Maharashtra Board Class 12 Maths Part 2 Chapter 3 Linear Regression 3.3 Solutions

Question 1. From the two regression equations find r, \( \bar{x} \) and \( \bar{y} \).
4y = 9x + 15 and 25x = 4y + 17
Answer:
Solution:
Given 4y = 9x + 15 and 25x = 4y + 17
From 4y = 9x + 15,
\( y = \frac{9x}{4} + \frac{15}{4} \)
So, \( b_{yx} = \frac{9}{4} \)
From 25x = 4y + 17,
\( x = \frac{4y}{25} + \frac{17}{25} \)
So, \( b_{xy} = \frac{4}{25} \)
Now, \( b_{yx} \cdot b_{xy} = \frac{9}{4} \times \frac{4}{25} = \frac{9}{25} \)
which belongs to \( [0, 1] \)
Our assumption is correct.
\( \therefore r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{9}{25} \)
\( r = \pm \frac{3}{5} \)
Since \( b_{yx} \) and \( b_{xy} \) are positive,
\( \therefore r = \frac{3}{5} = 0.6 \)
\( (\bar{x}, \bar{y}) \) is the point of intersection of the regression lines
\( 9x - 4y = -15 \quad \ldots(i) \)
\( 25x - 4y = 17 \quad \ldots(ii) \)
Subtracting equation (i) from equation (ii):
\( (25x - 4y) - (9x - 4y) = 17 - (-15) \)
\( 16x = 32 \)
\( x = 2 \)
\( \therefore \bar{x} = 2 \)
Substituting \( x = 2 \) in equation (i)
\( 9(2) - 4y = -15 \)
\( 18 - 4y = -15 \)
\( 18 + 15 = 4y \)
\( 33 = 4y \)
\( y = \frac{33}{4} = 8.25 \)
\( \therefore \bar{y} = 8.25 \)
In simple words: This problem asks us to find the correlation coefficient 'r' and the mean values of X and Y, denoted as \( \bar{x} \) and \( \bar{y} \), from two given regression equations. We first rearrange the equations to find the regression coefficients, then calculate 'r' using their product. Finally, \( \bar{x} \) and \( \bar{y} \) are found by solving the two regression equations simultaneously, as they intersect at the point of means.

🎯 Exam Tip: Always verify that the product of the regression coefficients (\( b_{yx} \cdot b_{xy} \)) lies between 0 and 1. If not, your assumption about which equation represents Y on X or X on Y is incorrect. The sign of 'r' must match the sign of both regression coefficients.

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.
Answer:
Solution:
Given, \( \sigma_x^2 = 9, \sigma_x = 3 \)
(i) \( (\bar{x}, \bar{y}) \) is the point of intersection of the regression lines
From 8x – 10y + 66 = 0, multiply by 5: \( 40x - 50y + 330 = 0 \)
\( 40x - 50y = -330 \quad \ldots(i) \)
From 40x – 18y = 214,
\( 40x - 18y = 214 \quad \ldots(ii) \)
Subtracting equation (i) from (ii):
\( (40x - 18y) - (40x - 50y) = 214 - (-330) \)
\( 32y = 544 \)
\( y = 17 \)
\( \therefore \bar{y} = 17 \)
Substituting \( y = 17 \) in equation \( 8x - 10y + 66 = 0 \)
\( 8x - 10(17) + 66 = 0 \)
\( 8x - 170 + 66 = 0 \)
\( 8x = 104 \)
\( x = 13 \)
\( \therefore \bar{x} = 13 \)
(ii) For the correlation coefficient, we find \( b_{yx} \) and \( b_{xy} \).
From 8x – 10y + 66 = 0, assuming Y on X:
\( 10y = 8x + 66 \)
\( y = \frac{8x}{10} + \frac{66}{10} \)
\( \therefore b_{yx} = \frac{8}{10} = \frac{4}{5} \)
From 40x – 18y = 214, assuming X on Y:
\( 40x = 18y + 214 \)
\( x = \frac{18y}{40} + \frac{214}{40} \)
\( b_{xy} = \frac{18}{40} = \frac{9}{20} \)
Check assumption: \( b_{yx} \cdot b_{xy} = \frac{4}{5} \times \frac{9}{20} = \frac{36}{100} = \frac{9}{25} \)
which belongs to \( [0, 1] \). Our assumption is correct.
\( r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{9}{25} \)
\( r = \pm \frac{3}{5} \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = \frac{3}{5} \)
(iii) Standard deviation of Y.
We know \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \)
\( \frac{4}{5} = \frac{3}{5} \times \frac{\sigma_y}{3} \)
\( \frac{4}{5} = \frac{\sigma_y}{5} \)
\( \sigma_y = 4 \)
In simple words: This problem requires us to retrieve lost statistical information from partial data. We find the average values of X and Y by solving the two regression equations. Then, we extract the regression coefficients from these equations to calculate the correlation coefficient 'r'. Finally, we use the regression coefficient formula relating 'r' and standard deviations to determine the standard deviation of Y.

🎯 Exam Tip: When given two regression equations, always solve them simultaneously to find the mean values \( (\bar{x}, \bar{y}) \). This is a foundational step. Remember that the product of regression coefficients, \( b_{yx} \cdot b_{xy} \), must be between 0 and 1 (inclusive), and the sign of 'r' must match the sign of both coefficients.

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 \( (\frac{9}{16})^{th} \) of the variance of marks in accountancy. Find the mean in statistics and the correlation coefficient between marks in two subjects.
Answer:
Solution:
Given, n = 50, \( \bar{y} = 44 \)
\( \sigma_x^2 = \frac{9}{16} \sigma_y^2 \)
Taking square root on both sides:
\( \frac{\sigma_x}{\sigma_y} = \frac{3}{4} \)
Since \( (\bar{x}, \bar{y}) \) is the point intersection of the regression line.
\( \therefore (\bar{x}, \bar{y}) \) satisfies the regression equation.
Given regression equation of X on Y is \( 3y - 5x + 180 = 0 \)
Substitute \( \bar{x} \) and \( \bar{y} \):
\( 3\bar{y} - 5\bar{x} + 180 = 0 \)
\( 3(44) - 5\bar{x} + 180 = 0 \)
\( 132 - 5\bar{x} + 180 = 0 \)
\( \therefore 5\bar{x} = 132 + 180 \)
\( 5\bar{x} = 312 \)
\( \bar{x} = \frac{312}{5} = 62.4 \)
\( \therefore \) Mean marks in statistics is 62.4
Regression equation of X on Y is \( 3y - 5x + 180 = 0 \)
To find \( b_{xy} \):
\( 5x = 3y + 180 \)
\( x = \frac{3}{5}y + \frac{180}{5} \)
\( \therefore b_{xy} = \frac{3}{5} \)
Also, we know \( b_{xy} = r \frac{\sigma_x}{\sigma_y} \)
\( \frac{3}{5} = r \frac{3}{4} \)
\( r = \frac{3}{5} \times \frac{4}{3} \)
\( r = \frac{4}{5} = 0.8 \)
In simple words: This question involves finding the mean marks in Statistics and the correlation coefficient between Statistics and Accountancy marks. We are given the mean of Accountancy marks and a relationship between their variances. Using the fact that the mean values satisfy the regression equation, we first solve for the mean of Statistics marks. Then, by extracting the regression coefficient from the given equation and using the standard deviation ratio, we calculate the correlation coefficient.

🎯 Exam Tip: Remember that the point \( (\bar{x}, \bar{y}) \) always lies on both regression lines. This property is crucial for finding unknown mean values. Also, pay attention to the formula relating regression coefficients, correlation coefficient, and standard deviations; ensure you use \( \frac{\sigma_x}{\sigma_y} \) for \( b_{xy} \) and \( \frac{\sigma_y}{\sigma_x} \) for \( b_{yx} \).

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.
Answer:
Solution:
Given, \( b_{yx} = 0.4, b_{xy} = 0.9, \sigma_x^2 = 9, \sigma_x = 3 \)
We know that \( r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = 0.4 \times 0.9 \)
\( r^2 = 0.36 \)
\( r = \pm \sqrt{0.36} \)
\( r = \pm 0.6 \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = 0.6 \)
Also, we know \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \)
\( 0.4 = 0.6 \times \frac{\sigma_y}{3} \)
\( \frac{4}{10} = \frac{6}{10} \times \frac{\sigma_y}{3} \)
\( 4 = 2 \sigma_y \)
\( \sigma_y = 2 \)
\( \therefore \sigma_y^2 = 4 \)
Thus, Variance of Y is 4.
In simple words: This problem asks us to find the variance of Y given both regression coefficients and the variance of X. We first calculate the correlation coefficient 'r' using the product of the regression coefficients. Then, we use the formula that relates the regression coefficient of Y on X (byx) with 'r' and the standard deviations of X and Y to solve for the standard deviation of Y, and finally its variance.

🎯 Exam Tip: Remember the relationship \( r^2 = b_{yx} \cdot b_{xy} \). Also, choose the correct regression coefficient formula based on what you need to find. If you need \( \sigma_y \), using \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \) is usually more direct than using \( b_{xy} \).

Question 5. The equation of two regression lines are 2x + 3y – 6 = 0 and 3x + 2y – 12 = 0
Find (i) Correlation coefficient (ii) \( \frac{\sigma_x}{\sigma_y} \)
Answer:
Solution:
(i) Let's assume the first equation, \( 2x + 3y - 6 = 0 \), is Y on X.
\( 3y = -2x + 6 \)
\( y = -\frac{2}{3}x + 2 \)
So, \( b_{yx} = -\frac{2}{3} \)
Let's assume the second equation, \( 3x + 2y - 12 = 0 \), is X on Y.
\( 3x = -2y + 12 \)
\( x = -\frac{2}{3}y + 4 \)
So, \( b_{xy} = -\frac{2}{3} \)
Check assumption: \( b_{yx} \cdot b_{xy} = (-\frac{2}{3}) \times (-\frac{2}{3}) = \frac{4}{9} \)
\( \frac{4}{9} \in [0, 1] \). Our assumption is correct.
\( \therefore r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{4}{9} \)
\( r = \pm \sqrt{\frac{4}{9}} \)
\( r = \pm \frac{2}{3} \)
Since \( b_{yx} \) and \( b_{xy} \) are negative, \( \therefore r = -\frac{2}{3} \)
(ii) To find \( \frac{\sigma_x}{\sigma_y} \), we can use the formula for \( b_{xy} \).
\( b_{xy} = r \frac{\sigma_x}{\sigma_y} \)
\( -\frac{2}{3} = (-\frac{2}{3}) \frac{\sigma_x}{\sigma_y} \)
\( \therefore \frac{\sigma_x}{\sigma_y} = 1 \)
In simple words: This problem requires us to find the correlation coefficient 'r' and the ratio of standard deviations \( \frac{\sigma_x}{\sigma_y} \) from two given regression equations. We first determine the regression coefficients by rearranging the equations and confirm our assumption by checking their product. Then, 'r' is calculated from the product of the regression coefficients, taking its sign from the coefficients. Finally, the ratio of standard deviations is derived using the formula for one of the regression coefficients.

🎯 Exam Tip: When both regression coefficients are negative, the correlation coefficient 'r' must also be negative. The relationship \( \frac{\sigma_x}{\sigma_y} \) can be found using either \( b_{xy} \) or \( b_{yx} \), but ensure you use the correct standard deviation ratio for each coefficient (i.e., \( b_{xy} = r \frac{\sigma_x}{\sigma_y} \) and \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \)).

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.
Answer:
Solution:
The regression equation of Y on X is given by:
\( (Y - \bar{y}) = b_{yx} (X - \bar{x}) \)
Substitute the given values:
\( (Y - 28) = -1.5 (50 - 53) \)
\( Y - 28 = -1.5 (-3) \)
\( Y - 28 = 4.5 \)
\( Y = 4.5 + 28 \)
\( Y = 32.5 \)
In simple words: This problem asks us to predict the value of Y for a given X, using the regression equation of Y on X. We are provided with the means of X and Y, the regression coefficient of Y on X, and a specific value for X. We simply plug these values into the Y on X regression line formula to find the estimated Y.

🎯 Exam Tip: To estimate Y for a given X, always use the regression equation of Y on X. Similarly, to estimate X for a given Y, use the regression equation of X on Y. Ensure you correctly substitute the mean values \( \bar{x} \) and \( \bar{y} \).

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.
Answer:
Solution:
Since \( (\bar{x}, \bar{y}) \) is the point of intersection of the regression lines.
Given equations:
\( x - 4y = 5 \quad \ldots(i) \)
\( -x + 16y = 64 \quad \ldots(ii) \)
Adding equation (i) and (ii):
\( (x - 4y) + (-x + 16y) = 5 + 64 \)
\( 12y = 69 \)
\( y = \frac{69}{12} = 5.75 \)
Substituting \( y = 5.75 \) in equation (i):
\( x - 4(5.75) = 5 \)
\( x - 23 = 5 \)
\( x = 28 \)
\( \therefore \bar{x} = 28, \bar{y} = 5.75 \)
Now, let's find the regression coefficients.
From \( x - 4y = 5 \), assuming X on Y:
\( x = 4y + 5 \)
\( \therefore b_{xy} = 4 \)
From \( 16y - x = 64 \), assuming Y on X:
\( 16y = x + 64 \)
\( y = \frac{1}{16}x + \frac{64}{16} \)
\( y = \frac{1}{16}x + 4 \)
\( b_{yx} = \frac{1}{16} \)
Check assumption: \( b_{yx} \cdot b_{xy} = \frac{1}{16} \times 4 = \frac{4}{16} = \frac{1}{4} \)
\( \frac{1}{4} \in [0, 1] \). Our assumption is correct.
\( \therefore r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{1}{4} \)
\( r = \pm \sqrt{\frac{1}{4}} \)
\( r = \pm \frac{1}{2} \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = \frac{1}{2} = 0.5 \)
In simple words: This problem asks for the mean values of X and Y and their correlation coefficient from two regression equations. We begin by solving the equations simultaneously to find the point of intersection, which represents the means. Then, we rearrange each equation to identify the regression coefficients. Finally, the correlation coefficient 'r' is calculated using the square root of the product of these regression coefficients, ensuring its sign matches that of the coefficients.

🎯 Exam Tip: Always solve the regression equations simultaneously to find \( (\bar{x}, \bar{y}) \) accurately. For the correlation coefficient 'r', remember that \( r^2 = b_{xy} \cdot b_{yx} \), and the sign of 'r' must align with the signs of \( b_{xy} \) and \( b_{yx} \).

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.
Answer:
Solution:
Given \( \sigma_x^2 = 25, \therefore \sigma_x = 5 \)
(i) Since \( (\bar{x}, \bar{y}) \) is the point of intersection of regression lines
Given equations:
Regression of Y on X: \( -x + 5y = 22 \quad \ldots(i) \)
Regression of X on Y: \( 64x - 45y = 22 \quad \ldots(ii) \)
Multiply equation (i) by 9:
\( -9x + 45y = 198 \quad \ldots(iii) \)
Add equation (ii) and (iii):
\( (64x - 45y) + (-9x + 45y) = 22 + 198 \)
\( 55x = 220 \)
\( x = 4 \)
Substituting \( x = 4 \) in equation (i):
\( -4 + 5y = 22 \)
\( 5y = 26 \)
\( \therefore y = 5.2 \)
\( \therefore \bar{x} = 4, \bar{y} = 5.2 \)
(ii) To find standard deviation of Y, we first need regression coefficients.
From Regression equation of Y on X: \( 5y - x = 22 \)
\( 5y = x + 22 \)
\( y = \frac{1}{5}x + \frac{22}{5} \)
\( b_{yx} = \frac{1}{5} \)
From Regression equation of X on Y: \( 64x - 45y = 22 \)
\( 64x = 45y + 22 \)
\( x = \frac{45}{64}y + \frac{22}{64} \)
\( b_{xy} = \frac{45}{64} \)
(iii) Coefficient of correlation between X and Y.
\( r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{1}{5} \times \frac{45}{64} \)
\( r^2 = \frac{9}{64} \)
\( r = \pm \sqrt{\frac{9}{64}} \)
\( r = \pm \frac{3}{8} \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = \frac{3}{8} = 0.375 \)
Now find \( \sigma_y \). We know \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \)
\( \frac{1}{5} = \frac{3}{8} \times \frac{\sigma_y}{5} \)
\( \frac{1}{5} = \frac{3 \sigma_y}{40} \)
\( 8 = 3 \sigma_y \)
\( \sigma_y = \frac{8}{3} \)
In simple words: This problem asks us to find the mean values of X and Y, the standard deviation of Y, and the correlation coefficient, given the variance of X and two regression equations. First, we solve the two regression equations simultaneously to find the means. Then, we extract the regression coefficients from these equations. The correlation coefficient is calculated using the product of the regression coefficients. Finally, we use the formula linking the regression coefficient of Y on X, the correlation coefficient, and the standard deviations of X and Y to find \( \sigma_y \).

🎯 Exam Tip: Always clearly identify which regression equation is Y on X and which is X on Y to correctly extract \( b_{yx} \) and \( b_{xy} \). Ensure that the signs of the regression coefficients determine the sign of the correlation coefficient 'r'. Double-check your algebraic steps, especially when solving simultaneous equations and rearranging formulas for standard deviations.

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 \( \sqrt{0.375} = 0.61 \)]
Answer:
Solution:
Since \( (\bar{x}, \bar{y}) \) is the point of intersection of the regression line
Given equations:
\( 2x = y + 15 \implies 2x - y = 15 \quad \ldots(i) \)
\( 4y - 3x + 25 = 0 \implies -3x + 4y = -25 \quad \ldots(ii) \)
Multiplying equation (i) by 4:
\( 8x - 4y = 60 \quad \ldots(iii) \)
Add equation (ii) and (iii):
\( (-3x + 4y) + (8x - 4y) = -25 + 60 \)
\( 5x = 35 \)
\( x = 7 \)
(i) \( \bar{x} = 7 \)
Substituting \( x = 7 \) in equation (i):
\( 2(7) - y = 15 \)
\( 14 - y = 15 \)
\( -y = 1 \)
\( y = -1 \)
(ii) \( \bar{y} = -1 \)
For regression coefficients:
From \( 2x = y + 15 \) (X on Y):
\( x = \frac{1}{2}y + \frac{15}{2} \)
(iv) \( b_{xy} = \frac{1}{2} \)
From \( 4y = 3x - 25 \) (Y on X, from \( -3x+4y=-25 \)):
\( y = \frac{3}{4}x - \frac{25}{4} \)
(iii) \( b_{yx} = \frac{3}{4} \)
Check assumption: \( b_{xy} \cdot b_{yx} = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} \)
\( \frac{3}{8} \in [0, 1] \). Our assumption is correct.
(v) Coefficient of correlation 'r':
\( r^2 = b_{xy} \cdot b_{yx} \)
\( r^2 = \frac{3}{8} \)
\( r^2 = 0.375 \)
\( r = \pm \sqrt{0.375} \)
Given \( \sqrt{0.375} = 0.61 \)
\( r = \pm 0.61 \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = 0.61 \)
In simple words: This problem asks for the mean values of X and Y, both regression coefficients, and the correlation coefficient 'r'. We first solve the given regression equations simultaneously to find the mean values. Then, we rearrange each equation to isolate X and Y respectively to identify \( b_{xy} \) and \( b_{yx} \). Finally, 'r' is calculated from the product of these coefficients, considering their signs.

🎯 Exam Tip: When the problem explicitly states which equation is "X on Y" or "Y on X", use that information to directly identify \( b_{xy} \) and \( b_{yx} \) without needing to test assumptions. However, still verify that their product is between 0 and 1. Pay close attention to the algebraic manipulation of the equations.

Question 10. The two regression equation are 5x – 6y + 90 = 0 and 15x – 8y – 130 = 0. Find \( \bar{x}, \bar{y} \), r.
Answer:
Solution:
Since \( (\bar{x}, \bar{y}) \) is the point of intersection of the regression lines
Given equations:
\( 5x - 6y + 90 = 0 \implies 5x - 6y = -90 \quad \ldots(i) \)
\( 15x - 8y - 130 = 0 \implies 15x - 8y = 130 \quad \ldots(ii) \)
Multiply equation (i) by 3:
\( 15x - 18y = -270 \quad \ldots(iii) \)
Subtract equation (iii) from (ii):
\( (15x - 8y) - (15x - 18y) = 130 - (-270) \)
\( 10y = 400 \)
\( y = 40 \)
Substituting \( y = 40 \) in equation (i):
\( 5x - 6(40) = -90 \)
\( 5x - 240 = -90 \)
\( 5x = 150 \)
\( x = 30 \)
\( \therefore \bar{x} = 30, \bar{y} = 40 \)
For regression coefficients:
From \( 5x - 6y = -90 \), assuming Y on X:
\( 6y = 5x + 90 \)
\( y = \frac{5}{6}x + 15 \)
\( b_{yx} = \frac{5}{6} \)
From \( 15x - 8y = 130 \), assuming X on Y:
\( 15x = 8y + 130 \)
\( x = \frac{8}{15}y + \frac{130}{15} \)
\( b_{xy} = \frac{8}{15} \)
Check assumption: \( b_{yx} \cdot b_{xy} = \frac{5}{6} \times \frac{8}{15} = \frac{40}{90} = \frac{4}{9} \)
\( \frac{4}{9} \in [0, 1] \). Our assumption is correct.
Correlation coefficient 'r':
\( r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{4}{9} \)
\( \therefore r = \pm \sqrt{\frac{4}{9}} \)
\( r = \pm \frac{2}{3} \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = \frac{2}{3} \)
In simple words: This problem asks us to find the mean values of X and Y, and the correlation coefficient 'r', from two given regression equations. First, we solve the system of equations to find the point of intersection, which gives us \( \bar{x} \) and \( \bar{y} \). Then, we rearrange each regression equation to determine the regression coefficients \( b_{yx} \) and \( b_{xy} \). Finally, 'r' is calculated using the formula \( r = \sqrt{b_{yx} \cdot b_{xy}} \), making sure its sign aligns with the coefficients.

🎯 Exam Tip: When deciding which equation is Y on X or X on Y, check if the product of the resulting regression coefficients is between 0 and 1. If not, switch your assumptions. Ensure to simplify fractions for regression coefficients for clarity and easier calculation of 'r'.

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.
Answer:
Solution:
Let's test assumptions for regression coefficients.
Assume \( 10x + 3y - 62 = 0 \) is X on Y:
\( 10x = -3y + 62 \)
\( x = -\frac{3}{10}y + \frac{62}{10} \)
\( b_{xy} = -\frac{3}{10} \)
Assume \( 6x + 5y - 50 = 0 \) is Y on X:
\( 5y = -6x + 50 \)
\( y = -\frac{6}{5}x + 10 \)
\( b_{yx} = -\frac{6}{5} \)
Check assumption: \( b_{xy} \cdot b_{yx} = (-\frac{3}{10}) \times (-\frac{6}{5}) = \frac{18}{50} = \frac{9}{25} \)
\( \frac{9}{25} \in [0, 1] \). Our assumption is correct.
Therefore, the regression equation of X on Y is \( 10x + 3y - 62 = 0 \).
Now, find \( \bar{x}, \bar{y} \) by solving the two equations simultaneously:
\( 10x + 3y = 62 \quad \ldots(i) \)
\( 6x + 5y = 50 \quad \ldots(ii) \)
Multiply equation (i) by 5 and equation (ii) by 3:
\( 50x + 15y = 310 \quad \ldots(iii) \)
\( 18x + 15y = 150 \quad \ldots(iv) \)
Subtract equation (iv) from (iii):
\( (50x + 15y) - (18x + 15y) = 310 - 150 \)
\( 32x = 160 \)
\( x = 5 \)
Substituting \( x = 5 \) in equation (i):
\( 10(5) + 3y = 62 \)
\( 50 + 3y = 62 \)
\( 3y = 12 \)
\( \therefore y = 4 \)
\( \therefore \bar{x} = 5, \bar{y} = 4 \)
For correlation coefficient 'r':
\( r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{9}{25} \)
\( r = \pm \sqrt{\frac{9}{25}} \)
\( r = \pm \frac{3}{5} \)
Since \( b_{yx} \) and \( b_{xy} \) are negative, \( \therefore r = -\frac{3}{5} = -0.6 \)
In simple words: This problem asks us to identify the X on Y regression equation, then find the means of X and Y, and the correlation coefficient 'r'. We test two assumptions for the regression coefficients and choose the one whose product lies between 0 and 1. Once the correct regression equations are identified, we solve them simultaneously to find the mean values. Finally, the correlation coefficient 'r' is determined from the square root of the product of the regression coefficients, ensuring its sign matches.

🎯 Exam Tip: When it's not specified which equation is X on Y or Y on X, you must test assumptions. The crucial test is \( 0 \le b_{yx} \cdot b_{xy} \le 1 \). If the product falls outside this range, your assumption for the coefficients is incorrect. Remember that if both \( b_{yx} \) and \( b_{xy} \) are negative, 'r' must also be negative.

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.
Answer:
Solution:
Let's test assumptions for regression coefficients.
Assume \( 10y = 3x + 170 \) is Y on X:
\( y = \frac{3}{10}x + \frac{170}{10} \)
\( y = \frac{3}{10}x + 17 \)
\( b_{yx} = \frac{3}{10} \)
Assume \( 5x + 70 = 6y \) (which is \( 6y = 5x + 70 \)) is X on Y:
\( 5x = 6y - 70 \)
\( x = \frac{6}{5}y - \frac{70}{5} \)
\( x = \frac{6}{5}y - 14 \)
\( b_{xy} = \frac{6}{5} \)
Check assumption: \( b_{yx} \cdot b_{xy} = \frac{3}{10} \times \frac{6}{5} = \frac{18}{50} = \frac{9}{25} \)
\( \frac{9}{25} \in [0, 1] \). Our assumption is correct.
For correlation coefficient 'r':
\( r^2 = b_{yx} \cdot b_{xy} \)
\( r^2 = \frac{9}{25} \)
\( r = \pm \sqrt{\frac{9}{25}} \)
\( r = \pm \frac{3}{5} \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( r = \frac{3}{5} = 0.6 \)
Now, find \( \bar{x}, \bar{y} \) by solving the two equations simultaneously:
From \( 10y = 3x + 170 \implies 3x - 10y = -170 \quad \ldots(i) \)
From \( 5x + 70 = 6y \implies 5x - 6y = -70 \quad \ldots(ii) \)
Multiply equation (i) by 3 and equation (ii) by 5:
\( 9x - 30y = -510 \quad \ldots(iii) \)
\( 25x - 30y = -350 \quad \ldots(iv) \)
Subtract equation (iii) from (iv):
\( (25x - 30y) - (9x - 30y) = -350 - (-510) \)
\( 16x = 160 \)
\( x = 10 \)
Substituting \( x = 10 \) in equation (i):
\( 3(10) - 10y = -170 \)
\( 30 - 10y = -170 \)
\( -10y = -200 \)
\( y = 20 \)
\( \therefore \bar{x} = 10, \bar{y} = 20 \)
In simple words: This problem requires finding the correlation coefficient 'r' and the mean values of X and Y from two given regression equations. We first rearrange the equations to identify the regression coefficients and confirm our assumption by verifying that their product is within the valid range. From this product, 'r' is calculated. Then, the mean values of X and Y are found by solving the two regression equations simultaneously, as their intersection point represents the means.

🎯 Exam Tip: Always state your assumption for which equation represents Y on X and X on Y, and then validate it by checking \( 0 \le b_{yx} \cdot b_{xy} \le 1 \). If the product is not in this range, reverse your assumption. Remember to present both 'r' and the mean values as requested.

Question 13. Regression equation of two series are 2x – y – 15 = 0 and 3x- 4y + 25 = 0. Find \( \bar{x}, \bar{y} \) and regression coefficients, Also find coefficients of correlation. [Given \( \sqrt{0.375} = 0.61 \)]
Answer:
Solution:
Since \( (\bar{x}, \bar{y}) \) is the point of intersection of the regression line
Given equations:
\( 2x - y - 15 = 0 \implies 2x - y = 15 \quad \ldots(i) \)
\( 3x - 4y + 25 = 0 \implies 3x - 4y = -25 \quad \ldots(ii) \)
Multiply equation (i) by 4:
\( 8x - 4y = 60 \quad \ldots(iii) \)
Subtract equation (ii) from (iii):
\( (8x - 4y) - (3x - 4y) = 60 - (-25) \)
\( 5x = 85 \)
\( x = 17 \)
Substituting \( x = 17 \) in equation (i):
\( 2(17) - y = 15 \)
\( 34 - y = 15 \)
\( y = 34 - 15 \)
\( y = 19 \)
\( \therefore \bar{x} = 17, \bar{y} = 19 \)
For regression coefficients:
Assume \( 2x - y = 15 \) is X on Y:
\( 2x = y + 15 \)
\( x = \frac{1}{2}y + \frac{15}{2} \)
\( b_{xy} = \frac{1}{2} \)
Assume \( 3x - 4y = -25 \) is Y on X:
\( 4y = 3x + 25 \)
\( y = \frac{3}{4}x + \frac{25}{4} \)
\( b_{yx} = \frac{3}{4} \)
Check assumption: \( b_{xy} \cdot b_{yx} = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} \)
\( \frac{3}{8} \in [0, 1] \). Our assumption is correct.
Coefficient of correlation 'r':
\( r^2 = b_{xy} \cdot b_{yx} \)
\( r^2 = \frac{3}{8} \)
\( r^2 = 0.375 \)
\( r = \pm \sqrt{0.375} \)
Given \( \sqrt{0.375} = 0.61 \)
\( r = \pm 0.61 \)
Since \( b_{yx} \) and \( b_{xy} \) are positive, \( \therefore r = 0.61 \)
In simple words: This problem asks for the mean values of X and Y, both regression coefficients, and the correlation coefficient 'r'. We start by solving the two regression equations simultaneously to find the means, \( \bar{x} \) and \( \bar{y} \). Then, we rearrange each equation to identify \( b_{xy} \) and \( b_{yx} \), ensuring their product is valid. Finally, 'r' is calculated from the square root of the product of these coefficients, considering their positive signs.

🎯 Exam Tip: When calculating 'r' from \( r^2 = b_{xy} \cdot b_{yx} \), always check the signs of \( b_{xy} \) and \( b_{yx} \). If both are positive, 'r' is positive; if both are negative, 'r' is negative. Never just take the positive square root blindly.

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.
Answer:
Solution:
Since \( (\bar{x}, \bar{y}) \) is the point of intersection of the regression lines
Given equations:
\( 4y - 15x + 500 = 0 \implies -15x + 4y = -500 \implies 15x - 4y = 500 \quad \ldots(i) \)
\( 20x - 3y - 900 = 0 \implies 20x - 3y = 900 \quad \ldots(ii) \)
Multiply equation (i) by 3 and equation (ii) by 4:
\( 45x - 12y = 1500 \quad \ldots(iii) \)
\( 80x - 12y = 3600 \quad \ldots(iv) \)
Subtract equation (iii) from (iv):
\( (80x - 12y) - (45x - 12y) = 3600 - 1500 \)
\( 35x = 2100 \)
\( x = 60 \)
Substituting \( x = 60 \) in equation (i):
\( 15(60) - 4y = 500 \)
\( 900 - 4y = 500 \)
\( -4y = -400 \)
\( y = 100 \)
\( \therefore \bar{x} = 60, \bar{y} = 100 \)
To estimate Y (weight) for a given X (height), we need the regression equation of Y on X.
From \( 4y - 15x + 500 = 0 \):
\( 4y = 15x - 500 \)
\( Y = \frac{15}{4}x - \frac{500}{4} \)
\( Y = \frac{15}{4}x - 125 \)
So, \( b_{yx} = \frac{15}{4} \)
Now, let's find \( b_{xy} \) from the second equation to verify consistency.
From \( 20x - 3y - 900 = 0 \):
\( 20x = 3y + 900 \)
\( x = \frac{3}{20}y + \frac{900}{20} \)
\( x = \frac{3}{20}y + 45 \)
So, \( b_{xy} = \frac{3}{20} \)
Check assumption: \( b_{yx} \cdot b_{xy} = \frac{15}{4} \times \frac{3}{20} = \frac{45}{80} = \frac{9}{16} \)
\( \frac{9}{16} \in [0, 1] \). Our assumption is correct.
Estimate the weight of a girl whose height is 70 inches.
Using the regression equation of Y on X:
\( Y = \frac{15}{4}x - 125 \)
When \( x = 70 \):
\( Y = \frac{15}{4} \times 70 - 125 \)
\( Y = 15 \times 17.5 - 125 \)
\( Y = 262.5 - 125 \)
\( Y = 137.5 \text{ kg} \)
In simple words: This problem asks for the average height and weight, and to estimate a girl's weight given her height, using two regression lines. First, we find the mean height and weight by solving the two regression equations simultaneously. To estimate weight from height, we identify the regression equation of Y on X (weight on height), and then substitute the given height value into this equation to find the estimated weight.

🎯 Exam Tip: When estimating a value (e.g., Y from X), always use the correct regression line (Y on X). It's good practice to derive both regression coefficients and check their product to ensure consistency, even if not explicitly asked for 'r', as it validates your equations.

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