(n)`
D. `(nbarX-x_(2)+lambda)/(n)`

Answer

D

Explanation

Mean of series `(x_(1),x_(2),x_(3)……….x_(n))`
`barx=(x_(1)+x_(2)+x_(3)+………..x_(n))/(n)`
`rArr x_(1)+x_(2)+x_(3)+…….x_(n)=nbarx`
Now will replace `x_(2)` by `lambda` so no. of elements in series will bot change.
New series will include `lambda` and exclude `x_(2)`
Hence new series sum :
`(x_(1)+x_(2)+..........x_(n))-x_(2)+lambda=nbarx+lambda-x_(2)`
Now new mean`=(n barx+lambda-x_(2))/(n)=(nbarx-x_(2)+lambda)/(n)`