Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
Comprehensive Concepts & Principles: Lines of Regression
The lines of regression are algebraic best-fit lines derived under the principle of least squares to describe how the conditional average of one variable changes in response to changes in another. In CBSE Class 11 Applied Mathematics, two distinct regression equations are formulated depending on which variable is designated as the independent predictor: the regression line of $Y$ on $X$ and the regression line of $X$ on $Y$.
Standard Algebraic Formulations:
1. Line of Regression of $Y$ on $X$ (Estimating $Y$ from $X$): $$Y - \bar{Y} = b_{yx}(X - \bar{X})$$
where $b_{yx}$ is the regression coefficient of $Y$ on $X$, defined as:
$$b_{yx} = r \frac{\sigma_y}{\sigma_x} = \frac{\sum (x_i - \bar{X})(y_i - \bar{Y})}{\sum (x_i - \bar{X})^2} = \frac{N\sum XY - (\sum X)(\sum Y)}{N\sum X^2 - (\sum X)^2}$$
2. Line of Regression of $X$ on $Y$ (Estimating $X$ from $Y$): $$X - \bar{X} = b_{xy}(Y - \bar{Y})$$
where $b_{xy}$ is the regression coefficient of $X$ on $Y$, defined as:
$$b_{xy} = r \frac{\sigma_x}{\sigma_y} = \frac{\sum (x_i - \bar{X})(y_i - \bar{Y})}{\sum (y_i - \bar{Y})^2} = \frac{N\sum XY - (\sum X)(\sum Y)}{N\sum Y^2 - (\sum Y)^2}$$
3. Normal Equations of Least Squares: • For $Y = a + bX$:
(i) $\sum Y = n a + b\sum X$
(ii) $\sum XY = a\sum X + b\sum X^2$
• For $X = c + dY$:
(i) $\sum X = n c + d\sum Y$
(ii) $\sum XY = c\sum Y + d\sum Y^2$
Step-by-Step Prediction Protocol
1. Determine Means: $\bar{X} = \frac{\sum X}{N}$ and $\bar{Y} = \frac{\sum Y}{N}$.
2. Compute Slope $b_{yx}$ using deviations or raw moments.
3. Formulate $Y$ on $X$ equation: $Y = \bar{Y} + b_{yx}(X - \bar{X})$.
4. Formulate $X$ on $Y$ equation: $X = \bar{X} + b_{xy}(Y - \bar{Y})$.
5. Substitute the known value into the corresponding equation to forecast the target variable.
Worked Benchmark Examples
Example 1 (Deriving Both Regression Lines from Summary Stats):
From the following summary data for 100 corporate employees:
$\bar{X} = 40 \text{ years (Age)}, \quad \bar{Y} = ₹60,000 \text{ (Monthly Salary)}$
$\sigma_x = 5 \text{ years}, \quad \sigma_y = ₹12,000, \quad r = 0.8$
(a) Find the regression equation of $Y$ on $X$.
(b) Estimate the monthly salary of an employee aged 45 years.
(c) Find the regression equation of $X$ on $Y$, and estimate the age of an employee earning ₹84,000.
Graded textbook exercises on Formulating Lines of Regression and Predictive Forecasting from CBSE Class 11 Applied Mathematics.
Problem 1Regression Line of Y on X
Given $\bar{X} = 18, \bar{Y} = 100, \sigma_x = 14, \sigma_y = 20, r = 0.7$.
(a) Find the regression equation of $Y$ on $X$.
(b) Estimate the value of $Y$ when $X = 70$.
Using the same data from Problem 1 ($ar{X} = 18, ar{Y} = 100, sigma_x = 14, sigma_y = 20, r = 0.7$):
(a) Find the regression equation of $X$ on $Y$.
(b) Estimate the value of $X$ when $Y = 90$.
For 10 pairs of observations:
$\sum x = 0, \quad \sum y = 0, \quad \sum x^2 = 60, \quad \sum y^2 = 120, \quad \sum xy = 72$.
Formulate both lines of regression.
Show Step-by-Step Solution
• Since $\sum x = 0$ and $\sum y = 0$, deviations are already centered at means $\bar{X} = 0, \bar{Y} = 0$.
• $b_{yx} = \frac{\sum xy}{\sum x^2} = \frac{72}{60} = \mathbf{1.2}$.
Equation of $Y$ on $X$: $Y - 0 = 1.2(X - 0) \implies \mathbf{Y = 1.2X}$.
• $b_{xy} = \frac{\sum xy}{\sum y^2} = \frac{72}{120} = \mathbf{0.6}$.
Equation of $X$ on $Y$: $X - 0 = 0.6(Y - 0) \implies \mathbf{X = 0.6Y}$.
Problem 4Slope & Mean Intersection Check
A researcher computes two regression lines as $Y = 2X + 3$ and $X = 3Y + 2$. Verify whether these can represent authentic lines of regression.
Show Step-by-Step Solution
• Slope of $Y$ on $X$ is $b_{yx} = 2$.
• Slope of $X$ on $Y$ is $b_{xy} = 3$.
• Recall the fundamental property: $r = \pm \sqrt{b_{yx} \cdot b_{xy}}$.
• Here $b_{yx} \cdot b_{xy} = 2 \times 3 = 6 > 1$, which implies $r = \sqrt{6} \approx 2.45 > 1$!
• Since correlation coefficient cannot exceed 1 ($|r| \le 1$), these cannot be authentic lines of regression.
Problem 5Commercial Forecast Drill
In a study of factory floor temperature ($X$ in °C) and defective items per shift ($Y$), the regression line is $Y = 1.5X - 10$. If the air conditioning fails and the floor temperature hits 36°C, forecast the expected defective output.
Show Step-by-Step Solution
• Substitute $X = 36$ into the regression line of $Y$ on $X$:
$$\hat{Y} = 1.5(36) - 10 = 54 - 10 = \mathbf{44 \text{ defective items}}$$
Comprehensive Practice Worksheet: Lines of Regression
Examination problems classified into 1-Mark, 2-Mark, and 4-Mark sections according to CBSE marking guidelines.
Q4: State the two normal equations used to determine the constants $a$ and $b$ for the regression line $Y = a + bX$.
Show Answer & Scheme
Normal Equations [2 Marks]:
(1) $\sum Y = na + b\sum X$
(2) $\sum XY = a\sum X + b\sum X^2$
SECTION C: 4-Mark Comprehensive Numerical Drill
Q5: The following data relates to heights of fathers ($X$ in inches) and their sons ($Y$ in inches) across 6 pairs:
$X$: $65, \quad 66, \quad 67, \quad 67, \quad 68, \quad 69$
$Y$: $67, \quad 68, \quad 65, \quad 68, \quad 72, \quad 72$
(i) Find the mean height of fathers $\bar{X}$ and sons $\bar{Y}$.
(ii) Find the regression coefficient $b_{yx}$ and the regression line of $Y$ on $X$.
(iii) Estimate the height of a son whose father's height is 70 inches.
Show Step-by-Step Solution & Marking Scheme
Part (i) Means [1 Mark]:
• $\sum X = 65+66+67+67+68+69 = 402 \implies \bar{X} = 402/6 = \mathbf{67 \text{ inches}}$.
• $\sum Y = 67+68+65+68+72+72 = 412 \implies \bar{Y} = 412/6 = 68.67 \text{ inches}$.
Part (iii) Prediction for $X=70$ [1 Mark]:
$$\hat{Y} = 1.4(70) - 25.13 = 98 - 25.13 = \mathbf{72.87 \text{ inches}}$$
High-Yield Quick Revision: Lines of Regression
Summary of equations, slopes, and predictive rules for Chapter 12.2.
Master Formula Reference
Line of Regression
Equation
Slope Formula
Target Prediction
$Y$ on $X$
$Y - \bar{Y} = b_{yx}(X - \bar{X})$
$b_{yx} = r\frac{\sigma_y}{\sigma_x}$
Estimate $Y$ for given $X$
$X$ on $Y$
$X - \bar{X} = b_{xy}(Y - \bar{Y})$
$b_{xy} = r\frac{\sigma_x}{\sigma_y}$
Estimate $X$ for given $Y$
Denominator Memory Anchor
Notice the symmetry:
• In $b_{yx}$ ($Y$ on $X$), the denominator is $\sigma_x^2$ or $\sum x^2$ (the independent variable $X$).
• In $b_{xy}$ ($X$ on $Y$), the denominator is $\sigma_y^2$ or $\sum y^2$ (the independent variable $Y$).
10-Question Diagnostic Board Mock Examination: Lines of Regression
Standard timed test: 4 Objective MCQs / Assertion-Reason with instant automatic grading, plus 6 multi-mark subjective problems with CBSE marking rubrics.
PART 1: Objective Section (4 Marks)
Q1: In the regression equation $Y - \bar{Y} = b_{yx}(X - \bar{X})$, the slope coefficient $b_{yx}$ represents:
Solution: (B) $b_{yx} = \frac{\Delta Y}{\Delta X}$ gives the expected marginal change in $Y$ per unit increase in $X$.
Q2: If $\sigma_x = 4, \sigma_y = 5,$ and $r = -0.6$, the value of $b_{xy}$ is:
Q3: If the regression line of $Y$ on $X$ is $Y = 3X + 5$ and $\bar{X} = 2$, then $\bar{Y}$ must be:
Solution: (B) Since the regression line passes through $(\bar{X}, \bar{Y})$: $\bar{Y} = 3(2) + 5 = 6 + 5 = 11$.
Q4 (Assertion-Reason): Assertion (A): If the regression line of $Y$ on $X$ is $Y = 2X + 4$, the regression line of $X$ on $Y$ cannot be $X = 3Y - 1$. Reason (R): The product of the two regression coefficients $b_{yx} \cdot b_{xy}$ equals $r^2$, which cannot exceed 1.
Solution: (A) Here $b_{yx} = 2$ and $b_{xy} = 3 \implies b_{yx} b_{xy} = 6 > 1$, which violates $r^2 \le 1$. Assertion and Reason are true.
PART 2: Subjective Section (Q5 - Q10)
Q52 Marks
If $\bar{X} = 30, \bar{Y} = 40, b_{yx} = 0.8,$ and $b_{xy} = 0.45$, write both regression equations.
Given the regression line $2X - 5Y + 10 = 0$, express $Y$ in terms of $X$ and state the regression coefficient $b_{yx}$.
Show Step-by-Step Solution & Marking Scheme
• $5Y = 2X + 10 \implies Y = \frac{2}{5}X + 2 = \mathbf{0.4X + 2}$ [1 Mark].
• The slope coefficient is $b_{yx} = \frac{2}{5} = \mathbf{0.4}$ [1 Mark].
Q73 Marks
Show that the arithmetic mean of the two regression coefficients is always greater than or equal to the correlation coefficient $r$ ($\frac{b_{yx} + b_{xy}}{2} \ge r$).
Show Step-by-Step Solution & Marking Scheme
• By the AM-GM inequality, for any two positive real numbers: $\frac{a + b}{2} \ge \sqrt{ab}$ [1 Mark].
• Let $a = b_{yx}$ and $b = b_{xy}$.
$$\frac{b_{yx} + b_{xy}}{2} \ge \sqrt{b_{yx} \cdot b_{xy}} \quad [1 \text{ Mark}]$$
• Since $r = \sqrt{b_{yx} \cdot b_{xy}}$:
$$\mathbf{\frac{b_{yx} + b_{xy}}{2} \ge r} \quad [1 \text{ Mark}]$$
• Equality holds if and only if $b_{yx} = b_{xy}$ (i.e. $\sigma_x = \sigma_y$).
Q83 Marks
In an evaluation of advertising spend ($X$ in ₹ Lakhs) and sales ($Y$ in ₹ Crores):
$\bar{X} = 10, \bar{Y} = 90, \sigma_x = 3, \sigma_y = 12, r = 0.8$.
(a) Find the line of $Y$ on $X$.
(b) Predict sales if advertising spend is ₹15 Lakhs.
(c) What is the marginal increase in sales per ₹1 Lakh increase in advertising?
A corporate human resource directorate studies sales targets met ($X$ in %) and quarterly bonus granted ($Y$ in ₹ Thousands) across 50 sales executives:
$\bar{X} = 100\%, \quad \bar{Y} = 50, \quad \sigma_x = 15\%, \quad \sigma_y = 10, \quad r = 0.75$.
Q9 (2 Marks): Formulate the regression line of bonus ($Y$) on performance ($X$). Q10 (2 Marks): What bonus should an executive achieve who meets 130% of their sales target?
Solution to Q10 [2 Marks]:
• For $X = 130\%$:
$$\hat{Y} = 0.5(130) = \mathbf{₹65 \text{ Thousands}} \quad (₹65,000)$$
• The executive should receive a bonus of ₹65,000.