If `bar X_1 and bar X_2` are the means of two series such that `bar X_1 lt bar X_2 and bar X` is the mean of the combined series, then

Options

A. `barX lt barX_(1)`
B. `barX gt barX_(2)`
C. `barX = (barX_(1) + barX_(2))/(2)`
D. `barX_(1) lt barX lt barX_(2)`

Answer

D

Explanation

Let `n_(1)` and `n_(2)` be the number of observatios in two groups having means `barX_(1)` and `barX_(2)` respectively . Then
`barX = (n_(1) barX_(1) + n_(2)barX_(2))/(n_(1) + n_(2))`
Now , `barX - barX_(1) = (n_(1) barX_(1) + n_(2)barX_(2))/(n_(1) + n_(2)) - barX_(1)`
= `
(n_(2)(barX_(2) - barX_(1)))/(n_(1) + n_(2)) gt 0 [ because barX_(2) gt barX_(1)]`
`implies barX gt barX_(1) " " .....(1)`
And , `barX - barX_(2) = (n(barX_(1) - barX_(2)))/(n_(1) + n_(2)) gt 0 [because barX_(2) gt barX_(1)]`
`implies barX lt barX_(2)`
From (1) and (2)
`barX_(1) lt barX lt barX_(2)`.
Hence (4) is the correct answer.