Statistics
Mean and Variance — Finding β/α
nta_pyq_2024_jan
Grade 11
Question:
Consider 10 observations $x_1,x_2,\ldots,x_{10}$ such that $\displaystyle\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\displaystyle\sum_{i=1}^{10}(x_i-\beta)^2=40$, where $\alpha,\beta$ are positive integers. Let the mean and the variance of the observations be $\dfrac{6}{5}$ and $\dfrac{84}{25}$ respectively. Then $\dfrac{\beta}{\alpha}$ is equal to:
2
$\dfrac{3}{2}$
$\dfrac{5}{2}$
1
Step-by-Step Solution
Key Concept: $\sum(x_i-\alpha)=2\Rightarrow\sum x_i=10\alpha+2=12$ (using mean $6/5$). So $\alpha=1$. $\sigma^2=\frac{1}{10}\sum(x_i-\beta)^2-\left(\frac{\sum(x_i-\beta)}{10}\right)^2=\frac{84}{25}$. Solve for $\beta$.
$\alpha=1$, $\beta=2$. $\beta/\alpha=2$.
Correct Answer: 1