Question

Let \(X_1,X_2,\dots,X_{18}\) be eighteen observations such that
\(\sum_{i=1}^{18}(X_i-\alpha)=36\)
and
\(\sum_{i=1}^{18}(X_i-\beta)^2=90\)
, where \(\alpha\) and \(\beta\) are distinct real numbers. If the standard deviation of these observations is 1, then the value of \(|\alpha-\beta|\) is ____________.

Hint:

Find the mean from the first sum, then use \(\sum(X_i-\beta)^2=\sum(X_i-\bar X)^2+18(\bar X-\beta)^2\).

Answer 1

Correct Answer: 4

\(\sum(X_i-\alpha)=36\Rightarrow18\bar X-18\alpha=36\Rightarrow\bar X=\alpha+2\).
SD \(=1\) with \(n=18\): \(\sum(X_i-\bar X)^2=18\).
\(\sum(X_i-\beta)^2=\sum(X_i-\bar X)^2+18(\bar X-\beta)^2\Rightarrow90=18+18(\bar X-\beta)^2\Rightarrow(\bar X-\beta)^2=4\).
\(\bar X-\beta=\pm2\Rightarrow\alpha+2-\beta=\pm2\Rightarrow\alpha-\beta=0\) or \(-4\). Since \(\alpha\ne\beta\), \(|\alpha-\beta|=4\).
Answer: 4