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

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

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

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

Question 4.
If the line of regression of x on y is 3x + 2y – 5 = 0, then the value of b_xy 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 = \(-\frac{2}{3} y+\frac{5}{3}\)
∴ b_xy = – \(\frac{2}{3}\)

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 b_yx is
(a) 0.75
(b) 0.48
(c) – 0.48
(d) – 0.75
Solution:
(c) – 0.48

Given r = – 0.6 ;
σ_x = 5
and σ_y² = 16
⇒ σ_y = 4
∴ b_yx = 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 b_xy = – 1.5 and b_xy = – 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 b_yx = – 1.5 ;
b,_xy = – 0.5
since b_yx < 0
and b_xy < 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 b_xy 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
∴ b_xy = \(\frac{n \Sigma x y-\Sigma x \Sigma y}{n \Sigma y^2-(\Sigma y)^2}\)
= \(\frac{5 \times 83-15 \times 25}{5 \times 135-(25)^2}\)
= \(\frac{415-375}{675-625}\)
= \(\frac{40}{50}=\frac{4}{5}\)
= 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 = \(\frac{2}{9} y+\frac{38}{9}\)
∴ b_xy = \(\frac{2}{9}\) < 1
Now b_yx b_xy = 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 b_yx = 1.6 and b_xy = 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 b_yx = 1.6 ;
b_xy = 0.4
We know that,
tan θ = \(\left|\frac{1-r^2}{b_{x y}+b_{y x}}\right|\)
[∵ b_xy, b_yx > 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 b_yx = \(\frac{n \Sigma x y-\Sigma r \Sigma y}{n \Sigma x^2-(\Sigma x)^2}\)
= \(\frac{25 \times 1500-200 \times 150}{25 \times 3000-(200)^2}\)
= \(\frac{37500-30000}{75000-40000}\)
= \(\frac{7500}{35000}=\frac{3}{14}\)
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.