Question

Let $X_1,X_2.....X_{18}$ be eighteen observations 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 $\alpha \text{and}\beta$ are distinct real numbers.

If the standard deviation of these observations is $1$, then the value of $\left|\alpha -\beta \right|$ is

Answer 1

Determining the value of $\left|\alpha -\beta \right|$.

It is given

$$\begin{gathered} \sum _{i=1}^{18}\left(X_i-\alpha \right)=36 \\ \Rightarrow \sum _{i=1}^{18}{x}_i=18(\alpha +2)...\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \sum _{i=1}^{18}{\left(X_i-\beta \right)}^2=90 \\ \underset{i=1}{\overset{18}{\Rightarrow \sum }}{x_i}^2-2\beta \sum _{i=1}^{18}{x}_i+18{\beta }^2=90 \\ \Rightarrow \sum {x_i}^2=90-18{\beta }^2+36\beta (\alpha +2)...[from(1\left)\right] \end{gathered}$$

We have been given that the standard deviation of these observations is one

$$\begin{gathered} \therefore \frac{\sum _{}{x_i}^2}{18}-{\left(\frac{\sum _{}{x}_i}{18}\right)}^2=1 \\ \Rightarrow 90-18{\beta }^2+36\beta (\alpha +2)-18{(\alpha +2)}^2=18\Rightarrow 5-{\beta }^2+2\alpha \beta +4\beta -{\alpha }^2-4\alpha -4=1 \\ \Rightarrow {(\alpha -\beta )}^2+4(\alpha -\beta )=0 \\ \Rightarrow \left|\alpha -\beta \right|=0or4 \\ \text{As}\alpha \text{and}\beta \text{are distinct}|\alpha -\beta |=4 \end{gathered}$$

Hence, the value of $\left|\alpha -\beta \right|=4$.