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) = m_{1} = b_{yx}

slope of line (2) = m_{2}

= \(\frac{1}{b_{x y}}\)

Since two lines are coincides

∴ m_{1} = m_{2}

⇒ 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) = m_{1} = b_{yx}

slope of line (2) = m_{2} = \(\frac{1}{b_{x y}}\)

The two regression lines are ⊥.

iff m_{1} m_{2} = – 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_{y}^{2} = 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, Σy^{2} = 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, Σy^{2} = 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, Σx^{2} = 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,

Σx^{2} = 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.