Mathematics - Statistics Question with Solution | TestHub

$\sum _{i=1}^{18}\left(x_i-\alpha \right)=36,\sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90$

$\Rightarrow \sum _{i=1}^{18}{x}_i=18\left(\alpha +2\right),\sum _{i=1}^{18}{x}_i^2-2\beta \sum _{i=1}^{18}{x}_i+18{\beta }^2=90$

Hence $\sum x_i^2=90-18{\beta }^2+36\beta \left(\alpha +2\right)$

Given $\frac{\sum x_i^2}{18}-{\left(\frac{\sum x_i}{18}\right)}^2=1$

$\Rightarrow 90-18{\beta }^2+36\beta \left(\alpha +2\right)-18{\left(\alpha +2\right)}^2=18$

$\Rightarrow 5-{\beta }^2+2\alpha \beta +4\beta -{\alpha }^2-4\alpha -4=1$

$\Rightarrow {\left(\alpha -\beta \right)}^2+4\left(\alpha -\beta \right)=0\Rightarrow \left|\alpha -\beta \right|=0$ or $4$

As $\alpha$ and $\beta$ are distinct $|\alpha -\beta |=4$