Well-structured ISC Maths Class 12 Solutions Chapter 2 Linear Regression MCQs facilitate a deeper understanding of mathematical principles.

## ML Aggarwal Class 12 Maths Solutions Section C Chapter 2 Linear Regression MCQs

Choose the correct answer from the given four options in questions (1 to 10) :

Question 1.
If the lines of regression are parallel to the coordinate axes, then the coefficient of correlation is
(a) 1
(b) 0
(c) – 1
(d) $$\frac{1}{2}$$
Solution:
(b) 0

The two lines of regression are given by
y – $$\bar{y}$$ = r $$\frac{\sigma_y}{\sigma_x}(x-\bar{x})$$ ………………(1)
and x – $$\bar{x}$$ = r $$\frac{\sigma_x}{\sigma_y}(y-\bar{y})$$ ………………….(2)
Two lines (1)and (2) are given to be parallel to coordinate axes which is only possible.
When r = 0
[In this case, y = $$\bar{y}$$ and x = $$\bar{x}$$]

Question 2.
If the lines of regression coincide, then the coefficient of correlation Is
(a) 0
(b) 1 only
(c) – 1 only
(d) 1 or – 1
Solution:
(d) 1 or – 1

The regression line ofy on x be given by
y – $$\bar{y}$$ = byx (x – $$\bar{x}$$) …………………(1)
The regression line of x ony be given by
x – $$\bar{x}$$ = bxy (y – $$\bar{y}$$) …………………(2)
∴ slope of line (1) = m1 = byx
slope of line (2) = m2
= $$\frac{1}{b_{x y}}$$
Since two lines are coincides
∴ m1 = m2
⇒ byx = $$\frac{1}{b_{x y}}$$
⇒ bxy byx = 1
⇒ r = ± $$\sqrt{b_{x y} b_{y x}}$$
= ± √1
= ± 1

Question 3.
The two lines of regression are perpendicular if
(a) byx . bxy = 1
(b) byx . bxy = – 1
(c) bxy – byx = 0
(d) bxy + byx = 0
Solution:
(d) bxy + byx = 0

The regression line ofy on x be given by
y – $$\bar{y}$$ = byx (x – $$\bar{x}$$) …………………(1)
The regression line of x on y be given by
x – $$\bar{x}$$ = bxy (y – $$\bar{y}$$) …(2)
slope of line (1) = m1 = byx
slope of line (2) = m2 = $$\frac{1}{b_{x y}}$$
The two regression lines are ⊥.
iff m1 m2 = – 1
⇒ byx × $$\frac{1}{b_{x y}}$$ = – 1
⇒ byx + bxy = 0

Question 4.
If the line of regression of x on y is 3x + 2y – 5 = 0, then the value of bxy is
(a) – $$\frac{3}{2}$$
(b) $$\frac{2}{3}$$
(c) – $$\frac{2}{3}$$
(d) $$\frac{3}{2}$$
Solution:
(c) – $$\frac{2}{3}$$

The regression line of x on y is given by
3x + 2y – 5 = 0
⇒ 3x = – 2y + 5
⇒ x = $$-\frac{2}{3} y+\frac{5}{3}$$
∴ bxy = – $$\frac{2}{3}$$

Question 5.
If coefficient of correlation is – 0.6, standard deviation of x is 5 and variance of y is 16, then the value of byx is
(a) 0.75
(b) 0.48
(c) – 0.48
(d) – 0.75
Solution:
(c) – 0.48

Given r = – 0.6 ;
σx = 5
and σy2 = 16
⇒ σy = 4
∴ byx = r $$\frac{\sigma_y}{\sigma_x}$$
= $$-\frac{0.6 \times 4}{5}$$
= – $$\frac{2.4}{5}$$ = – 0.48

Question 6.
In a bivariate data, if bxy = – 1.5 and bxy = – 0.5, then the coefficient of correlation is
(a) $$\frac{\sqrt{3}}{2}$$
(b) $$\frac{3}{4}$$
(c) – $$\frac{3}{4}$$
(d) – $$\frac{\sqrt{3}}{2}$$
Solution:
(d) – $$\frac{\sqrt{3}}{2}$$

Given byx = – 1.5 ;
b,xy = – 0.5
since byx < 0
and bxy < 0
∴ r < 0
and r = – $$\sqrt{b_{x y} b_{y x}}$$
i.e. r = – $$\sqrt{(-1.5)(-0.5)}$$
= $$-\sqrt{0.75}$$
= $$-\sqrt{\frac{75}{100}}$$
= $$-\frac{\sqrt{3}}{2}$$

Question 7.
If for 5 observations of pairs (x, y), Σx = 15, Σy = 25, Σy2 = 135 and Σxy = 83, then the value of bxy is
(a) 0.8
(b) 1.25
(c) – 0.8
(d) 1
Solution:
(a) 0.8

Given n = 5 ;
Σx = 15, Σy = 25, Σy2 = 135 and Σxy = 83
∴ bxy = $$\frac{n \Sigma x y-\Sigma x \Sigma y}{n \Sigma y^2-(\Sigma y)^2}$$
= $$\frac{5 \times 83-15 \times 25}{5 \times 135-(25)^2}$$
= $$\frac{415-375}{675-625}$$
= $$\frac{40}{50}=\frac{4}{5}$$
= 0.8

Question 8.
If the two lines of regression are 2x – y – 4 = 0 and 9x – 2y – 38 = 0, then the means of x and y variates respectively are
(a) 8, 6
(b) 6, 8
(c) 7, 10
(d) 5, 6
Solution:
(b) 6, 8

Let us assume 2x – y – 4 = 0 be the regression line of on x.
⇒ y = 2 – 4
⇒ b xy– 2 > 1
Then 9x – 2y – 38 = 0 be the regression line of x on y
∴ 9x = 2y + 38
⇒ x = $$\frac{2}{9} y+\frac{38}{9}$$
∴ bxy = $$\frac{2}{9}$$ < 1
Now byx bxy = 2 × $$\frac{2}{9}$$
= $$\frac{4}{9}$$ < 1 which is possible
Hence our assumption is correct.
We know that ($$\bar{x}, \bar{y}$$) is the common point of two regression lines.
Now, 2x – y – 4 = 0 ………………….(1)
9x – 2y – 8 = 0 ………………(2)
eqn.(2) – 2 × eqn. (1) ; we have
5x – 30 = 0
⇒ x = 6
∴ from (1) ;
12 – y – 4 = 0
⇒ y = 8
Thus $$\bar{x}$$ = 6, $$\bar{x}$$ = 8

Question 9.
If the regression coefficients byx = 1.6 and bxy = 0.4, and θ is the angle between the two lines of regression, then the value of tan θ is
(a) 0.36
(b) 0.72
(c) 0.18
(d) 0.64
Solution
(c) 0.18

Given byx = 1.6 ;
bxy = 0.4
We know that,
tan θ = $$\left|\frac{1-r^2}{b_{x y}+b_{y x}}\right|$$
[∵ bxy, byx > 0
⇒ r > 0
∴ r = $$\sqrt{b_{x y} b_{y x}}$$
= $$\sqrt{1.6 \times 0.4}$$
⇒ r = $$\sqrt{0.64}$$ = 0.8
∴ tan θ = $$\left|\frac{1-(0.8)^2}{1.6+0.4}\right|$$
= $$\left|\frac{1-0.64}{2.0}\right|=\frac{0.36}{2}$$
= 0.18

Question 10.
If for 25 observations of pairs (x, y), Σx = 200, Σy = 150, Σx2 = 3000 and Σxy = 1500, then the equation of line of regression of y on x is
(a) 3x + 14y = 60
(b) 3x – 14y + 60 = 0
(c) 14x – 3y = 60
(d) 3x – 14y = 60
Solution:
(b) 3x – 14y + 60 = 0

Given n = 25 ;
Σx = 200,
Σy = 150,
Σx2 = 3000
and Σxy = 1500
∴ x = $$\frac{\Sigma x}{n}$$
= $$\frac{200}{25}$$ = 8
and $$\bar{y}=\frac{\Sigma y}{n}$$
= $$\frac{150}{25}$$ = 6
and byx = $$\frac{n \Sigma x y-\Sigma r \Sigma y}{n \Sigma x^2-(\Sigma x)^2}$$
= $$\frac{25 \times 1500-200 \times 150}{25 \times 3000-(200)^2}$$
= $$\frac{37500-30000}{75000-40000}$$
= $$\frac{7500}{35000}=\frac{3}{14}$$
Thus eqn. of regression line of y on x be given by
y – $$\bar{y}$$ = b (x – $$\bar{x}$$)
⇒ y – 6 = $$\frac{3}{14}$$ (x – 8)
⇒ 14y – 84 = 3x – 24
⇒ 3x – 14y + 60 = 0.