Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.3 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.3

Question 1.

From the two regression equations find r, \(\bar{x}\) and \(\bar{y}\).

4y = 9x + 15 and 25x = 4y + 17

Solution:

Given 4y = 9x + 15 and 25x = 4y + 17

Since b_{yx} and b_{xy} are positive.

∴ r = \(\frac{3}{5}\) = 0.6

(\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines

9x – 4y = -15 …….(i)

25x – 4y = 17 ……….(ii)

-16x = -32

x = 2

∴ \(\bar{x}\) = 2

Substituting x = 2 in equation (i)

9(2) – 4y = -15

18 + 15 = 4y

33 = 4y

y = 33/4 = 8.25

∴ \(\bar{y}\) = 8.25

Question 2.

In a partially destroyed laboratory record of an analysis of regression data, the following data are legible:

Variance of X = 9

Regression equations:

8x – 10y + 66 = 0 And 40x – 18y = 214.

Find on the basis of the above information

(i) The mean values of X and Y.

(ii) Correlation coefficient between X and Y.

(iii) Standard deviation of Y.

Solution:

Given, \(\sigma_{x}{ }^{2}=9, \sigma_{x}=3\)

(i) (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines

40x – 50y = -330 …….(i)

40x – 50y = +214 ………(ii)

-32y = -544

y = 17

∴ \(\bar{y}\) = 17

8x – 10(17) + 66 = 0

8x = 104

x = 13

∴ \(\bar{x}\) = 13

Question 3.

For 50 students of a class, the regression equation of marks in statistics (X) on the marks in Accountancy (Y) is 3y – 5x + 180 = 0. The mean marks in accountancy is 44 and the variance of marks in statistics \(\left(\frac{9}{16}\right)^{t h}\) of the variance of marks in accountancy. Find the mean in statistics and the correlation coefficient between marks in two subjects.

Solution:

Given, n = 50, \(\bar{y}\) = 44

\(\sigma_{x}^{2}=\frac{9}{16} \sigma_{y}^{2}\)

∴ \(\frac{\sigma_{x}}{\sigma_{x}}=\frac{3}{4}\)

Since (\(\bar{x}\), \(\bar{y}\)) is the point intersection of the regression line.

∴ (\(\bar{x}\), \(\bar{y}\)) satisfies the regression equation.

3\(\bar{y}\) – 5\(\bar{x}\) + 180 = 0

3(44) – 5\(\bar{x}\) + 180 = 0

∴ 5\(\bar{x}\) = 132 + 180

\(\bar{x}\) = \(\frac{312}{5}\) = 62.4

∴ Mean marks in statistics is 62.4

Regression equation of X on Y is 3y – 5x + 180 = 0

∴ 5x = 3y + 180

Question 4.

For bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of the variance of Y if the variance of X is 9.

Solution:

Question 5.

The equation of two regression lines are 2x + 3y – 6 = 0 and 3x + 2y – 12 = 0

Find (i) Correlation coefficient (ii) \(\frac{\sigma_{x}}{\sigma_{y}}\)

Solution:

Question 6.

For a bivariate data \(\bar{x}\) = 53, \(\bar{y}\) = 28, b_{yx} =-1.5 and b_{xy} = -0.2. Estimate Y when X = 50.

Solution:

Regression equation of Y on X is

(Y – \(\bar{y}\)) = b_{yx} (X – \(\bar{x}\))

(Y – 28) = -1.5(50 – 53)

Y – 28 = -1.5(-3)

Y – 28 = 4.5

Y = 32.5

Question 7.

The equation of two regression lines are x – 4y = 5 and 16y – x = 64. Find means of X and Y. Also, find the correlation coefficient between X and Y.

Solution:

Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines.

x – 4y = 5 …..(i)

-x + 16y = 64 …….(ii)

12y = 69

y = 5.75

Substituting y = 5.75 in equation (i)

x – 4(5.75) = 5

x – 23 = 5

x = 28

∴ \(\bar{x}\) = 28, \(\bar{y}\) = 5.75

x – 4y = 5

x = 4y + 5

∴ b_{xy} = 4

16y – x = 64

16y = x + 64

y = \(\frac{1}{16}\)x + 4

b_{yx} = \(\frac{1}{16}\)

b_{yx} . b_{xy} = \(\frac{1}{16}\) × 4 = \(\frac{1}{4}\) ∈ [0, 1]

∴ Our assumption is correct

∴ r^{2} = b_{yx} . b_{xy}

r^{2} = \(\frac{1}{4}\)

r = ±\(\frac{1}{2}\)

Since b_{yx} and b_{xy} are positive,

∴ r = \(\frac{1}{2}\) = 0.5

Question 8.

In partially destroyed record, the following data are available variance of X = 25. Regression equation of Y on X is 5y – x = 22 and Regression equation of X on Y is 64x – 45y = 22 Find

(i) Mean values of X and Y.

(ii) Standard deviation of Y.

(iii) Coefficient of correlation between X and Y.

Solution:

Given \(\sigma_{x}^{2}\) = 25, ∴ σ_{x} = 5

(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression lines

-x + 5y = 22 …….(i)

64x – 45y = 22 ………..(ii)

equation (i) becomes

-9x + 45y = 198

64y – 45y = 22

55x = 220

x = 4

Substituting x = 4 in equation (i)

-4 + 5y = 22

5y = 26

∴ y = 5.2

∴ \(\bar{x}\) = 4, \(\bar{y}\) = 5.2

Regression equation of X on Y is

64x – 45y – 22

64x = 45y + 22

x = \(\frac{45}{64} y+\frac{22}{64}\)

bxy = \(\frac{45}{64}\)

(ii) Regression equation of Y on X is

5y – x = 22

5y = x + 22

Question 9.

If the two regression lines for a bivariate data are 2x = y + 15 (x on y) and 4y – 3x + 25 (y on x) find

(i) \(\bar{x}\)

(ii) \(\bar{y}\)

(iii) b_{yx}

(iv) b_{xy}

(v) r [Given √0.375 = 0.61]

Solution:

Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression line

2x = y + 15

4y = 3x + 25

2x – y = 15 …….(i)

3x – 4y = -25 ……..(ii)

Multiplying equation (i) by 4

8x – 4y = 60

3x – 4y = -25

on Subtracting,

5x = 85

∴ x = 17

Substituting x in equation (i)

2(17) – y = 15

34 – 15 = y

∴ y = 15

Since b_{yx} and b_{xy} are positive, ∴ r = 0.61

Question 10.

The two regression equation are 5x – 6y + 90 = 0 and 15x – 8y – 130 = 0. Find \(\bar{x}\), \(\bar{y}\), r.

Solution:

Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines

5x – 6y + 90 = 0 ……(i)

15x – 8y – 130 = 0

15x – 18y + 270 = 0

15x – 8y – 130 = 0

on subtracting,

-10y + 400 = 0

y = 40

Substituting y = 40 in equation (i)

5x – 6(40) + 90 = 0

5x = 150

x = 30

∴ \(\bar{x}\) = 30, \(\bar{y}\) = 40

Since b_{yx} and b_{xy} are positive

∴ r = \(\frac{2}{3}\)

Question 11.

Two lines of regression are 10x + 3y – 62 = 0 and 6x + 5y – 50 = 0 Identify the regression equation equation of x on y. Hence find \(\bar{x}\), \(\bar{y}\), and r.

Solution:

∴ Our assumption is correct.

∴ Regression equation of X on Y is 10x + 3y – 62 = 0

r^{2} = b_{yx} . b_{xy}

r^{2} = \(\frac{9}{25}\)

r = ±\(\frac{3}{5}\)

Since, byx and bxy are negative, r = –\(\frac{3}{5}\) = -0.6

Also (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines

50x + 15y = 310

18x + 15y = 150

on subtracting

32x = 160

x = 5

Substituting x = 5 in 10x + 3y = 62

10(5) + 3y = 62

3y = 12

∴ y = 4

∴ \(\bar{x}\) = 5, \(\bar{y}\) = 4

Question 12.

For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean values of X and Y.

Solution:

Since byx and bxy are positive,

r = \(\frac{3}{5}\) = 0.6

Since, (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines

3x – 10y = -170 …….(i)

5x – 6y = -70 ………(ii)

9x – 30y = -510

25x – 30y = -350

on subtracting

-16x = -160

x = 10

Substituting x = 10 in equation (i)

3(10) – 10y = -170

30 + 170 = 10y

200 = 10y

y = 20

∴ \(\bar{x}\) = 10, \(\bar{y}\) = 20

Question 13.

Regression equation of two series are 2x – y – 15 = 0 and 4y + 25 = 0 and 3x- 4y + 25 = 0. Find \(\bar{x}\), \(\bar{y}\) and regression coefficients, Also find coefficients of correlation. [Given √0.375 = 0.61]

Solution:

Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression line

2x = y + 15

4y = 3x + 25

2x – y = 15 ……(i)

3x – 4y = -15 ……..(ii)

Multiply equation (i) by 4

8x – 4y = 60

3x – 4y = -25

on subtracting,

5x = 85

x = 17

Substituting x in equation (i)

2(17) – y = 15

34 – 15 = y

y = 15

∴ \(\bar{x}\) = 17, \(\bar{y}\) = 19

∴ Our assumption is correct

r^{2} = b_{xy} . b_{yx}

r^{2} = \(\frac{3}{8}\) = 0.375

r = ±√o.375 = ±0.61

Since, b_{yx} and b_{xy} are positive, ∴ r = 0.61

Question 14.

The two regression lines between height (X) in includes and weight (Y) in kgs of girls are 4y – 15x + 500 = 0 and 20x – 3y – 900 = 0. Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.

Solution:

Since (\(\bar{x}\), \(\bar{y}\)) is the point intersection of the regression lines

15x – 4y = 500 ……(i)

20x – 3y = 900 …….(ii)

60x – 16y – 2000

60x – 9y = 2700

on subtracting,

-7y = -700

y = 100

Substituting y = 100 in equation (i)

15x – 4(100) = 500

15x = 900

x = 60

∴ Our assumption is correct

∴ Regression equation of Y on X is

Y = \(\frac{15}{4}\)x – 125

When x = 70

Y = \(\frac{15}{4}\) × 70 = -125

= 262.5 – 125

= 137.5 kg