The mean of $5$ observations is $5$ and their variance is $124$. If three of the observations are $1, 2$ and $6$ ; then the mean deviation from the mean of the data is :
$\overline{ x }=\frac{ x _{1}+ x _{2}+ x _{3}+ x _{4}+ x _{5}}{5}=5$ $\displaystyle\sum_{ i =1}^{5} x _{ i }=25 \ldots \ldots \ldots . .( i )$ Also $ \sigma^{2}=124$ $\Rightarrow \frac{\sum x _{1}^{2}}{5}-(\overline{ x })^{2}=124$ $\Rightarrow \frac{\sum x _{1}^{2}}{5}=124+25=149$ $\Rightarrow \left(x_{1}^{2}+x_{2}^{2}+\ldots . .+x_{5}^{2}\right)=745 $ $\Rightarrow x_{1}^{2}+x_{2}^{2}=704 \ldots \ldots \ldots (ii) $ by (i) } $ x_{1}+x_{2}=16 \ldots \ldots \ldots \ldots . (iii) $ $2 x_{1} x_{2}+704=256$ $x_{1} x_{2}=\frac{256-704}{2}$ $x_{1} x_{2}=128-352=-224 \ldots \ldots \ldots \ldots$ (iv) Now $\frac{\sum\left|x_{1}-5\right|}{5}-\frac{\left|x_{1}-5\right|+\left|x_{2}-5\right|+4+3+1}{5}$ $=\frac{8+\left|x_{1}-5\right|+\left|11-x_{1}\right|}{5}$ $=\frac{8+6}{5}=2.8$ Ans
