# Solve the Following Questions

Question:

Let $\mathrm{X}_{1}, \mathrm{X}_{2}, \ldots, \mathrm{X}_{18}$ be eighteen observations such that $\sum_{i=1}^{18}\left(X_{i}-\alpha\right)=36 \quad$ and $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90$, where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of $|\alpha-\beta|$ is

Solution:

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

$\Rightarrow \sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}=18(\alpha+2), \sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}^{2}-2 \beta \sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}+18 \beta^{2}=90$

Hence $\sum x_{i}^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$

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

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

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

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

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