Exercise 11.3: Karl Pearson’s Coefficient of Correlation
Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
Comprehensive Concepts & Principles: Karl Pearson's Coefficient of Correlation
Karl Pearson's Coefficient of Correlation ($r$), also known as the Pearson product-moment correlation coefficient, is the mathematical standard for quantifying the strength and direction of linear association between two quantitative continuous variables. It standardizes the covariance of $X$ and $Y$ by the product of their individual standard deviations, yielding a pure, dimensionless index bounded strictly between $-1$ and $+1$.
• Property 1 (Range): $-1 \le r \le +1$. (Proven via Cauchy-Schwarz Inequality).
• Property 2 (Dimensionless): $r$ is completely independent of the physical units of measurement.
• Property 3 (Symmetry): $r_{xy} = r_{yx}$.
• Property 4 (Invariance of Origin & Scale): If $u = \frac{X - a}{h}$ and $v = \frac{Y - b}{k}$:
- If $h$ and $k$ share the same sign ($h \cdot k > 0$), then $r_{uv} = r_{xy}$.
- If $h$ and $k$ have opposite signs ($h \cdot k < 0$), then $r_{uv} = -r_{xy}$.
• $N = 6$.
$\sum X = 2+4+5+6+8+11 = 36 \implies \bar{X} = 36/6 = \mathbf{6}$.
$\sum Y = 18+12+10+8+7+5 = 60 \implies \bar{Y} = 60/6 = \mathbf{10}$.
• Compute deviations $x = X - 6$ and $y = Y - 10$:
- Point 1: $x = -4, y = +8 \implies x^2 = 16, y^2 = 64, xy = -32$
- Point 2: $x = -2, y = +2 \implies x^2 = 4, y^2 = 4, xy = -4$
- Point 3: $x = -1, y = 0 \implies x^2 = 1, y^2 = 0, xy = 0$
- Point 4: $x = 0, y = -2 \implies x^2 = 0, y^2 = 4, xy = 0$
- Point 5: $x = +2, y = -3 \implies x^2 = 4, y^2 = 9, xy = -6$
- Point 6: $x = +5, y = -5 \implies x^2 = 25, y^2 = 25, xy = -25$
• Sums: $\sum x^2 = 16+4+1+0+4+25 = \mathbf{50}$.
$\sum y^2 = 64+4+0+4+9+25 = \mathbf{106}$.
$\sum xy = -32 - 4 + 0 + 0 - 6 - 25 = \mathbf{-67}$.
$$r = \frac{\sum xy}{\sqrt{\sum x^2 \cdot \sum y^2}} = \frac{-67}{\sqrt{50 \times 106}} = \frac{-67}{\sqrt{5300}} = \frac{-67}{72.80} = \mathbf{-0.9203}$$ Answer: Karl Pearson's correlation coefficient is -0.92, indicating a very high negative linear correlation.
Example 2 (Change of Origin and Scale Property):
If the correlation coefficient between $X$ and $Y$ is $r_{xy} = 0.65$, find the correlation coefficient between:
(a) $u = X + 10$ and $v = Y - 25$
(b) $u = \frac{X - 50}{5}$ and $v = \frac{Y - 30}{10}$
(c) $u = \frac{X - 10}{-2}$ and $v = \frac{Y - 20}{4}$
Show Step-by-Step Solution
• (a) Origin shifts (+10, -25) have zero effect on correlation: $r_{uv} = r_{xy} = \mathbf{0.65}$.
• (b) Here $h = 5 > 0$ and $k = 10 > 0$. Since $h \cdot k = 50 > 0$, both scale factors are positive: $r_{uv} = r_{xy} = \mathbf{0.65}$.
• (c) Here $h = -2 < 0$ and $k = 4 > 0$. The product $h \cdot k = -8 < 0$ (opposite signs!). Therefore:
$$r_{uv} = -r_{xy} = \mathbf{-0.65}$$
CHECK YOUR PROGRESS 11.3
Graded textbook calculation exercises on Karl Pearson's Coefficient of Correlation from CBSE Class 11 Applied Mathematics.
Problem 1Direct Evaluation Drill
Calculate Karl Pearson's $r$ from the following data:
$N = 10, \quad \sum X = 140, \quad \sum Y = 150, \quad \sum (X - 14)^2 = 180, \quad \sum (Y - 15)^2 = 215, \quad \sum (X - 14)(Y - 15) = 60$.
Given 5 pairs of observations $(X, Y)$:
$(1, 2), \quad (2, 5), \quad (3, 3), \quad (4, 8), \quad (5, 7)$
Calculate Karl Pearson's correlation coefficient using the raw data formula.
The covariance between two variables $X$ and $Y$ is $-16.5$. The variance of $X$ is 25 and the variance of $Y$ is 16. Find the correlation coefficient between $X$ and $Y$.
Interpret the practical commercial meaning of each of the following correlation coefficients:
(a) $r = +0.94$ between R&D budget and patent filings.
(b) $r = -0.89$ between car fuel efficiency and carbon emissions per km.
(c) $r = +0.03$ between employee height and sales targets achieved.
Show Step-by-Step Solution
• (a) $r = +0.94$: Extremely strong positive linear correlation. Increasing R&D expenditure consistently leads to higher patent output.
• (b) $r = -0.89$: Strong negative linear correlation. Cars with higher fuel efficiency emit substantially lower carbon per kilometre.
• (c) $r = +0.03$: Negligible / near-zero correlation. Employee height has no linear bearing whatsoever on sales achievements.
Problem 5Opposite Signs Scale Property
If $r(X, Y) = 0.72$, determine the correlation coefficient between $U = -3X + 5$ and $V = 2Y - 8$.
Q1: State the mathematical boundaries between which Karl Pearson's correlation coefficient must always lie.
Show Answer & Scheme
$$\mathbf{-1 \le r \le +1} \quad [1 \text{ Mark}]$$
Q2: Is Karl Pearson's correlation coefficient affected by change of origin? State yes or no with reason.
Show Answer & Scheme
Answer [1 Mark]:No. Adding or subtracting a constant to $X$ or $Y$ shifts the origin but leaves the deviations $(x_i - \bar{X})$ and standard deviations completely unchanged.
SECTION B: 2-Mark Short Answer Problems
Q3: For 10 pairs of observations on $X$ and $Y$, the sum of products of deviations from their respective means is 120. If $\sigma_x = 4$ and $\sigma_y = 5$, find $r$.
Q4: State two critical assumptions underlying Karl Pearson's correlation coefficient.
Show Answer & Scheme
Assumptions [2 Marks]:
(1) Linearity: The relationship between the two variables must be linear.
(2) Normality: The variables should follow an approximately bivariate normal distribution, with neither variable being heavily distorted by extreme outliers.
SECTION C: 4-Mark Comprehensive Numerical Drill
Q5: An economist investigates the relationship between the price of crude oil ($X$ in $ per barrel) and the price of aviation jet fuel ($Y$ in cents per gallon) over 6 quarters:
$X$: $60, \quad 65, \quad 70, \quad 75, \quad 80, \quad 85$
$Y$: $150, \quad 165, \quad 170, \quad 185, \quad 200, \quad 210$
(i) Using Assumed Means $A = 70$ and $B = 180$, compute $u = X - 70$ and $v = Y - 180$.
(ii) Construct the short-cut table showing $u, v, u^2, v^2, uv$.
(iii) Calculate Karl Pearson's correlation coefficient and interpret the result.
Show Step-by-Step Solution & Marking Scheme
Deviations & Table [2 Marks]:
• Point 1: $(60, 150) \implies u = -10, v = -30 \implies u^2 = 100, v^2 = 900, uv = 300$
• Point 2: $(65, 165) \implies u = -5, v = -15 \implies u^2 = 25, v^2 = 225, uv = 75$
• Point 3: $(70, 170) \implies u = 0, v = -10 \implies u^2 = 0, v^2 = 100, uv = 0$
• Point 4: $(75, 185) \implies u = +5, v = +5 \implies u^2 = 25, v^2 = 25, uv = 25$
• Point 5: $(80, 200) \implies u = +10, v = +20 \implies u^2 = 100, v^2 = 400, uv = 200$
• Point 6: $(85, 210) \implies u = +15, v = +30 \implies u^2 = 225, v^2 = 900, uv = 450$
• Sums: $\sum u = 15, \quad \sum v = 0, \quad \sum u^2 = 475, \quad \sum v^2 = 2550, \quad \sum uv = 1050$. $N = 6$.
If $u = ax + b$ and $v = cy + d$:
• If $a \cdot c > 0$ (same signs): $r_{uv} = r_{xy}$.
• If $a \cdot c < 0$ (opposite signs): $r_{uv} = -r_{xy}$.
Exam Trap: Be vigilant when one variable is multiplied by a negative number; it flips the sign of $r$!
10-Question Diagnostic Board Mock Examination: Karl Pearson's Correlation
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: Karl Pearson's coefficient of correlation $r$ always satisfies:
Solution: (B) Pearson's correlation coefficient is strictly bounded: $-1 \le r \le +1$.
Q2: If $\text{Cov}(X, Y) = 24$, $\sigma_x = 6$, and $\sigma_y = 5$, then $r(X, Y)$ is:
Q3: If $r(X, Y) = 0.5$, then the correlation coefficient between $U = 2X + 3$ and $V = -4Y + 1$ is:
Solution: (B) The scale factors are $+2$ and $-4$. Since $(+2)(-4) = -8 < 0$, the sign flips: $r(U, V) = -0.5$.
Q4 (Assertion-Reason): Assertion (A): Karl Pearson's correlation coefficient is a pure, dimensionless number without any physical units. Reason (R): The units of the numerator (covariance) are the product of units of $X$ and $Y$, which cancel exactly with the product of the units of $\sigma_x$ and $\sigma_y$ in the denominator.
Solution: (A) Both Assertion and Reason are true and Reason provides the precise dimensional analysis.
State whether the correlation coefficient changes when temperature is converted from Celsius to Fahrenheit ($F = \frac{9}{5}C + 32$). Justify mathematically.
Show Step-by-Step Solution & Marking Scheme
• Answer [1 Mark]: The correlation coefficient remains completely unchanged.
• Justification [1 Mark]: The conversion $F = \frac{9}{5}C + 32$ is a positive linear transformation with scale factor $a = \frac{9}{5} > 0$. Pearson's $r$ is invariant under change of origin and positive scale changes ($r_{F, Y} = r_{C, Y}$).
Q73 Marks
Prove using the Cauchy-Schwarz inequality that Karl Pearson's correlation coefficient cannot exceed 1 in magnitude (i.e. $|r| \le 1$).
Show Step-by-Step Solution & Marking Scheme
• Let $a_i = x_i - \bar{X}$ and $b_i = y_i - \bar{Y}$.
• By Cauchy-Schwarz Inequality for real vectors:
$$\left(\sum_{i=1}^n a_i b_i\right)^2 \le \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right) \quad [1.5 \text{ Marks}]$$
• Dividing both sides by the right-hand term:
$$\frac{\left(\sum a_i b_i\right)^2}{\left(\sum a_i^2\right)\left(\sum b_i^2\right)} \le 1 \implies r^2 \le 1 \quad [1 \text{ Mark}]$$
• Taking the square root: $|r| \le 1 \implies \mathbf{-1 \le r \le +1}$ [0.5 Mark].
Q83 Marks
If $r(X, Y) = 0.8$, $\sigma_x = 2$, $\sigma_y = 3$, and $\text{Cov}(X, Y)$ is calculated, what happens to the covariance and correlation if all $X$ values are doubled and all $Y$ values are multiplied by $-3$?
Show Step-by-Step Solution & Marking Scheme
• Original $\text{Cov}(X, Y) = r \cdot \sigma_x \cdot \sigma_y = 0.8 \times 2 \times 3 = \mathbf{4.8}$ [1 Mark].
• Let $X' = 2X$ and $Y' = -3Y$.
• $\text{Cov}(X', Y') = (2)(-3) \text{Cov}(X, Y) = -6 \times 4.8 = \mathbf{-28.8}$ [1 Mark].
• New correlation: Scale multipliers are $a = 2 > 0$ and $b = -3 < 0$. Since $ab = -6 < 0$, the sign reverses: $r' = -r = \mathbf{-0.8}$ [1 Mark].
Q9 & Q10Case Study (4 Marks Total)
Case Context: Portfolio Asset Diversification
An investment manager analyzes the monthly returns of a Tech Stock ETF ($X$) and a Treasury Bond ETF ($Y$) over 5 months:
$X$: $4\%, \quad 6\%, \quad 8\%, \quad 10\%, \quad 12\%$ (Mean = 8%)
$Y$: $9\%, \quad 8\%, \quad 6\%, \quad 5\%, \quad 2\%$ (Mean = 6%)
Q9 (2 Marks): Calculate Karl Pearson's correlation coefficient $r$ between the Tech Stock ETF and Treasury Bond ETF. Q10 (2 Marks): From a portfolio risk reduction perspective, explain why combining assets with this correlation value is highly advantageous for an investor.
Solution to Q10 [2 Marks]:
• Portfolio Diversification: A near-perfect negative correlation ($r \approx -0.98$) means when the stock market plunges, the bond asset predictably surges in value, offsetting losses.
• This powerful hedge dramatically dampens overall portfolio volatility without sacrificing long-term capital preservation.