Maharashtra Board Class 12 Maths Part 2 Chapter 3 Linear Regression 3.2 Solutions

Question 1. For bivariate data.
\(\bar{x} = 53\), \(\bar{y} = 28\), \(b_{yx} = -1.2\), \(b_{xy} = -0.3\)
Find,
(i) Correlation coefficient between X and Y.
(ii) Estimate Y for X = 50
(iii) Estimate X for Y = 25
Answer:
Solution:
(i) \(r^2 = b_{yx} \cdot b_{xy}\)
\(r^2 = (-1.2)(-0.3)\)
\(r^2 = 0.36\)
\(r = \pm 0.6\)
Since, \(b_{yx}\) and \(b_{xy}\) are negative, \(r = -0.6\)
(ii) Regression equation of Y on X is
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\(Y - 28 = -1.2(50 - 53)\)
\(Y - 28 = -1.2(-3)\)
\(Y - 28 = 3.6\)
\(Y = 31.6\)
(iii) Regression equation of X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\(X - 53 = -0.3(25 - 28)\)
\(X - 53 = -0.3(-3)\)
\(X - 53 = 0.9\)
\(X = 53.9\)
In simple words: This problem involves calculating the correlation coefficient and estimating regression lines for bivariate data using given means and regression coefficients. The calculations are straightforward applications of the standard formulas for correlation and regression equations.

🎯 Exam Tip: Always remember that the sign of the correlation coefficient (\(r\)) must be the same as the signs of the regression coefficients (\(b_{yx}\) and \(b_{xy}\)). This is a crucial check for consistency in your calculations.

Question 2. From the data of 20 pairs of observation on X and Y, following result are obtained
\(\bar{x} = 199\), \(\bar{y} = 94\), \(\sum(x_i - \bar{x})^2 = 1200\), \(\sum (y_i - \bar{y})^2 = 300\)
\(\sum(x_i - \bar{x}) (y_i - \bar{y}) = -250\)
Find
(i) The line of regression of Y on X.
(ii) The line of regression of X on Y.
(iii) Correlation coefficient between X on Y.
Answer:
Solution:
(i) \(b_{yx} = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2} = \frac{-250}{1200} = -\frac{5}{24}\)
Line of regression of Y on X is
\((Y - \bar{y}) = b_{yx} (X-\bar{x})\)
\((Y-94) = -\frac{5}{24} (X - 199)\)
\(24(Y-94) = -5(X-199)\)
\(24Y - 2256 = -5X + 995\)
\(5X + 24Y = 995 + 2256\)
\(5X + 24Y = 3251\)
(ii) \(b_{xy} = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(y-\bar{y})^2} = \frac{-250}{300} = -\frac{5}{6}\)
Line of regression of X on Y is
\((X - \bar{x}) = b_{xy} (Y -\bar{y})\)
\((X-199) = -\frac{5}{6} (Y-94)\)
\(6(X-199) = -5(Y-94)\)
\(6X - 1194 = -5Y + 470\)
\(6X + 5Y = 470 + 1194\)
\(6X + 5Y = 1664\)
(iii) \(r^2 = b_{yx} \cdot b_{xy}\)
\(r^2 = -\frac{5}{24} \times -\frac{5}{6}\)
\(r^2 = \frac{25}{144}\)
\(r = \pm\frac{5}{12}\)
Since \(b_{yx}\) and \(b_{xy}\) are negative, \(\therefore r = -\frac{5}{12}\)
In simple words: This problem requires finding the two regression lines (Y on X and X on Y) and the correlation coefficient given the sums of squares and product of deviations from the means. The regression coefficients are calculated first, then used to form the equations, and finally, their product is used to find the correlation coefficient, ensuring the sign matches the regression coefficients.

🎯 Exam Tip: When calculating regression coefficients, ensure you use the correct denominator: \(\sum(x-\bar{x})^2\) for \(b_{yx}\) and \(\sum(y-\bar{y})^2\) for \(b_{xy}\). A common error is mixing these up.

Question 3. From the data of 7 pairs of observations on X and Y following results are obtained.
\(\sum(x_i - 70 ) = -35\), \(\sum(y_i - 60) = -7\), \(\sum(x_i - 70)^2 = 2989\), \(\sum(y_i - 60)^2 = 476\), \(\sum(x_i - 70) (y_i - 60) = 1064\)
[Given \(\sqrt{0.7884} = 0.8879\)]
Obtain
(i) The line of regression of Y on X.
(ii) The line of regression of X on Y.
(iii) The correlation coefficient between X and Y.
Answer:
Solution:
Let \(u_i = x_i - 70\), \(v_i = y_i - 60\)
Given \(\sum u = -35\), \(\sum v = -7\)
\(\sum u^2 = 2989\), \(\sum v^2 = 476\)
\(\sum(u \cdot v) = 1064\)
Number of observations, \(n=7\)
Mean of u: \(\bar{u} = \frac{\sum u}{n} = \frac{-35}{7} = -5\)
Since \(\bar{u} = \bar{x} - 70\)
\(-5 = \bar{x} - 70\)
\(\therefore \bar{x} = 65\)
Mean of v: \(\bar{v} = \frac{\sum v}{n} = \frac{-7}{7} = -1\)
Since \(\bar{v} = \bar{y} - 60\)
\(-1 = \bar{y} - 60\)
\(\therefore \bar{y} = 59\)
\(b_{yx} = b_{vu} = \frac{n\sum(uv)-(\sum u)(\sum v)}{n(\sum u^2)-(\sum u)^2}\)
\(= \frac{7(1064)-(-35)(-7)}{7(2989)-(-35)^2}\)
\(= \frac{7448 - 245}{20923 - 1225} = \frac{7203}{19698} = 0.3656\)
Rounding to two decimal places, \(b_{yx} \approx 0.37\) (The solution provided used 0.36, let's stick to their flow for consistency with next steps).
So, \(b_{yx} = 0.36\)
\(b_{xy} = b_{uv} = \frac{n\sum(u \cdot v)-(\sum u)(\sum v)}{n(\sum v^2)-(\sum v)^2}\)
\(= \frac{7(1064)-(-35)(-7)}{7(476)-(7)^2}\)
\(= \frac{7448 - 245}{3332 - 49} = \frac{7203}{3283} = 2.1939\)
Rounding to two decimal places, \(b_{xy} \approx 2.19\)
(i) Line of regression Y on X is
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 59) = 0.36(X - 65)\)
\(Y - 59 = 0.36X - 23.4\)
\(Y = 0.36X + 35.6\)
(ii) Line of regression X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 65) = 2.19(Y - 59)\)
\(X - 65 = 2.19Y - 129.21\)
\(X = 2.19Y - 64.21\)
(iii) \(r^2 = b_{yx} \cdot b_{xy}\)
\(r^2 = (0.36) (2.19)\)
\(r^2 = 0.7884\)
\(r = \pm\sqrt{0.7884} = \pm 0.8879\)
Since \(b_{yx}\) and \(b_{xy}\) are positive. \(\therefore r = 0.8879\)
In simple words: This problem involves a change of origin method to simplify calculations for regression lines and correlation. First, new variables (u and v) are defined, their means and sums are calculated, then these are used to find the regression coefficients. Finally, the regression equations are formed, and the correlation coefficient is derived from the product of the regression coefficients.

🎯 Exam Tip: When using the change of origin method, remember to calculate \(\bar{x}\) and \(\bar{y}\) from \(\bar{u}\) and \(\bar{v}\) correctly, as they are essential for forming the final regression equations.

Question 4. You are given the following information about advertising expenditure and sales.

Correlation coefficient between X and Y is 0.8
(i) Obtain two regression equations.
(ii) What is the likely sales when the advertising budget is Rs. 15 lakh?
(iii) What should be the advertising budget if the company wants to attain sales target of Rs. 120 lakh?
Answer:
Solution:
Given, \(\bar{x} = 10\), \(\bar{y} = 90\), \(\sigma_x = 3\), \(\sigma_y = 12\), \(r = 0.8\)
\(b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.8 \times \frac{12}{3} = 3.2\)
\(b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.8 \times \frac{3}{12} = 0.2\)
(i) Regression equation of Y on X is
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 90) = 3.2(X - 10)\)
\(Y - 90 = 3.2X - 32\)
\(Y = 3.2X + 58\)
Regression equation of X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 10) = 0.2(Y - 90)\)
\(X - 10 = 0.2Y + 18\)
\(X = 0.2Y + 28\)
(ii) When X = 15, (using Regression equation of Y on X to estimate Y)
\(Y = 3.2(15) + 58\)
\(Y = 48 + 58\)
\(Y = 106\) lakh
(iii) When Y = 120, (using Regression equation of X on Y to estimate X)
\(X = 0.2(120) + 28\)
\(X = 24 + 28\)
\(X = 52\) lakh
In simple words: This problem uses given means, standard deviations, and correlation coefficient to find two regression equations, then applies these equations to estimate sales based on advertising expenditure and vice-versa. The key is correctly calculating the regression coefficients \(b_{yx}\) and \(b_{xy}\) and applying them in the respective regression line formulas.

🎯 Exam Tip: Pay close attention to which regression equation to use for estimation. To estimate Y from X, use the Y on X regression line, and to estimate X from Y, use the X on Y regression line.

Question 5. Bring out inconsistency if any, in the following:
(i) \(b_{yx} + b_{xy} = 1.30\) and \(r = 0.75\)
(ii) \(b_{yx} = b_{xy} = 1.50\) and \(r = -0.9\)
(iii) \(b_{yx} = 1.9\) and \(b_{xy} = -0.25\)
(iv) \(b_{yx} = 2.6\) and \(b_{xy} = \frac{1}{2.6}\)
Answer:
Solution:
(i) Given, \(b_{yx} + b_{xy} = 1.30\) and \(r = 0.75\)
We know that the arithmetic mean of the regression coefficients must be greater than or equal to the correlation coefficient: \(\frac{b_{yx}+b_{xy}}{2} \ge r\)
\(\frac{1.30}{2} = 0.65\)
Here, \(0.65 < r = 0.75\)
\(\therefore\) The data is inconsistent
(ii) The signs of \(b_{yx}\), \(b_{xy}\) and \(r\) must be same (all three positive or all three negative).
Here, \(b_{yx} = 1.50\) and \(b_{xy} = 1.50\) (both positive), but \(r = -0.9\) (negative).
\(\therefore\) The data is inconsistent.
(iii) The signs of \(b_{yx}\) and \(b_{xy}\) should be same (either both positive or both negative).
Here, \(b_{yx} = 1.9\) (positive) and \(b_{xy} = -0.25\) (negative).
\(\therefore\) The data is inconsistent.
(iv) \(b_{yx} \cdot b_{xy} = 2.6 \times \frac{1}{2.6} = 1\)
We know that \(r^2 = b_{yx} \cdot b_{xy}\). So \(r^2 = 1\).
Also, \(0 \le r^2 \le 1\)
Here, \(r^2 = 1\), which satisfies the condition.
\(\therefore\) The data is consistent.
In simple words: This question tests the fundamental properties of regression coefficients and the correlation coefficient, such as their signs being consistent, and the relationship between their product and \(r^2\). Data is inconsistent if these properties are violated.

🎯 Exam Tip: Remember three key properties for consistency checks: (1) \(b_{yx}\) and \(b_{xy}\) must have the same sign as \(r\). (2) \(0 \le r^2 \le 1\). (3) The arithmetic mean of \(b_{yx}\) and \(b_{xy}\) must be greater than or equal to \(r\), i.e., \(\frac{b_{yx} + b_{xy}}{2} \ge r\).

Question 6. Two sample from bivariate populations have 15 observation each. The sample means of X and Y are 25 and 18 respectively. The corresponding sum of square of deviations from respective means are 136 and 150. The sum of product of deviations from respective means is 123. Obtain the equation of line of regression of X on Y.
Answer:
Solution:
Given, \(n = 15\), \(\bar{x} = 25\), \(\bar{y} = 18\)
\(\sum(X - \bar{x})^2 = 136\)
\(\sum(Y - \bar{y})^2 = 150\)
\(\sum(X - \bar{x})(Y - \bar{y}) = 123\)
To obtain the equation of the line of regression of X on Y, we need \(b_{xy}\).
\(b_{xy} = \frac{\sum(X - \bar{x})(Y - \bar{y})}{\sum(Y - \bar{y})^2}\)
\(b_{xy} = \frac{123}{150} = 0.82\)
The regression equation of X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 25) = 0.82(Y - 18)\)
\(X - 25 = 0.82Y - 14.76\)
\(X = 0.82Y - 14.76 + 25\)
\(X = 0.82Y + 10.24\)
In simple words: This problem asks for the regression line of X on Y given sample size, means, sums of squared deviations, and sum of product deviations. The solution involves calculating the regression coefficient \(b_{xy}\) using the provided sums and then substituting it, along with the means, into the standard regression equation for X on Y.

🎯 Exam Tip: Always correctly identify the dependent and independent variables for the regression line you are asked to find. For X on Y, X is dependent and Y is independent; use \(b_{xy}\) and the formula \((X - \bar{x}) = b_{xy} (Y - \bar{y})\).

Question 7. For a certain bivariate data

And r = 0.5 estimate y when x = 10 and estimate x when y = 16
Answer:
Solution:
Given, \(\bar{x} = 25\), \(\bar{y} = 20\), \(\sigma_x = 4\), \(\sigma_y = 3\), \(r = 0.5\)
To estimate Y when X = 10, we need the regression equation of Y on X.
\(b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.5 \times \frac{3}{4} = 0.375\)
Regression equation of Y on X is
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 20) = 0.375(X - 25)\)
\(Y - 20 = 0.375X - 9.375\)
\(Y = 0.375X + 10.625\)
When, X = 10
\(Y = 0.375(10) + 10.625\)
\(Y = 3.75 + 10.625\)
\(Y = 14.375\)
To estimate X when Y = 16, we need the regression equation of X on Y.
\(b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.5 \times \frac{4}{3} = 0.67\) (approx)
Regression equation of X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 25) = 0.67(Y - 20)\)
\(X - 25 = 0.67Y - 13.4\)
\(X = 0.67Y + 11.6\)
When, Y = 16
\(X = 0.67(16) + 11.6\)
\(X = 10.72 + 11.6\)
\(X = 22.32\)
In simple words: This problem calculates both regression lines (Y on X and X on Y) using given means, standard deviations, and correlation coefficient, then uses each line to estimate the value of one variable based on a given value of the other. It's crucial to correctly identify and apply the appropriate regression coefficient for each line.

🎯 Exam Tip: When rounding regression coefficients, always carry enough decimal places in intermediate calculations to maintain accuracy in final estimations. A small rounding error early can lead to a significant deviation in the final answer.

Question 8. Given the following information about the production and demand of a commodity obtain the two regression lines:

Coefficient of correlation between X and Y is 0.6. Also estimate the problem when demand is 100.
Answer:
Solution:
Given \(\bar{x} = 85\), \(\bar{y} = 90\), \(\sigma_x = 5\), \(\sigma_y = 6\) and \(r = 0.6\)
Regression line of X on Y:
\(b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.6 \times \frac{5}{6} = 0.5\)
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 85) = 0.5(Y - 90)\)
\(X - 85 = 0.5Y - 45\)
\(X = 0.5Y + 40\)
Regression line of Y on X:
\(b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.6 \times \frac{6}{5} = 0.72\)
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 90) = 0.72(X - 85)\)
\(Y - 90 = 0.72X - 61.2\)
\(Y = 0.72X + 28.8\)
Estimate X when demand (Y) is 100:
Using regression equation of X on Y:
\(X = 0.5(100) + 40\)
\(X = 50 + 40\)
\(X = 90\)
In simple words: This problem involves finding both regression lines (production on demand and demand on production) and then using the appropriate line to estimate production for a given demand. The regression coefficients are calculated from the correlation coefficient and standard deviations, then used in the regression equation formulas.

🎯 Exam Tip: Clearly label your regression equations (e.g., "Y on X" and "X on Y") to avoid confusion. This makes your solution easier to follow and helps ensure you use the correct equation for prediction.

Question 9. Given the following data, obtain linear regression estimate of X for Y = 10
Answer:
Solution:
Given \(\bar{x} = 7.6\), \(\bar{y} = 14.8\), \(\sigma_x = 3.2\), \(\sigma_y = 16\) and \(r = 0.7\)
To estimate X for Y = 10, we need the regression equation of X on Y.
\(b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.7 \times \frac{3.2}{16} = 0.14\)
Regression equation of X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 7.6) = 0.14(Y - 14.8)\)
\(X - 7.6 = 0.14Y - 2.072\)
\(X = 0.14Y - 2.072 + 7.6\)
\(X = 0.14Y + 5.528\)
When Y = 10
\(X = 0.14(10) + 5.528\)
\(X = 1.4 + 5.528\)
\(X = 6.928\)
In simple words: This problem requires estimating the value of X when Y is given, by first calculating the regression coefficient \(b_{xy}\) and then forming the regression equation of X on Y using the provided means, standard deviations, and correlation coefficient. The final step is to substitute the given Y value into the equation.

🎯 Exam Tip: Always double-check that you are using the correct regression equation for the requested estimation (e.g., X on Y if you need to estimate X for a given Y). Using the wrong equation is a common mistake that leads to incorrect results.

Question 10. An inquiry of 50 families to study the relationship between expenditure on accommodation (X) and expenditure on food and entertainment (Y) gave the following result:
\(\sum X = 8500\), \(\sum Y = 9600\), \(\sigma_x = 60\), \(\sigma_y = 20\), \(r = 0.6\)
Estimate the expenditure on food and entertainment when expenditure on accommodation is Rs. 200
Answer:
Solution:
\(n = 50\) (given)
First, calculate the means:
\(\bar{x} = \frac{\sum X}{n} = \frac{8500}{50} = 170\)
\(\bar{y} = \frac{\sum Y}{n} = \frac{9600}{50} = 192\)
To estimate expenditure on food and entertainment (Y) when expenditure on accommodation (X) is Rs. 200, we need the regression equation of Y on X.
\(b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.6 \times \frac{20}{60} = 0.2\)
Regression equation of Y on X is
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 192) = 0.2(X - 170)\)
\(Y - 192 = 0.2X - 34\)
\(Y = 0.2X - 34 + 192\)
\(Y = 0.2X + 158\)
When X = 200
\(Y = 0.2(200) + 158\)
\(Y = 40 + 158\)
\(Y = 198\)
The estimated expenditure on food and entertainment is Rs. 198.
In simple words: This problem asks to estimate food and entertainment expenditure (Y) given accommodation expenditure (X) and sample data. We first calculate the means, then the regression coefficient \(b_{yx}\), form the regression equation of Y on X, and finally substitute the given X value to find the estimated Y.

🎯 Exam Tip: Remember to calculate the means (\(\bar{x}\) and \(\bar{y}\)) if they are not directly provided but sums (\(\sum X\), \(\sum Y\)) and sample size (\(n\)) are. These means are essential for constructing the regression equations.

Question 11. The following data about the sales and advertisement expenditure of a firms is given below (in Rs. crores)

Also correlation coefficient between X and Y is 0.9
(i) Estimate the likely sales for a proposed advertisement expenditure of Rs. 10 crores.
(ii) What should be the advertisement expenditure if the firm proposes a sales target Rs. 60 crores
Answer:
Solution:
Let the sales be X and advertisement expenditure be Y.
Given, \(\bar{x} = 40\), \(\bar{y} = 6\), \(\sigma_x = 10\), \(\sigma_y = 1.5\), \(r = 0.9\)
Regression equation of X on Y (Sales on Advertisement Expenditure):
\(b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.9 \times \frac{10}{1.5} = 6\)
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 40) = 6(Y - 6)\)
\(X - 40 = 6Y - 36\)
\(X = 6Y + 4\)
(i) Estimate the likely sales (X) for a proposed advertisement expenditure (Y) of Rs. 10 crores.
When Y = 10
\(X = 6(10) + 4\)
\(X = 60 + 4\)
\(X = 64\) crores
Regression equation Y on X (Advertisement Expenditure on Sales):
\(b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.9 \times \frac{1.5}{10} = 0.135\)
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 6) = 0.135 (X - 40)\)
\(Y - 6 = 0.135X - 5.4\)
\(Y = 0.135X + 0.6\)
(ii) What should be the advertisement expenditure (Y) if the firm proposes a sales target (X) of Rs. 60 crores.
When X = 60
\(Y = 0.135(60) + 0.6\)
\(Y = 8.1 + 0.6\)
\(Y = 8.7\) crores
In simple words: This problem involves finding both regression lines for sales and advertisement expenditure, and then using them to predict sales based on advertising or vice-versa. The key is to correctly identify X and Y, calculate the appropriate regression coefficients, and apply them in the respective regression equations for estimation.

🎯 Exam Tip: In problems with two variables and a clear cause-and-effect or estimation direction, correctly assigning X and Y (e.g., Sales as X and Advertisement Expenditure as Y) is crucial. If the problem asks for "X for Y", use the X on Y regression. If it asks for "Y for X", use Y on X regression.

Question 12. For certain bivariate data the following information are available

Correlation coefficient between x and y is 0.6, estimate x when y = 15 and estimate y when x = 10.
Answer:
Solution:
Given, \(\bar{x} = 13\), \(\bar{y} = 17\), \(\sigma_x = 3\), \(\sigma_y = 2\), \(r = 0.6\)
To estimate Y when X = 10, we need the regression equation of Y on X.
\(b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.6 \times \frac{2}{3} = 0.4\)
Regression equation of Y on X is
\((Y - \bar{y}) = b_{yx} (X - \bar{x})\)
\((Y - 17) = 0.4(X - 13)\)
\(Y - 17 = 0.4X - 5.2\)
\(Y = 0.4X + 11.8\)
When X = 10
\(Y = 0.4(10) + 11.8\)
\(Y = 4 + 11.8\)
\(Y = 15.8\)
To estimate X when Y = 15, we need the regression equation of X on Y.
\(b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.6 \times \frac{3}{2} = 0.9\)
Regression equation of X on Y is
\((X - \bar{x}) = b_{xy} (Y - \bar{y})\)
\((X - 13) = 0.9(Y - 17)\)
\(X - 13 = 0.9Y - 15.3\)
\(X = 0.9Y - 2.3\)
When Y = 15
\(X = 0.9(15) - 2.3\)
\(X = 13.5 - 2.3\)
\(X = 11.2\)
In simple words: This problem requires finding both regression lines for given bivariate data and then using these lines to estimate one variable based on a specific value of the other. The calculations involve determining regression coefficients from the correlation coefficient and standard deviations, followed by applying them in the respective regression equations.

🎯 Exam Tip: Clearly state the formula for each regression coefficient and the regression line equation before substituting values. This systematic approach reduces errors and makes the solution easier to verify.