Question

Calculate the mean deviation about the mean of the set of firts n natural numbers when n is an odd number.

Answer 1

First n natural numbers are :
1,2,3,4, ... n (odd)

Let n = 2m + 1, \( m \in \mathbb{N} \)

\( \therefore Mean (\overline x ) = \dfrac{1+2+3+ ... +(2m+1)}{(2m+1)} \)

\( \Rightarrow \overline x = \dfrac{1}{(2m+1)} \left [ \dfrac{(2m+1)(2m+1)+1}{2} \right ] \)

\( \left [ \because \text{Sum of first n natural numbers } S_{n} = \dfrac{n(n+1)}{2} \right ] \)

\( \Rightarrow \overline x = m+1 \)

\( \Rightarrow \overline x = m+1 \)
Deviation about mean

\( = \sum_{i=1}^{i=(2m+1)} \left |d_{i}\right | \)

\( = \sum_{i=1}^{i=(2m+1)} \left |x_{i}-\overline x\right | \)

\( = \sum_{i=1}^{i=m} \left | \overline x-x_{i}\right | + \sum_{i=m+1}^{i=(2m+1)} \left |x_{i}-\overline x\right | \)

\( = \left [ m(m+1) - \dfrac{m(m+1)}{2} \right ]+ \left [ \dfrac{m+1}{2}([m+1] +[2m+1]) -(m+1)(m+1) \right ] \)

\( \left [ \because S_{n(A.P.)}= \dfrac{n}{2} (a+l) \right ] \)


\( = \sum_{i=1}^{i=(2m+1)} \left |d_{i}\right | = (m+1) \left [m-\dfrac{m}{2}+\dfrac{1}{2}(3m+2)-(m+1) \right ] \)

\( = (m+1) \left [\dfrac{m}{2}+\dfrac{3m}{2}+1-m-1 \right ] \)

\( = (m+1)[2m-m] \)

\(= (m+1)m \)

Now, mean deviation about mean

\( M.D. = \dfrac{\sum_{i=1}^{i=(2m+1)} \left |d_{i}\right |}{2m+1} \)

\( M.D. = \dfrac{(m)(m+1)}{2m+1} \)


\( M.D. = \dfrac{(m)(m+1)}{2m+1} \)
Putting back n = 2m+1, we get

\( M.D. = \dfrac{\left ( \dfrac{n-1}{2} \right ) \left ( \dfrac{n-1}{2}+1 \right )}{n} \)


\( M.D. = \dfrac{n^{2}-1}{4n} \)