Maharashtra Board Class 12 Maths Part 2 Chapter 3 Linear Regression 3.1 Solutions

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.

(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?
Answer: Solution:

Given, n = 7
\[\therefore \bar{x} = \frac{\Sigma X_i}{n} = \frac{56}{7} = 8\] \[\bar{y} = \frac{\Sigma Y_i}{n} = \frac{56}{7} = 8\] \[b_{yx} = \frac{Cov(X, Y)}{\sigma^2_x} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2}\] \[\therefore b_{yx} = \frac{21}{28} = \frac{3}{4} = 0.75\] (i) Regression equation of Y on X is \( (Y - \bar{y}) = b_{yx} (x - \bar{x}) \)
\( (Y - 8) = 0.75(x - 8) \)
\( Y = 0.75x + 2 \)
(ii) When x = 13
\( Y = 0.75(13) + 2 = 11.75 \)
Recommended income for the person is Rs.11750.
In simple words: The regression equation helps predict income based on years of service. For a new applicant with 13 years of service, the recommended monthly income is Rs. 11,750, calculated using the derived regression line.

🎯 Exam Tip: Ensure all calculations for means and sums of squares are accurate, as they form the foundation for the regression coefficient and subsequent predictions.

Question 2. Calculate the regression equations of X on Y and Y on X from the following date:

Answer: Solution:

Given, n = 5
\[\bar{x} = \frac{\Sigma X}{n} = \frac{70}{5} = 14\] \[\bar{y} = \frac{\Sigma Y_i}{n} = \frac{40}{5} = 8\] \[b_{xy} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(y - \bar{y})^2} = \frac{40}{40} = 1\] \[b_{yx} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2} = \frac{40}{46} = 0.87\] 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 \( (Y - \bar{y}) = b_{yx} (X - \bar{x}) \)
\( (Y - 8) = 0.87(X - 14) \)
\( Y - 8 = 0.87X - 12.18 \)
\( Y = 0.87X - 4.18 \)
In simple words: We calculated two regression equations: one for X on Y, which is \(X = Y + 6\), and one for Y on X, which is \(Y = 0.87X - 4.18\). These equations allow us to predict one variable's value based on the other.

🎯 Exam Tip: Remember to clearly distinguish between \(b_{xy}\) and \(b_{yx}\) calculations, as an error in one will affect the corresponding regression equation. Pay attention to the denominator in each coefficient formula.

Question 3. For a certain bivariate data on 5 pairs of observations given
\( \Sigma x = 20, \Sigma y = 20, \Sigma x^2 = 90, \Sigma y^2 = 90, \Sigma xy = 76 \)
Calculate (i) cov(x, y), (ii) \(b_{yx}\) and \(b_{xy}\), (iii) r
Answer: Solution:
\[Cov(x, y) = \frac{\Sigma(x \cdot y)}{n} - \bar{x} \cdot \bar{y}\]
\[= \frac{\Sigma(x \cdot y)}{n} - \left(\frac{\Sigma x}{n}\right) \left(\frac{\Sigma y}{n}\right)\]
\[= \frac{76}{5} - \left(\frac{20}{5}\right) \left(\frac{20}{5}\right)\]
\[= 15.2 - 16\]
\[= -0.8\]
\[b_{yx} = \frac{n \Sigma(x \cdot y) - (\Sigma x)(\Sigma y)}{n \Sigma x^2 - (\Sigma x)^2}\]
\[= \frac{5(76) - (20)(20)}{5(90) - (20)^2} = \frac{380 - 400}{450 - 400} = \frac{-20}{50} = -0.4\]
\[b_{xy} = \frac{n \Sigma(x \cdot y) - (\Sigma x)(\Sigma y)}{n \Sigma y^2 - (\Sigma y)^2}\]
\[= \frac{5(76) - (20)(20)}{5(90) - (20)^2} = \frac{380 - 400}{450 - 400} = \frac{-20}{50} = -0.4\]
Now, \( r^2 = b_{yx} \cdot b_{xy} \)
\( = (-0.4) (-0.4) \)
\( = 0.16 \)
\( \therefore r = \pm 0.4 \)
Sine \(b_{yx}\) and \(b_{xy}\) are negative, r = -0.4
In simple words: We calculated the covariance between x and y as -0.8, indicating a weak inverse relationship. Both regression coefficients, \(b_{yx}\) and \(b_{xy}\), were -0.4, leading to a correlation coefficient (r) of -0.4, confirming a negative linear relationship.

🎯 Exam Tip: When calculating 'r', remember that its sign must match the signs of \(b_{yx}\) and \(b_{xy}\). If the signs differ, recheck your calculations for the regression coefficients.

Question 4. From the following data estimate y when x = 125

Answer: Solution:
Let u = x - 122, v = y - 14

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{729}{6} = 121.5\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{81}{6} = 13.5\] \[b_{yx} = b_{vu} = \frac{n \Sigma(u \cdot v) - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2}\]
\[= \frac{6(-16) - (-3)(-3)}{6(83) - (-3)^2} = \frac{-96 - 9}{498 - 9} = \frac{-105}{489} = -0.21\] 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
\( Y = -0.21(125) + 39.02 \)
\( = -26.25 + 39.02 \)
\( = 12.77 \)
In simple words: To estimate y when x = 125, we first simplified the data using transformations u = x-122 and v = y-14. We then calculated the regression coefficient \(b_{yx}\) using these transformed variables, obtaining -0.21. Substituting this into the regression equation \(Y = -0.21x + 39.02\), we estimated y to be 12.77 when x is 125.

🎯 Exam Tip: Using assumed mean (u and v) can simplify calculations for larger numbers. Remember to convert back to the original variables if the question requires values for x and y, though here the prediction is directly for y.

Question 5. The following table gives the aptitude test scores and productivity indices of 10 works selected at workers selected randomly.

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.
Answer: Solution:

\[\bar{x} = \frac{\Sigma X_i}{n} = \frac{650}{10} = 65\] \[\bar{y} = \frac{\Sigma Y_i}{n} = \frac{650}{10} = 65\] \[b_{yx} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2} = \frac{1044}{894} = 1.16\] \[b_{xy} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(y - \bar{y})^2} = \frac{1044}{1752} = 0.59\] Regression equation of Y on X,
\( (Y - \bar{y}) = b_{yx} (X - \bar{x}) \)
\( (Y - 65) = 1.16 (x - 65) \)
\( Y - 65 = 1.16x - 75.4 \)
\( Y = 1.16x - 10.4 \)
(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
\( x = 0.59(75) + 26.65 \)
\( = 44.25 + 26.65 \)
\( = 70.9 \)
In simple words: We established two regression lines: \(Y = 1.16X - 10.4\) for productivity based on test scores, and \(X = 0.59Y + 26.65\) for test scores based on productivity. Using these, a worker with a test score of 95 is estimated to have a productivity index of 99.8, and a worker with a productivity index of 75 is estimated to have a test score of 70.9.

🎯 Exam Tip: Clearly label and derive both regression equations separately. For prediction, ensure you use the correct regression line (e.g., Y on X to predict Y, X on Y to predict X).

Question 6. Compute the appropriate regression equation for the following data.

Answer: Solution:
Since x is the independent variable, and y is the dependent variable,
we need to find regression equation of y on x

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{36}{6} = 6\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{60}{6} = 10\] \[b_{yx} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2} = \frac{-67}{50} = -1.34\] Regression equation of y on x is \( (y - \bar{y}) = b_{yx} (x - \bar{x}) \)
\( (y - 10) = -1.34(x - 6) \)
\( y - 10 = -1.34x + 8.04 \)
\( y = -1.34x + 18.04 \)
In simple words: Given x as the independent variable and y as dependent, we calculated the regression equation of y on x. The resulting equation, \(y = -1.34x + 18.04\), shows a negative linear relationship, meaning as x increases, y tends to decrease.

🎯 Exam Tip: Identify the independent and dependent variables correctly to ensure you calculate the appropriate regression coefficient (\(b_{yx}\) for Y on X) and equation.

Question 7. The following are the marks obtained by the students in Economic (X) and Mathematics (Y)

Find the regression equation of Y and X.
Answer: Solution:
Let u = x - 61, v = y - 80

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{305}{5} = 61\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{402}{5} = 80.4\] \[b_{yx} = \frac{n \Sigma(u \cdot v) - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2}\]
\[= \frac{5(3) - (0)(2)}{5(10) - (0)^2} = \frac{15}{50} = 0.3\] Regression equation of Y on X is
\( (Y - \bar{y}) = b_{yx} (X - \bar{x}) \)
\( (Y - 80.4) = 0.3(x - 61) \)
\( Y - 80.4 = 0.3x - 18.3 \)
\( Y = 0.3x + 62.1 \)
In simple words: By using assumed means for simplification, we calculated the regression coefficient \(b_{yx}\) as 0.3. This allowed us to form the regression equation of Y on X as \(Y = 0.3x + 62.1\), which can be used to predict Mathematics marks based on Economics marks.

🎯 Exam Tip: The choice of assumed mean (u and v) should aim to simplify the calculations, especially when dealing with large numbers or decimals. Ensure to calculate \( \Sigma uv \), \( \Sigma u \), \( \Sigma v \), and \( \Sigma u^2 \) correctly.

Question 8. For the following bivariate data obtain the equation of two regressions lines:

Answer: Solution:

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{15}{5} = 3\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{45}{5} = 9\] \[b_{yx} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2} = \frac{20}{10} = 2\] \[b_{xy} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(y - \bar{y})^2} = \frac{20}{40} = 0.5\] 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
\( (X - \bar{x}) = b_{xy} (Y - \bar{y}) \)
\( (X - 3) = 0.5(y - 9) \)
\( X - 3 = 0.5y - 4.5 \)
\( X = 0.5y - 1.5 \)
In simple words: We calculated both regression lines from the given data. The regression equation of Y on X is \(Y = 2x + 3\), suggesting that Y increases significantly with X. The regression equation of X on Y is \(X = 0.5y - 1.5\), which can predict X based on Y.

🎯 Exam Tip: When finding two regression lines, ensure you calculate both \(b_{yx}\) and \(b_{xy}\) and apply them to the correct mean-centered equations to avoid swapping the lines.

Question 9. Find the following data obtain the equation of two regression lines:

Answer: Solution:

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{30}{5} = 6\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{40}{5} = 8\] \[b_{yx} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2} = \frac{-26}{40} = -0.65\] \[b_{xy} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(y - \bar{y})^2} = \frac{-26}{20} = -1.3\] 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
\( (X - \bar{x}) = b_{xy} (Y - \bar{y}) \)
\( (X - 6) = -1.3(y - 8) \)
\( X - 6 = -1.3y + 10.4 \)
\( X = -1.3y + 16.4 \)
In simple words: From the given data, we derived two regression lines: \(Y = -0.65x + 11.9\) for Y on X and \(X = -1.3y + 16.4\) for X on Y. Both show a negative relationship between the variables, meaning as one increases, the other tends to decrease.

🎯 Exam Tip: Negative regression coefficients indicate an inverse relationship. Double-check the signs in your calculations to ensure they are consistent across \(b_{yx}\), \(b_{xy}\), and the resulting equations.

Question 10. For the following data, find the regression line of Y on X

Hence find the most likely value of y when x = 4
Answer: Solution:

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{6}{3} = 2\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{9}{3} = 3\] \[b_{yx} = \frac{\Sigma(x - \bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^2} = \frac{4}{2} = 2\] Regression equation of Y on X is
\( (Y - \bar{y}) = b_{yx} (x - \bar{x}) \)
\( (Y - 3) = 2(x - 2) \)
\( Y - 3 = 2x - 4 \)
\( Y = 2x - 1 \)
When x = 4
\( Y = 2(4) - 1 \)
\( = 8 - 1 \)
\( = 7 \)
In simple words: We calculated the regression line of Y on X as \(Y = 2x - 1\). Using this equation, the most likely value of y when x is 4 is found to be 7, indicating a strong positive linear relationship.

🎯 Exam Tip: After finding the regression equation, ensure to substitute the given value of x correctly into the equation to predict y. A common mistake is arithmetic errors in the final substitution.

Question 11. Find the following data, find the regression equation of Y on X, and estimate Y when X = 10.

Answer: Solution:

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{21}{6} = 3.5\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{30}{6} = 5\] \[b_{yx} = \frac{n \Sigma(x \cdot y) - (\Sigma x)(\Sigma y)}{n \Sigma x^2 - (\Sigma x)^2}\]
\[= \frac{6(116) - (21)(30)}{6(91) - (21)^2}\]
\[= 0.63\] 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
\( Y = 0.63(10) + 2.8 \)
\( = 6.3 + 2.8 \)
\( = 9.1 \)
In simple words: We determined the regression equation of Y on X as \(Y = 0.63x + 2.8\). Using this equation, when X is 10, the estimated value of Y is 9.1, indicating a positive linear trend where Y generally increases with X.

🎯 Exam Tip: Ensure all summations are correct, especially \( \Sigma xy \) and \( \Sigma x^2 \), as these are crucial for accurately calculating \(b_{yx}\) using the direct formula.

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.

Obtain the line of regression of marks on hours of study.
Answer: Solution:
Let u = x - 5, v = y - 70

\[\bar{x} = \frac{\Sigma x_i}{n} = \frac{59}{12} = 4.92\] \[\bar{y} = \frac{\Sigma y_i}{n} = \frac{850}{12} = 70.83\] \[b_{yx} = b_{vu} = \frac{n \Sigma(u \cdot v) - (\Sigma u)(\Sigma v)}{n \Sigma u^2 - (\Sigma u)^2}\]
\[= \frac{12(190) - (-1)(10)}{12(29) - (-1)^2} = 6.6\]
\( \therefore \) Equation of marks on hours of study is
\( (Y - \bar{y}) = b_{yx} (X - \bar{x}) \)
\( (Y - 70.83) = 6.6(x - 4.92) \)
\( Y - 70.83 = 6.6x - 32.47 \)
\( \therefore Y = 6.6x + 38.36 \)
In simple words: To find the regression line of marks (Y) on hours of study (X), we used transformed variables (u and v) to simplify calculations. The regression coefficient \(b_{yx}\) was found to be 6.6, resulting in the regression equation \(Y = 6.6x + 38.36\), which shows a strong positive relationship between study hours and marks.

🎯 Exam Tip: When dealing with a large number of observations, using the assumed mean method for 'u' and 'v' can significantly reduce calculation complexity and potential errors. Remember to use `n` (number of observations) correctly in the formulas.

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