12th Commerce Maths 2 Chapter 3 Exercise 3.1 Answers Maharashtra Board

Linear Regression Class 12 Commerce Maths 2 Chapter 3 Exercise 3.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.1 Questions and Answers.

Std 12 Maths 2 Exercise 3.1 Solutions Commerce Maths

Question 1.
The HRD manager of the company wants to find a measure which he can use to fix the monthly income of persons applying for the job in the production department. As an experimental project. He collected data of 7 persons from that department referring to years of service and their monthly incomes.
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q1
(i) Find the regression equation of income on years of service.
(ii) What initial start would you recommend for a person applying for the job after having served in a similar capacity in another company for 13 years?
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q1.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q1.2
(i) Regression equation of Y on X is (Y – \(\bar{y}\)) = b yx (x – \(\bar{x}\))
(Y – 8) = 0.75(x – \(\bar{x}\))
Y = 0.75x + 2
(ii) When x = 13
Y = 0.75(13) + 2 = 11.75
Recommended income for the person is ₹ 11750.

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Question 2.
Calculate the regression equations of X on Y and Y on X from the following date:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q2
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q2.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q2.2
Regression equation of X on Y is (X – \(\bar{x}\)) = b xy (Y – \(\bar{y}\))
(X – 14) = 1(Y – 8)
X – 14 = Y – 8
X = Y + 6
Regression equation Y on X is (Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 8) = 0.87(X – 14)
Y – 8 = 0.87X – 12.18
Y = 0.87X – 4.18

Question 3.
For a certain bivariate data on 5 pairs of observations given
ÎŁx = 20, ÎŁy = 20, ÎŁx 2 = 90, ÎŁy 2 = 90, ÎŁxy = 76
Calculate (i) cov(x, y), (ii) b yx and b xy , (iii) r
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q3
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q3.1
Sine byx and bxy are negative, r = -0.4

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Question 4.
From the following data estimate y when x = 125
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q4
Solution:
Let u = x – 122, v = y – 14
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q4.1
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 13.5) = -0.21(x – 121.5)
Y – 13.5 = -0.21x + 25.52
Y = -0.21x + 39.02
When x = 125
Y = -0.21(125) + 39.02
= -26.25 + 39.02
= 12.77

Question 5.
The following table gives the aptitude test scores and productivity indices of 10 works selected at workers selected randomly.
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q5
Obtain the two regression equation and estimate
(i) The productivity index of a worker whose test score is 95.
(ii) The test score when productivity index is 75.
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q5.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q5.2
Regression equation of Y on X,
(Y – \(\bar{y}\)) = b yx (X – \(\bar{y}\))
(Y – 65) = 1.16 (x – 65)
Y – 65 = 1.16x – 75.4
Y = 1.16x – 10.4
(i) When x = 95
Y = 1.16(95) – 10.4
= 110.2 – 10.4
= 99.8
Regression equation of X on Y,
(X – \(\bar{x}\)) = b xy (Y – \(\bar{y}\))
(X – 65) = 0.59(y – 65)
(X – 65) = 0.59y – 38.35
X = 0.59y + 26.65
(ii) When y = 75
x = 0.59(75) + 26.65
= 44.25 + 26.65
= 70.9

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Question 6.
Compute the appropriate regression equation for the following data.
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q6
Solution:
Since x is the independent variable, and y is the dependent variable,
we need to find regression equation of y on x
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q6.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q6.2
Regression equation of y on x is (y – \(\bar{y}\)) = b yx (x – \(\bar{x}\))
(y – 10) = -13.4(x – 6)
y – 10 = -1.34x + 8.04
y = -1.34x + 18.04

Question 7.
The following are the marks obtained by the students in Economic (X) and Mathematics (Y)
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q7
Find the regression equation of Y and X.
Solution:
Let u = x – 61, v = y – 80
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q7.1
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 80.4) = 0.3(x – 61)
Y – 80.4 = 0.3x – 18.3
Y = 0.3x + 62.1

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Question 8.
For the following bivariate data obtain the equation of two regressions lines:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q8
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q8.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q8.2
Regression equation of Y on X
(Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 9) = 2(x – 3)
Y – 9 = 2x – 6
Y = 2x + 3
Regression equation of X on Y
(X – \(\bar{x}\)) = b xy (Y – \(\bar{y}\))
(X – 3) = 0.5(y – 9)
(X – 3) = 0.5y – 4.5
X = 0.5y – 1.5

Question 9.
Find the following data obtain the equation of two regression lines:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q9
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q9.1
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q9.2
Regression of Y on X,
(Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 8) = 0.65(x – 6)
Y – 8 = -0.65x + 3.9
Y = -0.65x + 11.9
Regression of X on Y
(X – \(\bar{x}\)) = b xy (Y – \(\bar{y}\))
(X – 6) = -1.3(y – 8)
(X – 6) = -1.3y + 10.4
X = -1.3y + 16.4

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Question 10.
For the following data, find the regression line of Y on X
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q10
Hence find the most likely value of y when x = 4
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q10.1
(Y – 3) = 2(x – 2)
Y – 3 = 2x – 4
Y = 2x – 1
When x = 4
Y = 2(4) – 1
= 8 – 1
= 7

Question 11.
Find the following data, find the regression equation of Y on X, and estimate Y when X = 10.
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q11
Solution:
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q11.1
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 5) = (0.63)(x – 3.5)
Y – 5 = 0.63x – 2.2
Y = 0.63x + 2.8
When x = 10
Y = 0.63(10) + 2.8
= 6.3 + 2.8
= 9.1

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Question 12.
The following sample gave the number of hours of study (X) per day for an examination and marks (Y) obtained by 12 students.
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q12
Obtain the line of regression of marks on hours of study.
Solution:
Let u = x – 5, v = y – 70
Maharashtra-Board-12th-Commerce-Maths-Solutions-Chapter-3-Linear-Regression-Ex-3.1-Q12.1
∴ Equation of marks on hours of study is
(Y – \(\bar{y}\)) = b yx (X – \(\bar{x}\))
(Y – 70.83) = 6.6(x – 4.92)
Y – 70.83 = 6.6x – 32.47
∴ Y = 6.6x + 38.36