Mathematics - Statistics Question with Solution | TestHub

Consider the data $1,2,4,5,x,y$.

So,

$\mathrm{Mean}=\frac{1+2+4+5+x+y}{6}$

$\Rightarrow 5=\frac{x+y+12}{6}$

$\Rightarrow x+y=18$

And,

$\mathrm{Variance}=\left(\frac{\sum _{}{x}_i^2}{n}\right)-{\left(\frac{\sum _{}{x}_i}{n}\right)}^2$

$\Rightarrow 10=\left(\frac{1^2+2^2+4^2+5^2+x^2+y^2}{6}\right)-{\left(5\right)}^2$

$\Rightarrow x^2+y^2=164$

We have $x+y=18$ and $x^2+y^2=164$.

Now solving above equations we get, $x=10, y=8$,

Now mean deviation about mean will be $=\frac{{\displaystyle \sum _{i=1}^6}\left|x_i-\overline{x}\right|}{6}$

$=\frac{\left|1-5\right|+\left|2-5\right|+\left|4-5\right|+\left|5-5\right|+\left|10-5\right|+\left|8-5\right|}{6}$

$=\frac{16}{6}$

$=\frac{8}{3}$

Therefore, the required answer is $\frac{8}{3}$