Question

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

Answer 1

$\underset{i=1}{\overset{18}{\Sigma}}\left( x _{ i }-\alpha\right)=36, \underset{i=1}{\overset{18}{\Sigma}}\left( x _{ i }-\beta\right)^{2}=90$
$\Rightarrow \underset{i=1}{\overset{18}{\Sigma}} x_{i}=18(\alpha+2),\underset{i=1}{\overset{18}{\Sigma}}x_{i}^{2}-2 \beta \underset{i=1}{\overset{18}{\Sigma}} x_{i}+18 \beta^{2}=90$
Hence $\sum x _{ i }^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$
Given $\frac{\sum x _{ i }^{2}}{18}-\left(\frac{\sum x _{ i }}{18}\right)^{2}=1$
$\Rightarrow 90-18 \beta^{2}+36 \beta(\alpha+2)-18(\alpha+2)^{2}=18$
$\Rightarrow 5-\beta^{2}+2 \alpha \beta+4 \beta-\alpha^{2}-4 \alpha-4=1$
$\Rightarrow (\alpha-\beta)^{2}+4(\alpha-\beta)=0 \Rightarrow |\alpha-\beta|=0$ or 4
As $\alpha$ and $\beta$ are distinct $|\alpha-\beta|=4$