JEE Main 2023 — Statistics Question with Solution

Given,

$a_1$  be any natural number 

So, $a_1,a_1+1,a_1+2,\dots .,a_1+99$ are values of $a_i$ or observation.

Now mean of the observation is given by,

$\overline{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\dots .+a_1+99}{100}$

$\Rightarrow \overline{x}=\frac{100a_1+(1+2+\dots ..+99)}{100}$

$\Rightarrow \overline{x}=a_1+\frac{99\times 100}{2\times 100}$

$\Rightarrow \overline{x}=a_1+\frac{99}{2}$

Mean deviation about mean$=\frac{{\sum }_{\mathrm{i}=1}^{100}\left|x_{\mathrm{i}}-\overline{\mathit{x}}\right|}{100}$

$=\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\dots .+\frac{1}{2}\right)}{100}$

$=\frac{1+3+\dots .+99}{100}$

$=\frac{\frac{50}{2}[1+99]}{100}$

$=25$

So, it is true for every natural no. ' $a_1$ ' as there is no term of $a_1$ in mean deviation.