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

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

Answers (1)

Set of first n natural numbers when n is an even 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 an even 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 = \left( \frac{1}{2}+\frac{3}{2}+ \ldots \ldots .+\frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2} \right) ~ \left( \frac{n}{2} \right)


\\\text{We know that sum of first n natural numbers }= \( n^{2} \) \\ \vspace{\baselineskip} \\ \text{ Therefore, mean deviation about the mean is }\frac{ \Sigma \vert x_{i}-\overline{x}~ \vert }{n}=\frac{ \left( \frac{1}{2}+\frac{3}{2}+ \ldots ..+\frac{n-1}{2} \right) \left( \frac{n}{2} \right) }{n}~ \\\\ \\ =\frac{ \left( \frac{n}{2} \right) ^{2}}{n}=\frac{n^{2}}{4n}=\frac{n}{4} \\\\

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