The mean deviation about the mean of the set of first ' \(n\) ' natural numbers, when ' \(n\) ' is an even…
The mean deviation about the mean of the set of first ' \(n\) ' natural numbers, when ' \(n\) ' is an even number is equal to
\(n\)
\(\frac{n}{2}\)
\(\frac{n}{3}\)
\(\frac{n}{4}\)
Solution
Mean \(\bar{x}=\frac{1+2+\ldots . n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2}\)
Here, \(n=\) even number
Now mean deviation is
\(\begin{aligned}
& M . D(\bar{x})=\left[\left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right| +\ldots .+\left|n-\frac{n+1}{2}\right|\right] \times \frac{1}{n} \\
&=\frac{2}{n}\left\{1+2+\ldots .+\left(\frac{n}{2}\right)\right\} \frac{n}{2} \text { terms } \\
&=\frac{2}{n}\left[\frac{\frac{n}{2}\left(\frac{n}{2}-1+1\right)}{2}\right]=\frac{n}{4}
\end{aligned}\)