Let X_1, X_2, , X_18 be eighteen observations, such that _i=1^18(X_i-)=36 and _i=1^18(X_i-…

Question

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

Answer: 4

Accepted Answer

From $\sum(X_i-\alpha)=36$: $\sum X_i = 36+18\alpha$ ...(i)

$\sum(X_i-\beta)^2 = \sum X_i^2+18\beta^2-2\beta\sum X_i = 90$

Using (i): $\sum X_i^2 = 90-18\beta^2+2\beta(36+18\alpha)$

S.D.$=1$ means $\sigma^2=1$:
$$\frac{\sum X_i^2}{18}-\left(\frac{\sum X_i}{18}\right)^2=1$$
$$\frac{1}{18}(90-18\beta^2+36\alpha\beta+72\beta)-(2+\alpha)^2=1$$
$$5-\beta^2+2\alpha\beta+4\beta-(2+\alpha)^2=1$$
$$-\beta^2+2\alpha\beta+4\beta-4-\alpha^2-4\alpha=0$$
$$(\alpha-\beta)^2+4(\alpha-\beta)=0 \implies (\alpha-\beta)(\alpha-\beta+4)=0$$

Since $\alpha\ne\beta$: $\alpha-\beta=-4 \implies |\alpha-\beta|=4$

Solutions · 1