Let `barx` be the mean of `x_(1), x_(2), ………, x_(n)` and `bary` be the mean of `y_(1), y_(2),……….,y_(n)`. If `barz` is the mean of `x_(1), x_(2), ……………..x_(n), y_(1), y_(2), …………,y_(n)`, then `barz` is equal to

Options

A. `bar(x)+bar(y)`
B. `(bar(x)+bar(y))/(2)`
C. `(bar(x)+bar(y))/(n)`
D. `(bar(x)+bar(y))/(2n)`

Answer

B

Explanation

We have `bar(x)` is the mean of `x_(1), x_(2),…., x_(n)` and `bar(y)` is the mean of `y_(1), y_(2),…. y_(n)`
` "So, " bar(x)=((x_(1)+ x_(2)+x_(3)+…+ x_(n)))/(n)`
`rArr x_(1)+ x_(2)+x_(3)+…+ x_(n)=nbar(x)`
`"and " bar(y)=((y_(1)+ y_(2)+y_(3)+…+ y_(n)))/(n)`
`rArr y_(1)+ y_(2)+y_(3)+…+ y_(n)=nbar(y)`
If `bar(z)` is the mean of `x_(1), x_(2),…., x_(n),y_(1), y_(2),…., y_(n),` then ,
`bar(z) = (nbar(x)+nbar(y))/(n+n) = (n(bar(x)+bar(y)))/(2n) = ((bar(x)+bar(y)))/(2)`
Hence, (b) is the correct answer.