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 ?

Correct answer: B. 11 x y x y x^2 -4 1 -4 16 -1 2 -2 1 2 7 14 4 3 1 3 9 aligned and =0, y=11, x y=11 x² and =30 x and =(0)/(4)=0, y =(11)/(4) b_ y x and =( x y-( x y)/(4))/( x^2-(( x)^2)/(4)) and =(11-0)/(30-0)=(11)/(30) aligned Equation of regression line of y on x will be array rlrl and and y- y and =b_ y x (x- x ) and y-(11)/(4) and =(11)/(30)(x-0) and y and =(11)/(30)x+(11)/(4) array aligned & =(11)/(4) and b=(11)/(30) 2 a+15 b & =2 × (11)/(4)+15 × (11)/(30)=11 aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Statistics

NDA 2024 · Mathematics · Statistics

From data (-4,1),(-1,2),(2,7) and (3,1) , the regression line of y...

From data

(-4,1),(-1,2),(2,7)

and

(3,1)

, the regression line of

y

x

is obtained as

y=a+b x

, then what is the value of

2 a+15 b

Options

Correct answer

B. 11

Step-by-step solution

x	y	x y	x^2
-4	1	-4	16
-1	2	-2	1
2	7	14	4
3	1	3	9

\begin{aligned} \Sigmax \text{ and } =0, \Sigma y=11, \Sigma x y=11 \\ \Sigma x² \text{ and } =30 \\ \bar{x} \text{ and } =\frac{0}{4}=0, \bar{y}=\frac{11}{4}\\ b_{y x} \text{ and } =\frac{\Sigma x y-\frac{\Sigma x \Sigma y}{4}}{\Sigma x^2-\frac{(\Sigma x)^2}{4}}\\ \text{ and } =\frac{11-0}{30-0}=\frac{11}{30}\end{aligned}

Equation of regression line of

y

x

will be

\begin{array}{rlrl} \text{ and } \text{ and } y-\bar{y}\text{ and } =b_{y x}(x-\bar{x}) \\ \Rightarrow \text{ and } y-\frac{11}{4} \text{ and } =\frac{11}{30}(x-0) \\ \Rightarrow \text{ and } y \text{ and } =\frac{11}{30}x+\frac{11}{4}\end{array}

\begin{aligned} \therefore \quada & =\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}

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