Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.

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

Answers (1)

Set of first n natural numbers when n is an odd number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is odd number.

So, mean is

\\ \\ \overline{x}=\frac{1+2+3+ \ldots ..+n}{n}=\frac{\frac{n \left( n+1 \right) }{2}}{n}=\frac{n+1}{2}~ \\\\ \\ \text{~~The deviations of numbers from the mean are as shown below, } \\\\ \\ ~ 1-\frac{n+1}{2},~ 2-\frac{n+1}{2} , \ldots \ldots \ldots \ldots \left( n-1 \right) -\frac{n-1}{2}~,~ n-\frac{n+1}{2}~ \\\\ \\ ~Or,\frac{2- \left( n+1 \right) }{2}~,\frac{4- \left( n+1 \right) }{2} \ldots \ldots \ldots . \frac{2 \left( n-1 \right) - \left( n+1 \right) }{2},~\frac{2n- \left( n+1 \right) }{2}~ \\\\


\\ \\ \text{~~~ Or, }\frac{1-n}{2},\frac{3-n}{2}~,~\frac{5-n}{2} \ldots \ldots .\frac{n-3}{2}~,\frac{n-1}{2}~~~ \\\\ \\ ~~\frac{- \left( n-1 \right) }{2},~\frac{- \left( n-3 \right) }{2},\frac{- \left( n-5 \right) }{2} , \ldots \ldots , \frac{n-5}{2},\frac{n-3}{2},\frac{n-1}{2} \\\\ \\ \text{~ So the absolute values of deviation from the mean is } \vert x_{i}-\overline{x}~ \vert =\frac{n-1}{2},~\frac{n-3}{2}~,\frac{n-5}{2}, \ldots \ldots .\frac{n-5}{2},\frac{n-3}{2},\frac{n-1}{2}~~ \\\\ \\ \text{~ The sum of absolute values of deviations from the mean, is } \Sigma \vert x_{i}-\overline{x}~ \vert =\frac{n-1}{2}+\frac{n-3}{2}+\frac{n-5}{2}+ \ldots \ldots \frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2}~~~~~~ \\\\


\\ \\ \Sigma \vert x_{i}-\overline{x}~ \vert =2 \left( 1+2+3+ \ldots \ldots .+\frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2} \right) ~ \\\\ \\ \text{~ That is 2 times the sum of }\frac{n-1}{2}\text{terms,~so it can be written as } \Sigma \vert x_{i}-\overline{x}~ \vert =2 \left( \frac{\frac{n-1}{2} \left( \frac{n-1}{2}+1 \right) }{2} \right) =n \left( \frac{n-1}{2} \right) \left( \frac{n+1}{2} \right) = \left( \frac{n^{2}-1}{4} \right) ~ \\\\ \\ \text{~~ Therefore, mean deviation about the mean is }\frac{ \Sigma \vert x_{i}-\overline{x}~ \vert }{n}=\frac{ \left( \frac{n^{2}-1}{4} \right) }{n}= \left( \frac{n^{2}-1}{4n} \right) ~~ \\\\

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question