Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Statistics
0 votes

Calculate the mean deviation about the mean of the set of first n natural numbers where numbers where n is an even number.

$\begin{array}{1 1}(A)\;\large\frac{n}{4}\\(B)\;(n+1)\\(C)\;0\\(D)\;\large\frac{n-1}{12}\end{array} $

Can you answer this question?

1 Answer

0 votes
Mean of first n natural numbers where n is an even number is $= \large\frac{1}{2} \bigg[ \frac{n}{2} +\frac{n+1}{2} \bigg]$
$\qquad= \large\frac{2n+1}{4}$
mean deviation $=\large\frac{\sum |x_i -x_n|}{n}$
$\qquad= \large\frac{1}{4n} $$ [1+3+5+9+.......+nth\; odd number]$
now sum of odd numbers $=n^2 $
$\qquad= \large\frac{n}{4}$
Hence A is the correct answer.
answered Jul 9, 2014 by meena.p
how can the mean of first n even natural numbers be (2n+1)/4.
eg. 1, 2, 3, 4, 5, 6 are first 6(even) natural numbers.
mean= (1+2+3+4+5+6)/6 =  21/6 = 3.5  -----------  (1)

now (2*6 + 1)/4 = 13/4 = 3.25   ----------------------- (2)

clearly 1&2 are not equal

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App