A) \[n\]
B) \[\frac{n}{2}\]
C) \[\frac{n}{4}\]
D) None of these
Correct Answer: B
Solution :
The mean of the observation \[1,2,.....,n\] is \[\bar{X}=\frac{1+2+....+n}{n}=\frac{n(n+1)}{n.2}\] \[\bar{X}=\frac{(n+1)}{2}\] Now, the mean deviation (MD) \[=\frac{\sum\limits_{i=1}^{n}{{{f}_{i}}|{{x}_{i}}-\bar{X}|}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\] Here, \[\sum\limits_{i=1}^{n}{{{f}_{i}}=(1+1+......n\,times=n}\] \[1.\left| 1-\frac{(n+1)}{2} \right|+1\left| 2-\frac{(n+1)}{2} \right|\] \[+1\left| 3-\frac{(n+1)}{2} \right|+....\] \[=\frac{\begin{align} & \left| \frac{1-n}{2} \right|+\left| \frac{3-n}{2} \right|+\left| \frac{5-n}{2} \right|+.....+\left| -\frac{1}{2} \right|+... \\ & +\left| \frac{n-3}{2} \right|+\left| \frac{n-1}{2} \right| \\ \end{align}}{n}\] \[=\frac{\left| \frac{n-1}{2} \right|+\left| \frac{n-3}{2} \right|+\left| \frac{1}{2} \right|+.....+\left| \frac{n-3}{2} \right|+\left| \frac{n-1}{2} \right|}{n}\] \[=\frac{2}{n}\,\,\left\{ \frac{(n-1)}{2}+\frac{(n-3)}{2}+....+\frac{1}{2} \right\}\] \[=\frac{1}{n}\,\{(n-1)+(n-3)+....+1\}\] \[=\frac{1}{n}[1+3+5+...+(n-1)]\] \[=\frac{1}{n}\times \left[ \frac{n}{2}\,(1+n-1) \right]\] \[=\frac{1}{n}\times \frac{{{n}^{2}}}{2}=\frac{n}{2}\]
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