Statistics
Statistics
nta_pyq_2025_jan
Grade 11
Question:
Let $x_{1},x_{2},\dots,x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_{i}-2)=30$, $\sum_{i=1}^{10}(x_{i}-\beta)^{2}=98,\ \beta>2$, and their variance is $\dfrac{4}{5}$. If $\mu$ and $\sigma^{2}$ are respectively the mean and the variance of $2(x_{1}-1)+4\beta,\ 2(x_{2}-1)+4\beta,\dots,2(x_{10}-1)+4\beta$, then $\dfrac{\beta\mu}{\sigma^{2}}$ is equal to:
Step-by-Step Solution
Key Concept: From the first sum find $\bar{x}=5$. Variance $\tfrac{4}{5}$ gives $\sum x_{i}^{2}=258.$ Then $\sum(x_{i}-\beta)^{2}=\sum x_{i}^{2}-2\beta\sum x_{i}+10\beta^{2}=98$ is a quadratic in $\beta.$ Linear-transformation rules: $y=2x+c\Rightarrow \mu_{y}=2\bar{x}+c,\ \sigma_{y}^{2}=4\sigma_{x}^{2}.$
$\sum(x_{i}-2)=\sum x_{i}-20=30\Rightarrow \sum x_{i}=50,\ \bar{x}=5.$
Variance $\dfrac{4}{5}=\dfrac{\sum x_{i}^{2}}{10}-25\Rightarrow \sum x_{i}^{2}=258.$
$\sum(x_{i}-\beta)^{2}=258-100\beta+10\beta^{2}=98\Rightarrow \beta^{2}-10\beta+16=0\Rightarrow \beta=2,\,8.$ Pick $\beta=8.$
Transformed data $y_{i}=2(x_{i}-1)+4\beta=2x_{i}+30$:
$\mu=2\bar{x}+30=40,\ \sigma^{2}=4\cdot\dfrac{4}{5}=\dfrac{16}{5}.$
$$\dfrac{\beta\mu}{\sigma^{2}}=\dfrac{8\cdot 40}{16/5}=\dfrac{320\cdot 5}{16}=100.$$
Correct Answer: 1