304 NCERT Exemplar (Class XI) Solutions

<math>M_{\rm e} = \left(\frac{20+1}{2}\right)</math>th item = <math>\left(\frac{21}{2}\right)</math> = 10.5th item

Now,

<math>M_{\rm e} = 12</math> ∴. <math>MD = \frac{\sum f_i d_i}{\sum f_i} = \frac{25}{20} = 1.25</math> ٠. <math>\mathbf{Q}</math>. 3 Calculate the mean deviation about the mean of the set of first n natural

numbers when n is an odd number.

Sol. Consider first natural number when n is an odd i.e., 1, 2, 3, 4, ...n, [odd].

Mean <math>\bar{x} = \frac{1+2+3+...+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}</math>

<math display="block">MD = \frac{\left| 1 - \frac{n+1}{2} \right| + \left| 2 - \frac{n+1}{2} \right| + \left| 3 - \frac{n+1}{2} \right| + \dots + \left| n - \frac{n+1}{2} \right|}{2}</math> : . <math>\left| -\frac{n+1}{2} \right| + \left| 2 - \frac{n+1}{2} \right| + \dots + \left| \frac{n-1}{2} - \frac{n+1}{2} \right|</math>

<math display="block">= \frac{+\left|\frac{n+1}{2} - \frac{n+1}{2}\right| + \left|\frac{n+3}{2} - \frac{n+1}{2}\right| + \dots + \left|\frac{2n-2}{2} - \frac{n+1}{2}\right| + \left|n - \frac{n+1}{2}\right|}{2}</math>

<math>=\frac{2}{n}\left[1+2+....+\frac{n-3}{2}+\frac{n-1}{2}\right]\left(\frac{n-1}{2}\right)</math> terms

<math display="block">= \frac{2}{n} \left| \frac{\binom{n}{2} \cdot \binom{n}{2} + 1}{2} \right| \qquad \left[ \because \text{sum of first } n \text{ natural numbers} = \frac{n(n+1)}{2} \right]</math>

<math display="block">= \frac{2}{n} \cdot \frac{1}{2} \left[ \left( \frac{n-1}{2} \right) \left( \frac{n+1}{2} \right) \right] = \frac{1}{n} \left( \frac{n^2 - 1}{4} \right) = \frac{n^2 - 1}{4n}</math>

<math>\boldsymbol{Q}_{\boldsymbol{\cdot}}</math> <math>\boldsymbol{4}</math> Calculate the mean of the set of first n natural

numbers when n is an even number.

Sol. Consider first n natural number, when n is even i.e., 1, 2, 3, 4, ......n.

[even]

Mean <math>\overline{x} = \frac{1+2+3+...+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}</math> ∴. <math display="block">MD = \frac{1}{n} \left[ \left| 1 - \frac{n+1}{2} \right| + \left| 2 - \frac{n+1}{2} \right| + \left| 3 - \frac{n+1}{2} \right| \right] + \left| \frac{n-2}{2} - \frac{n+1}{2} \right| + \left| \frac{n}{2} - \frac{n+1}{2} \right|</math>

<math>+\left|\frac{n+2}{2}-\frac{n+1}{2}\right|+...+\left|n-\frac{n+1}{2}\right|</math>

<math>=\frac{1}{n}\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+\dots+\left|\frac{-3}{2}\right|+\left|\frac{1}{2}\right|+\dots+\left|\frac{n-1}{2}\right|</math>

<math>=\frac{2}{n}\left[\frac{1}{2}+\frac{3}{2}+....+\frac{n-1}{2}\right]\left(\frac{n}{2}\right)</math> terms

<math>=\frac{1}{n}\cdot\left(\frac{n}{2}\right)^2</math>

[: sum of first n natural numbers = <math>n^2</math>]

<math display="block">=\frac{1}{n}\cdot\frac{n^2}{4}=\frac{n}{4}</math>