Calculate the mean deviation about the mean of the set of first `n` natural numbers when `n` is odd natural number.

Options

A. `(2n^(2)-1)/(4n)`
B. `(n^(2)-1)/(4n)`
C. `(n^(2)-2)/(4n)`
D. none of these

Answer

B

Explanation

Consider first natural number when n is an odd ie.1,2,3,4,…n [odd]
Mean `barx=(1+2+3+…….+n)/(n)=(n(n+1))/(2n)=(n+1)/(2)`
`therefore` MD=`(abs(1-(n+1)/2)+abs(2-(n+1)/(2))+abs(3-(n+1)/(2))+....+abs(n-(n+1)/(2)))/n`
`= (+abs((n+1)/(2)-(n+1)/(2))+abs((n+3)/(2)-(n+1)/(2))+....+abs((2n-2)/(2)-(n+1)/(2))+abs(n-(n+1)/(2)))/(n)`
`=(2)/(n)[1+2+....+(n-3)/(2)+(n-1)/(2)] ((n-1)/(2))` term
`=2/n[((n-1)/(2))((n-1)/(2)+1)] " " [because "sum of first n natural numbers" n(n+1)/(2)]`
`=(2)/(n).(1)/(n)[((n-1)/(2))((n+1)/(2))]=(1)/(n)((n^(2)-1)/(4))=(n^(2)-1)/(4n)`