Let ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\dots \dots {\mathrm{x}}_{18}$be observations such that $\sum _{\mathrm{i}=1}^{18} \left({\mathrm{x}}_{\mathrm{i}}-\mathrm{\alpha }\right)=36,\sum _{\mathrm{i}=1}^{18} {\left({\mathrm{x}}_{\mathrm{i}}-\mathrm{\beta }\right)}^2=90$ Where $\mathrm{\alpha },\mathrm{\beta }$are distinct real numbers if the S.D of these observations is 1 then the value of $|\mathrm{\alpha }-\mathrm{\beta }|$is 

$$\begin{gathered} \begin{array}{l}\sum _{\mathrm{i}=1}^{18} \left({\mathrm{x}}_{\mathrm{i}}-\mathrm{\alpha }\right)=36\Rightarrow \sum _{\mathrm{i}=1}^{18} {\mathrm{x}}_{\mathrm{i}}-18\mathrm{\alpha }=36 \\ \Rightarrow \sum _{\mathrm{i}=1}^{18} {\mathrm{x}}_{\mathrm{i}}=36+18\mathrm{\alpha }=18(2+\mathrm{\alpha }) \\ \sum _{\mathrm{i}=1}^{18} {\left({\mathrm{x}}_{\mathrm{i}}-\mathrm{\beta }\right)}^2=90\\ \Rightarrow \sum _{\mathrm{i}=1}^{18} \left({\mathrm{x}}_{\mathrm{i}}^2+{\mathrm{\beta }}^2-2{\mathrm{x}}_{\mathrm{i}}\mathrm{\beta }\right)=90\Rightarrow \sum _{\mathrm{i}=1}^{18} {\mathrm{x}}_{\mathrm{i}}^2+18{\mathrm{\beta }}^2-2\mathrm{\beta }\sum {\mathrm{x}}_{\mathrm{i}}=90 \\ \Rightarrow \sum _{\mathrm{i}=1}^{18} {\mathrm{x}}_{\mathrm{i}}^2=90-18{\mathrm{\beta }}^2+36\mathrm{\beta }(\mathrm{\alpha }+2)\end{array} \end{gathered}$$
$$\begin{gathered} \begin{array}{l}\Rightarrow \sum _{i=1}^{18} x_i^2=90-18{\beta }^2+36\beta \alpha +72\beta \text{ S.D }\left(\sigma \right)=1\Rightarrow {\sigma }^2=1 \\ \Rightarrow \frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2=1\\ \Rightarrow \frac{1}{18}\left(90-18{\beta }^2+36\beta \alpha +72\beta \right)-{\left(\frac{18(\alpha +2)}{18}\right)}^2=1 \\ \Rightarrow 90-18{\beta }^2+36\beta \alpha +72\beta -18(\alpha +2{)}^2=18\\ \Rightarrow 5-{\beta }^2+2\beta \alpha +4\beta -(\alpha +2{)}^2=1 \\ \Rightarrow -{\alpha }^2-{\beta }^2+2\beta \alpha -4\alpha +4\beta =0\\ \Rightarrow {\alpha }^2+{\beta }^2-2\beta \alpha +4\alpha -4\beta =0 \\ \Rightarrow (\alpha -\beta {)}^2+4(\alpha -\beta )=0\Rightarrow (\alpha -\beta \left)\right(\alpha -\beta +4)=0\\ \Rightarrow \alpha -\beta +4=0(\alpha \ne \beta )\Rightarrow \alpha -\beta =-4\Rightarrow |\alpha -\beta |=|-4|=4\end{array} \end{gathered}$$