# Let $X_1,X_2,......X_{18}$ be eighteen observation such that $\sum_{i=1}^{18}(X_{i}-\alpha )=36$ and $\sum_{i=1}^{18}(X_{i}-\beta )^2=90$,  where $\alpha \text{ and } \beta$ are distinct real numbers. If the standard deviation of these observation is 1,  then the value of $\left |\alpha - \beta \right |$ is ________ Option: 1 4 Option: 2 6 Option: 3 2 Option: 4 3

$\\\sum_{\mathrm{i}=1}^{18}\left(\mathrm{x}_{\mathrm{i}}-\alpha\right)=36, \sum_{\mathrm{i}=1}^{18}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^{2}=90 \\ \Rightarrow \sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}=18(\alpha+2), \\\sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}^{2}-2 \beta \sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}+18 \beta^{2}=90 \\ \text { Hence } \quad \sum \mathrm{x}_{\mathrm{i}}^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$

$\text { Given } \frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{18}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{18}\right)^{2}=1$

\begin{aligned} &\Rightarrow 90-18 \beta^{2}+36 \beta(\alpha+2)-18(\alpha+2)^{2}=18\\ &\Rightarrow 5-\beta^{2}+2 \alpha \beta+4 \beta-\alpha^{2}-4 \alpha-4=1\\ &\Rightarrow(\alpha-\beta)^{2}+4(\alpha-\beta)=0 \Rightarrow|\alpha-\beta|=0 \text { or } 4\\ &\text { As } \alpha \text { and } \beta \text { are distinct }|\alpha-\beta|=4 \end{aligned}

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