Prove the following

Question:

Let $X_{1}, \quad X_{2}, \ldots \ldots . . . X_{18}$ be eighteen observation such that $\sum_{i=1}^{18}\left(X_{i}-\alpha\right)=36$ 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:

Given, $\sum_{i=1}^{18}\left(X_{i}-\alpha\right)=36$

$\Rightarrow \sum x_{i}-18 \alpha=36$

$\Rightarrow \sum x_{i}-18(\alpha+2) \ldots(1)$

Also, $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90$

$\Rightarrow \sum x_{i}^{2}+18 \beta^{2}-2 \beta \sum x_{i}=90$

$\Rightarrow \sum x_{i}^{2}+18 \beta^{2}+2 \beta \times 18(\alpha+2)=90 \quad$ (using equation (1))

$\Rightarrow \sum \mathrm{x}_{\mathrm{i}}^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$

$\sigma^{2}=1 \Rightarrow \frac{1}{18} \sum x_{i}^{2}-\left(\frac{\sum x_{i}}{18}\right)^{2}=1 \quad(\because \sigma=1$, given $)$

$\Rightarrow \frac{1}{18}\left(90-18 \beta^{2}+36 \alpha \beta+72 \beta\right)-\left(\frac{18(\alpha+2)}{18}\right)^{2}=1$

$\Rightarrow 90-18 \beta^{2}+36 \alpha \beta+72 \beta-18(\alpha+2)^{2}=18$

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

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

$\Rightarrow \alpha^{2}-\beta^{2}+2 \alpha \beta+4 \beta-4 \alpha=0$

$\Rightarrow(\alpha-\beta)(\alpha-\beta+4)=0$

$\Rightarrow \alpha-\beta=-4$

$\therefore|\alpha-\beta|=4 \quad(\alpha \neq \beta)$