Statistics
Mean and variance conditions — finding alpha minus beta
MJAT_TS5_P2
Grade 12
Question:
**Paragraph:** Let $X_1,X_2,\ldots,X_{18}$ be 18 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:
A) 2\quad B) 4\quad C) 5\quad D) 8
Step-by-Step Solution
Key Concept: $\sum(X_i-\alpha)=36\Rightarrow\sum X_i-18\alpha=36\Rightarrow\bar{X}=\alpha+2$. Variance $=1$: $\frac{1}{18}\sum(X_i-\bar{X})^2=1$. Expand $\sum(X_i-\beta)^2=90$: $\sum X_i^2-2\beta\sum X_i+18\beta^2=90$.
$|\alpha-\beta|=\mathbf{4}$. Answer: B.
Correct Answer: B