Mathematics - Statistics Question with Solution | TestHub

Mean $\stackrel{-}{x}=\frac{1}{101}\left[1+\left(1+d\right)+\left(1+2d\right)\right]\dots .\left(1+100d\right)$

$=\frac{1}{101} \times \frac{101}{2} \left[1+\left(1+100d\right)\right]=1+50d$

Mean deviation from mean

$=\frac{1}{101}\left\{\left|1-\left(1+50d\right)\right|+\left|\left(1+d\right)-\left(1+50d\right)\right|\dots .\left|\left[1+100d\right]-\left(1+50d\right)\right|\right\}$

$=\frac{2\left|d\right|}{101} \left(1+2+3\dots .+50\right)$

$=\frac{2\left|d\right|}{101} \frac{\left(50\right) \left(50+1\right)}{2}$

$=\frac{2550}{101}\left|d\right|=255$

$\Rightarrow \left|d\right|=10.1$