Question

Let $X_1,X_2,\dots ,X_{18}$ be eighteen observations such that $\underset{i=1}{\stackrel{18}{\Sigma }}(X_i-\alpha )=36$ and $\underset{i=1}{\stackrel{18}{\Sigma }}{\left(X_i\beta \right)}^2=90,$ 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

Answer: 4

$\underset{i=1}{\stackrel{18}{\Sigma }}\left(x_i-\alpha \right)=36,\underset{i=1}{\stackrel{18}{\Sigma }}{\left(x_i-\beta \right)}^2=90$
$\Rightarrow \underset{i=1}{\stackrel{18}{\Sigma }}{x}_i=18(\alpha +2),\underset{i=1}{\stackrel{18}{\Sigma }}{x}_i^2-2\beta \underset{i=1}{\stackrel{18}{\Sigma }}{x}_i+18{\beta }^2=90$
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 $|\alpha -\beta |=4$