detailed solution

Correct option is D

We have,

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

$\Rightarrow \sum _{i=1}^{18} x_i=36+18\alpha$ and $\sum _{i=1}^{18} x_i^2-2\beta \sum _{i=1}^{18} x_i+18{\beta }^2=90$

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

$\Rightarrow \sum _{i=1}^{18} x_i^2=90+36\alpha \beta +72\beta -18{\beta }^2$ and $\sum _{i=1}^{18} x_i=36+18\alpha$

$\Rightarrow \frac{1}{18}\sum _{i=1}^{18} x_i=2+\alpha$ and $\frac{1}{18}\sum _{i=1}^{18} x_i^2=5+2\alpha \beta +4\beta -{\beta }^2$

Now, Variance= 1

$\begin{array}{l}\Rightarrow \frac{1}{18}\sum _{i=1}^{18} x_i^2-{\left(\frac{1}{18}\sum _{i=1}^{18} x_i\right)}^2=1\\ \Rightarrow 5+2\alpha \beta +4\beta -{\beta }^2-(2+\alpha {)}^2=1\\ \Rightarrow 4+2\alpha \beta +4\beta -{\beta }^2-4-4\alpha -{\alpha }^2=0\\ \Rightarrow {\alpha }^2+{\beta }^2-2\alpha \beta +4(\alpha -\beta )=0\\ \Rightarrow (\alpha -\beta {)}^2+4(\alpha -\beta )=0\\ \Rightarrow (\alpha -\beta )(\alpha -\beta +4)=0\Rightarrow \alpha -\beta =-4\Rightarrow |\alpha -\beta |=4\end{array}$

detailed solution

Correct answer is 4

We have,

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

$\Rightarrow \sum _{i=1}^{18} x_i=36+18\alpha$ and $\sum _{i=1}^{18} x_i^2-2\beta \sum _{i=1}^{18} x_i+18{\beta }^2=90$

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

$\Rightarrow \sum _{i=1}^{18} x_i^2=90+36\alpha \beta +72\beta -18{\beta }^2$ and $\sum _{i=1}^{18} x_i=36+18\alpha$

$\Rightarrow \frac{1}{18}\sum _{i=1}^{18} x_i=2+\alpha$ and $\frac{1}{18}\sum _{i=1}^{18} x_i^2=5+2\alpha \beta +4\beta -{\beta }^2$

Now, Variance= 1

$\begin{array}{l}\Rightarrow \frac{1}{18}\sum _{i=1}^{18} x_i^2-{\left(\frac{1}{18}\sum _{i=1}^{18} x_i\right)}^2=1\\ \Rightarrow 5+2\alpha \beta +4\beta -{\beta }^2-(2+\alpha {)}^2=1\\ \Rightarrow 4+2\alpha \beta +4\beta -{\beta }^2-4-4\alpha -{\alpha }^2=0\\ \Rightarrow {\alpha }^2+{\beta }^2-2\alpha \beta +4(\alpha -\beta )=0\\ \Rightarrow (\alpha -\beta {)}^2+4(\alpha -\beta )=0\\ \Rightarrow (\alpha -\beta )(\alpha -\beta +4)=0\Rightarrow \alpha -\beta =-4\Rightarrow |\alpha -\beta |=4\end{array}$