Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
Comprehensive Concepts & Principles: Properties of Regression Coefficients
The slopes of the two lines of regression, denoted by $b_{yx}$ (regression coefficient of $Y$ on $X$) and $b_{xy}$ (regression coefficient of $X$ on $Y$), are the fundamental governing parameters of linear regression analysis. They quantify the average marginal rate of change in the dependent variable per unit increase in the independent predictor, and they obey several profound algebraic and geometric theorems.
The 5 Golden Properties of Regression Coefficients:
1. Geometric Mean Property (Link to Correlation): The correlation coefficient $r$ is the geometric mean between the two regression coefficients:
$$r = \pm \sqrt{b_{yx} \cdot b_{xy}}$$
The sign of $r$ is identical to the sign of $b_{yx}$ and $b_{xy}$.
2. Same Sign Rule: Both regression coefficients $b_{yx}$ and $b_{xy}$, and the correlation coefficient $r$, MUST have the exact same algebraic sign (all positive or all negative). It is mathematically impossible for $b_{yx}$ and $b_{xy}$ to possess opposite signs because:
$$b_{yx} = r\frac{\sigma_y}{\sigma_x} \quad \text{and} \quad b_{xy} = r\frac{\sigma_x}{\sigma_y}, \quad \sigma_x > 0, \sigma_y > 0$$
3. Magnitude Bound Rule: If one regression coefficient is greater than 1 in absolute value, the other must be less than 1 in absolute value:
$$|b_{yx}| > 1 \implies |b_{xy}| < 1 \quad (\text{since } b_{yx} \cdot b_{xy} = r^2 \le 1)$$
They cannot simultaneously exceed unity.
4. Arithmetic Mean Property: The arithmetic mean of the two regression coefficients is always greater than or equal to the correlation coefficient $r$:
$$\frac{b_{yx} + b_{xy}}{2} \ge r \quad (\text{for positive } r)$$
5. Origin & Scale Transformation: • Regression coefficients are independent of change of origin (adding/subtracting constants leaves slopes unchanged).
• If $u = \frac{X - a}{h}$ and $v = \frac{Y - b}{k}$, then:
$$b_{yx} = \frac{k}{h} b_{vu} \quad \text{and} \quad b_{xy} = \frac{h}{k} b_{uv}$$
Protocol: Identifying Regression Lines from Unlabeled Equations
When given two linear equations without labels and asked to identify which is $Y$ on $X$ and which is $X$ on $Y$, follow this foolproof 3-step test:
1. Hypothesis: Arbitrarily assume Equation 1 is $Y$ on $X$ ($Y = a + b_{yx}X$) and Equation 2 is $X$ on $Y$ ($X = c + b_{xy}Y$).
2. Extract Slopes: Read off $b_{yx}$ and $b_{xy}$.
3. Validity Test: Compute $b_{yx} \cdot b_{xy}$.
• If $b_{yx} \cdot b_{xy} \le 1$: The assumption is CORRECT! Compute $r = \pm \sqrt{b_{yx} b_{xy}}$.
• If $b_{yx} \cdot b_{xy} > 1$: The assumption is FALSE. Reverse the roles: Equation 1 is $X$ on $Y$, and Equation 2 is $Y$ on $X$.
Worked Benchmark Examples
Example 1 (Identification & Complete Analysis):
Given the two regression equations:
$$8X - 10Y + 66 = 0 \quad \text{and} \quad 40X - 18Y = 214$$
(a) Identify which equation represents the regression line of $Y$ on $X$ and which represents $X$ on $Y$.
(b) Calculate the correlation coefficient $r$.
(c) Find the ratio of the standard deviations $\frac{\sigma_y}{\sigma_x}$.
Show Step-by-Step Solution
• Step 1: Test Assumption: Let (1) $8X - 10Y + 66 = 0$ be $Y$ on $X$:
$$10Y = 8X + 66 \implies Y = \frac{8}{10}X + 6.6 \implies b_{yx} = 0.8$$
Let (2) $40X - 18Y = 214$ be $X$ on $Y$:
$$40X = 18Y + 214 \implies X = \frac{18}{40}Y + 5.35 \implies b_{xy} = \frac{18}{40} = 0.45$$
• Step 2: Check Bound:
$$b_{yx} \cdot b_{xy} = 0.8 \times 0.45 = 0.36 \le 1$$
Since $0.36 \le 1$, our assumption is CORRECT! • $8X - 10Y + 66 = 0$ is the regression line of $Y$ on $X$.
• $40X - 18Y = 214$ is the regression line of $X$ on $Y$.
• (b) Correlation Coefficient $r$:
Since both $b_{yx}$ and $b_{xy}$ are positive, $r$ is positive:
$$r = +\sqrt{b_{yx} \cdot b_{xy}} = +\sqrt{0.36} = \mathbf{+0.60}$$
• (c) Ratio of Standard Deviations:
$$\frac{b_{yx}}{b_{xy}} = \frac{r(\sigma_y / \sigma_x)}{r(\sigma_x / \sigma_y)} = \frac{\sigma_y^2}{\sigma_x^2} = \frac{0.8}{0.45} = \frac{80}{45} = \frac{16}{9}$$
$$\frac{\sigma_y}{\sigma_x} = \sqrt{\frac{16}{9}} = \mathbf{\frac{4}{3}} \approx \mathbf{1.33}$$
Example 2 (Scale Transformation):
If the regression coefficient of $Y$ on $X$ is $b_{yx} = 1.6$, find the new regression coefficient $b_{vu}$ when variables are transformed by $u = 2X - 5$ and $v = 4Y + 10$.
Graded textbook exercises on Properties of Regression Coefficients from CBSE Class 11 Applied Mathematics.
Problem 1Correlation from Slopes
If the two regression coefficients are $b_{yx} = -0.9$ and $b_{xy} = -0.4$, find the correlation coefficient $r$.
Show Step-by-Step Solution
• Both regression coefficients are negative, so $r$ must be negative:
$$r = -\sqrt{b_{yx} \cdot b_{xy}} = -\sqrt{(-0.9)(-0.4)} = -\sqrt{0.36} = \mathbf{-0.60}$$ Answer: $r = -0.60$.
Problem 2AM vs Geometric Mean Check
If $b_{yx} = 1.2$ and $b_{xy} = 0.3$:
(a) Find the correlation coefficient $r$.
(b) Calculate the arithmetic mean of the regression coefficients and verify that $\frac{b_{yx} + b_{xy}}{2} \ge r$.
If $b_{yx} = 0.8$, $b_{xy} = 0.2$, and the variance of $X$ is 16, find the standard deviation of $Y$.
Show Step-by-Step Solution
• $\sigma_x^2 = 16 \implies \sigma_x = 4$.
• We know: $\frac{b_{yx}}{b_{xy}} = \frac{\sigma_y^2}{\sigma_x^2}$
$$\frac{0.8}{0.2} = \frac{\sigma_y^2}{16} \implies 4 = \frac{\sigma_y^2}{16} \implies \sigma_y^2 = 64$$
$$\sigma_y = \sqrt{64} = \mathbf{8}$$ Answer: The standard deviation of $Y$ is 8.
Problem 4Identification Test
Can $2X - 3Y + 1 = 0$ and $5X - Y + 4 = 0$ be the regression lines of $Y$ on $X$ and $X$ on $Y$ respectively? Justify.
Show Step-by-Step Solution
• If (1) is $Y$ on $X$: $3Y = 2X + 1 \implies b_{yx} = 2/3$.
• If (2) is $X$ on $Y$: $5X = Y - 4 \implies b_{xy} = 1/5$.
• $b_{yx} \cdot b_{xy} = \frac{2}{3} \times \frac{1}{5} = \frac{2}{15} \approx 0.133 \le 1$.
• Since both slopes are positive and their product is $\le 1$, YES, these can validly represent the regression lines with $r = \sqrt{2/15} \approx 0.365$.
Problem 5Impossible Coefficient Analysis
A student computes two regression coefficients as $b_{yx} = 1.8$ and $b_{xy} = -0.5$. Explain why this result is mathematically impossible.
Show Step-by-Step Solution
• Regression coefficients $b_{yx} = r(\sigma_y / \sigma_x)$ and $b_{xy} = r(\sigma_x / \sigma_y)$ share the sign of $r$ because standard deviations $\sigma_x, \sigma_y$ are strictly positive.
• It is impossible for $b_{yx}$ to be positive ($+1.8$) while $b_{xy}$ is negative ($-0.5$). Both must have the same sign.
Comprehensive Practice Worksheet: Properties of Regression Coefficients
Graded problems matching CBSE Class 11 Applied Mathematics examination blueprint.
Q2: Can both regression coefficients simultaneously be greater than 1? State yes or no with reason.
Show Answer & Scheme
Answer [1 Mark]:No. Because $b_{yx} \cdot b_{xy} = r^2 \le 1$. If both were greater than 1, their product would exceed 1, violating the bound $r^2 \le 1$.
SECTION B: 2-Mark Short Answer Problems
Q3: Prove that if the two regression coefficients are equal ($b_{yx} = b_{xy}$), then the standard deviations of both variables are equal ($\sigma_x = \sigma_y$).
Show Answer & Scheme
• $b_{yx} = r \frac{\sigma_y}{\sigma_x}$ and $b_{xy} = r \frac{\sigma_x}{\sigma_y}$ [1 Mark].
• $b_{yx} = b_{xy} \implies r \frac{\sigma_y}{\sigma_x} = r \frac{\sigma_x}{\sigma_y} \implies \frac{\sigma_y}{\sigma_x} = \frac{\sigma_x}{\sigma_y} \implies \sigma_y^2 = \sigma_x^2 \implies \mathbf{\sigma_x = \sigma_y}$ [1 Mark].
Q4: State the effect on $b_{yx}$ when: (i) 10 is added to all $X$ values, and (ii) All $Y$ values are multiplied by 3.
Show Answer & Scheme
• (i) No effect [1 Mark]: Adding a constant shifts origin, leaving regression coefficients unchanged.
• (ii) Multiplied by 3 [1 Mark]: $b_{yx}' = 3 \cdot b_{yx}$, because $b_{yx} = \Delta Y / \Delta X$.
SECTION C: 4-Mark Comprehensive Scenario Drill
Q5: The equations of two regression lines are given as:
$$3X + 2Y = 26 \quad \text{and} \quad 6X + Y = 31$$
(i) Which line is the regression line of $Y$ on $X$, and which is $X$ on $Y$? Prove using the product property.
(ii) Compute the correlation coefficient $r$.
(iii) Find the ratio of the variance of $Y$ to the variance of $X$ ($\sigma_y^2 / \sigma_x^2$).
Show Step-by-Step Solution & Marking Scheme
Part (i) Line Identification [2 Marks]:
• Let (1) $3X + 2Y = 26$ be $Y$ on $X$:
$$2Y = -3X + 26 \implies Y = -1.5X + 13 \implies b_{yx} = -1.5$$
• Let (2) $6X + Y = 31$ be $X$ on $Y$:
$$6X = -Y + 31 \implies X = -\frac{1}{6}Y + 5.17 \implies b_{xy} = -\frac{1}{6} \approx -0.167$$
• Check product: $b_{yx} \cdot b_{xy} = (-1.5) \times (-0.1667) = 0.25 \le 1$.
• Since $0.25 \le 1$, the assumption is CORRECT! - $3X + 2Y = 26$ is the line of $Y$ on $X$.
- $6X + Y = 31$ is the line of $X$ on $Y$.
Part (ii) Correlation Coefficient [1 Mark]:
• Since both slopes are negative, $r$ is negative:
$$r = -\sqrt{0.25} = \mathbf{-0.50}$$
Part (iii) Ratio of Variances [1 Mark]:
$$\frac{b_{yx}}{b_{xy}} = \frac{\sigma_y^2}{\sigma_x^2} = \frac{-1.5}{-1/6} = 1.5 \times 6 = \mathbf{9}$$
• The variance of $Y$ is 9 times the variance of $X$.
When $b_{yx}$ and $b_{xy}$ are both negative, never write $r$ as positive! Remember that $\sqrt{(-a)(-b)} = \sqrt{ab}$, but mathematically the sign of $r$ must match the negative slopes: $\mathbf{r = -\sqrt{b_{yx} b_{xy}}}$.
Q4 (Assertion-Reason): Assertion (A): The regression coefficients $b_{yx}$ and $b_{xy}$ cannot have opposite signs. Reason (R): The standard deviations $\sigma_x$ and $\sigma_y$ are strictly positive real numbers, so both $b_{yx} = r\frac{\sigma_y}{\sigma_x}$ and $b_{xy} = r\frac{\sigma_x}{\sigma_y}$ must take the sign of the correlation coefficient $r$.
Solution: (A) Both Assertion and Reason are true and Reason provides the exact algebraic justification.
PART 2: Subjective Section (Q5 - Q10)
Q52 Marks
If $b_{yx} = 0.85$ and $b_{xy} = 0.85$, find the correlation coefficient $r$ and compare $\sigma_x$ and $\sigma_y$.
If $b_{yx} = 1.5$ and $b_{xy} = 0.6$, check whether these can be true regression coefficients.
Show Step-by-Step Solution & Marking Scheme
• Check product bound: $b_{yx} \cdot b_{xy} = 1.5 \times 0.6 = 0.90$ [1 Mark].
• Since $0.90 \le 1$ and both slopes are positive, YES, these can be valid regression coefficients ($r = \sqrt{0.90} \approx 0.949$) [1 Mark].
Q73 Marks
Given two regression equations: $X + 2Y - 5 = 0$ and $2X + 3Y - 8 = 0$. Identify which equation is $Y$ on $X$ and which is $X$ on $Y$, and hence find $r$.
Case Context: Retail Chain Inventory Turnover & Profit
A retail conglomerate models inventory turnover ratio ($X$) and operating profit margin ($Y$ in %) across 40 store outlets:
$b_{yx} = 1.25, \quad b_{xy} = 0.20, \quad \bar{X} = 6 \text{ turns}, \quad \bar{Y} = 15\%$.
Q9 (2 Marks): Calculate the correlation coefficient $r$ between inventory turnover and operating profit margin. Q10 (2 Marks): If the standard deviation of inventory turnover is $\sigma_x = 2$ turns, compute the standard deviation of operating profit margin $\sigma_y$.
Show Step-by-Step Solution & Marking Scheme
Solution to Q9 [2 Marks]:
• Since both coefficients are positive:
$$r = +\sqrt{b_{yx} \cdot b_{xy}} = +\sqrt{1.25 \times 0.20} = +\sqrt{0.25} = \mathbf{+0.50}$$
Solution to Q10 [2 Marks]:
• We know $b_{yx} = r \frac{\sigma_y}{\sigma_x}$:
$$1.25 = 0.50 \times \frac{\sigma_y}{2} \implies 1.25 = 0.25 \sigma_y \implies \sigma_y = \frac{1.25}{0.25} = \mathbf{5\%}$$
• The standard deviation of profit margin is 5%.