Q253 / 270 · NDA 2024 (Phase 1) · Mathematics
From data $(-4,1),(-1,2),(2,7)$ and $(3,1)$, the regression line of $y$ on $x$ is obtained as $y=a+$ $b x$, then what is the value of $2 a+15 b$ ?
- A6
- Correct answerB11
- C17
- D21
Solution
$\begin{array}{|c|c|c|c|}\hline x & y & x y & x^2 \\\hline -4 & 1 & -4 & 16 \\\hline -1 & 2 & -2 & 1 \\\hline 2 & 7 & 14 & 4 \\\hline 3 & 1 & 3 & 9 \\\hline\end{array}$
$\begin{aligned} \Sigma x & =0, \Sigma y=11, \Sigma x y=11 \\ \Sigma x^2 & =30 \\ \bar{x} & =\frac{0}{4}=0, \bar{y}=\frac{11}{4} \\ b_{y x} & =\frac{\Sigma x y-\frac{\Sigma x \Sigma y}{4}}{\Sigma x^2-\frac{(\Sigma x)^2}{4}} \\ & =\frac{11-0}{30-0}=\frac{11}{30}\end{aligned}$
Equation of regression line of $y$ on $x$ will be
$\begin{array}{rlrl} & & y-\bar{y} & =b_{y x}(x-\bar{x}) \\ \Rightarrow & y-\frac{11}{4} & =\frac{11}{30}(x-0) \\ \Rightarrow & y & =\frac{11}{30} x+\frac{11}{4}\end{array}$
$\begin{aligned} \therefore \quad a & =\frac{11}{4} \text { and } b=\frac{11}{30} \\ 2 a+15 b & =2 \times \frac{11}{4}+15 \times \frac{11}{30}=11\end{aligned}$
Hence, the correct answer is option: 'B'.