Exam-useful questions from this chapter, grouped by marks, each with a full model answer. Try writing the answer yourself first, then check it against this one.
CBSE 2026-27 Class 11 (Part A, Unit 3) includes this whole chapter: meaning and properties, scatter diagram, Karl Pearson’s method (two variables, ungrouped data) and Spearman’s rank correlation (non-repeated and repeated ranks). In numericals write the formula, the substitution and the answer on separate lines, then a one-line interpretation.
1 1-Mark Questions
Q. Define correlation.
Answer: Correlation studies and measures the direction and intensity of relationship among variables.
Q. State the range of Karl Pearson’s coefficient of correlation.
Answer: −1 ≤ r ≤ +1.
Q. What does r = 0 mean?
Answer: There is no linear relation between the two variables; another (non-linear) relation may exist.
Q. Who developed rank correlation?
Answer: The British psychologist C.E. Spearman.
Q. What is the unit of the correlation coefficient?
Answer: It has no unit; r is a pure number.
Q. Write the formula of the correction factor for repeated ranks.
Answer: (m³ − m)/12, where m is the number of times a rank is repeated.
Q. What is a scatter diagram?
Answer: A technique of examining the relationship visually by plotting the values of two variables as points on a graph paper, without calculating any numerical value.
Q. Give two other names of Karl Pearson’s coefficient of correlation.
Answer: Product moment correlation coefficient and simple correlation coefficient.
Q. Which four questions does correlation analysis answer?
Answer: Is there any relationship between two variables; if one changes, does the other also change; do both move in the same direction; and how strong is the relationship.
2 2-Mark Questions
Q. Distinguish between positive and negative correlation with one example each.
Answer: Positive: both variables move in the same direction, e.g. when income rises, consumption rises. (1)
Negative: the variables move in opposite directions, e.g. when the price of apples falls, its demand increases. (1)
Q. What should ΣD be in a rank correlation calculation, and why?
Answer: ΣD should be 0. (1) The ranks of both variables add up to the same total (1 + 2 + … + n), so their differences add up to zero; this checks the working. (1)
Q. State two situations in which Spearman’s coefficient is used instead of Karl Pearson’s.
Answer: (1) When the variables are qualitative and cannot be measured, such as honesty or beauty. (1)
(2) When the data contain extreme values, because rank correlation is not affected by them. (1)
3 3-Mark Questions
Q. “Correlation measures covariation, not causation.” Explain with the book’s example.
Answer: Meaning: correlation only tells in which direction and how strongly two variables change together, not that one causes the other. (1)
Example: ice-cream sales and deaths by drowning are highly correlated, but people do not drown because they eat ice-cream. (1)
Reason: a third variable, temperature: in the heat more ice-cream is sold and more people go to swimming pools, so drownings rise. (1)
Q. Name the three techniques for measuring correlation and give one feature of each.
Answer: Scatter diagram: visual, no numerical value, shows any type of relationship. (1)
Karl Pearson’s coefficient: precise numerical value (r) of the linear relationship. (1)
Spearman’s rank correlation: based on ranks; useful for qualitative variables and extreme values. (1)
Q. Interpret the values of r as +1, −1 and 0.
Answer: r = +1: perfect positive correlation; all points on a rising straight line, an exact linear relation. (1)
r = −1: perfect negative correlation; all points on a falling straight line. (1)
r = 0: no linear relation; a non-linear relation may still exist. (1)
Q. During an epidemic, the number of doctors sent and the number of deaths were positively correlated. How would you interpret this?
Answer: Doctors should reduce deaths (negative correlation), so the positive correlation has other reasons. (1)
The data relate to a specific time period; many reported deaths may be terminal cases; the benefit of doctors becomes visible only after some time. (1)
The deaths may not be due to the epidemic at all, for example a tsunami. Lesson: statistical methods are no substitute for common sense. (1)
Q. Marks of 6 students in two tests: X: 45, 60, 35, 70, 55, 50; Y: 40, 55, 30, 50, 60, 45. Find the rank correlation coefficient.
Answer: Ranks (highest = 1): Rx = 5, 2, 6, 1, 3, 4; Ry = 5, 2, 6, 3, 1, 4 (1)
D: 0, 0, 0, −2, 2, 0; ΣD² = 8 (½)
Formula: rs = 1 − 6ΣD²/(n³ − n) = 1 − 48/(216 − 6) = 1 − 48/210 (½)
Answer: rs = 1 − 0.229 = 0.771, a strong positive rank correlation. (1)
Q. If Σxy = 42, Σx² = 112, Σy² = 38 (x, y being deviations from the mean), find r and interpret it.
Answer: Formula: r = Σxy/√(Σx² · Σy²) (1)
Substituting: r = 42/√(112 × 38) = 42/√4256 = 42/65.24 (1)
Answer: r = 0.644, a fairly strong positive correlation (book’s Example 1: years of schooling and yield). (1)
Q. “r is unaffected by change of origin and scale.” Explain this property and its use.
Answer: Property: if U = (X − A)/h and V = (Y − B)/k, with A, B assumed means and h, k common factors of the same sign, then r_XY = r_UV. (1)
Meaning: subtracting a number from all values (origin) or dividing them by a number (scale) does not change r. (1)
Use: with large values the step deviation method shortens the work; book’s Example 2 (price index and money supply, r ≈ 0.99). (1)
Q. Why should Karl Pearson’s r not be used for a non-linear relation?
Answer: r measures only the intensity of a linear relation. (1)
For a curved relation (Figs 6.6, 6.7) r can come out low or even zero although the relation is strong, so the result is misleading. (1)
That is why the scatter diagram should be examined before r; if the direction is clear but the relation is non-linear, Spearman’s coefficient can be used. (1)
4 4-Mark Questions
Q. Find Karl Pearson’s coefficient of correlation between X: 1, 2, 3, 4, 5 and Y: 2, 4, 5, 4, 5.
Answer: Means: X̄ = 15/5 = 3, Ȳ = 20/5 = 4 (½)
Deviations: x = −2, −1, 0, 1, 2; y = −2, 0, 1, 0, 1 (½)
Sums: Σxy = 4 + 0 + 0 + 0 + 2 = 6; Σx² = 10; Σy² = 4 + 0 + 1 + 0 + 1 = 6 (1)
Formula and substitution: r = Σxy/√(Σx² Σy²) = 6/√60 = 6/7.746 (1)
Answer: r = 0.77, a positive and fairly strong correlation. (1)
Q. Price of a good (X, ₹): 10, 12, 14, 16, 18 and demand (Y, units): 50, 44, 40, 36, 30. Find r and interpret it.
Answer: Means: X̄ = 70/5 = 14, Ȳ = 200/5 = 40 (½)
Deviations: x = −4, −2, 0, 2, 4; y = 10, 4, 0, −4, −10 (½)
Sums: Σxy = −40 − 8 + 0 − 8 − 40 = −96; Σx² = 40; Σy² = 232 (1)
Substituting: r = −96/√(40 × 232) = −96/√9280 = −96/96.33 (1)
Answer: r = −0.997, an almost perfect negative correlation: demand falls as price rises (law of demand). (1)
Q. Find the rank correlation coefficient between X: 80, 60, 60, 50, 40, 30 and Y: 70, 75, 55, 55, 55, 40.
Answer: Ranks: Rx = 1, 2.5, 2.5, 4, 5, 6; Ry = 2, 1, 4, 4, 4, 6 (1)
D: −1, 1.5, −1.5, 0, 1, 0; ΣD² = 1 + 2.25 + 2.25 + 0 + 1 + 0 = 6.5 (1)
Correction: X has m = 2 → 0.5; Y has m = 3 → 2; total 2.5 (1)
Answer: rs = 1 − 6(6.5 + 2.5)/(6 × 35) = 1 − 54/210 = 0.743, positive rank correlation. (1)
Q. What is a scatter diagram? Describe the scatter diagrams of positive, negative, zero and perfect positive correlation.
Answer: Meaning: the values of two variables plotted as points to examine the relation visually; the closeness and direction of the points show the relation. (1)
Positive: points around a rising line; negative: points around a falling line. (1)
Zero: points scattered, with no rising or falling line. (1)
Perfect positive: all points on a rising straight line, r = +1. (Draw sketches with axes X and Y.) (1)
Q. In which situations is Spearman’s rank correlation used?
Answer: (1) No measuring tools: in a remote village with neither measuring rods nor weighing machines, students can still be ranked by height and weight. (1)
(2) Qualitative variables: fairness, honesty, beauty, which can only be measured relatively. (1)
(3) Clear direction but a non-linear relation (Figs 6.6, 6.7). (1)
(4) Extreme values in the data, because it is not affected by them. (1)
Q. Give four differences between Karl Pearson’s and Spearman’s coefficients.
Answer: (1) Pearson is based on the actual values, Spearman on ranks. (1)
(2) Pearson needs precisely measured data; Spearman suits qualitative or imprecisely measured variables. (1)
(3) Pearson is affected by extreme values; Spearman is not. (1)
(4) Pearson is more accurate (all information used); Spearman is generally less, rk ≤ r. Both range from −1 to +1. (1)
Q. “Zero correlation does not mean independence.” Prove it using X: −3, −2, −1, 1, 2, 3 and Y: 9, 4, 1, 1, 4, 9.
Answer: Sums: ΣX = 0, ΣY = 28, ΣXY = −27 − 8 − 1 + 1 + 8 + 27 = 0 (1)
r: numerator = ΣXY − ΣXΣY/N = 0 − 0 = 0, so r = 0 (1)
But the relation: in every pair Y = X², so Y depends completely on X (1)
Conclusion: the relation is exact but not linear (a curve); r = 0 shows only the absence of a linear relation, not independence. (1)
Q. In the book’s Example 2, ΣU = 41, ΣV = 35, ΣU² = 423, ΣV² = 343, ΣUV = 378 and N = 5. Find r.
Answer: Numerator: ΣUV − ΣUΣV/N = 378 − (41 × 35)/5 = 378 − 287 = 91 (1)
Denominator parts: ΣU² − (ΣU)²/N = 423 − 336.2 = 86.8; ΣV² − (ΣV)²/N = 343 − 245 = 98 (1)
Substituting: r = 91/√(86.8 × 98) = 91/92.23 (1)
Answer: r = 0.987 ≈ 0.99: a strong positive correlation between price index and money supply. (1)
5 6-Mark Questions
Q. Find r between X: 100, 200, 300, 400, 500 and Y: 30, 40, 50, 60, 90 by the step deviation method (A = 300, h = 100, B = 50, k = 10).
Answer: Transformation: U = (X − 300)/100 = −2, −1, 0, 1, 2; V = (Y − 50)/10 = −2, −1, 0, 1, 4 (1)
Sums: ΣU = 0, ΣV = 2, ΣU² = 10, ΣV² = 4 + 1 + 0 + 1 + 16 = 22, ΣUV = 4 + 1 + 0 + 1 + 8 = 14 (1½)
Formula: r = [ΣUV − ΣUΣV/N]/[√(ΣU² − (ΣU)²/N) · √(ΣV² − (ΣV)²/N)] (1)
Substituting: r = (14 − 0)/[√10 · √(22 − 0.8)] = 14/√(10 × 21.2) = 14/14.56 (1½)
Answer: r = 0.96, a strong positive correlation; the raw X, Y give the same value because r is unaffected by origin and scale. (1)
Q. Three judges P, Q, R ranked 6 competitors: P: 1, 2, 3, 4, 5, 6; Q: 2, 1, 4, 3, 6, 5; R: 6, 4, 5, 3, 1, 2. Which two judges have the most similar tastes?
Answer: Formula: rs = 1 − 6ΣD²/(n³ − n); n = 6, n³ − n = 210 (1)
P and Q: D = −1, 1, −1, 1, −1, 1; ΣD² = 6; rs = 1 − 36/210 = 0.829 (1½)
P and R: D = −5, −2, −2, 1, 4, 4; ΣD² = 66; rs = 1 − 396/210 = −0.886 (1½)
Q and R: D = −4, −3, −1, 0, 5, 3; ΣD² = 60; rs = 1 − 360/210 = −0.714 (1)
Conclusion: P and Q have the highest rs (0.829), so their tastes are the most similar; P and R are almost opposite. (1)
Q. Write six properties of Karl Pearson’s coefficient of correlation with examples.
Answer: (1) No unit: a pure number; r between height (feet) and weight (kg), say 0.7. (1)
(2) Negative r = inverse relation: when the rate of interest rises, the demand for funds falls. (1)
(3) Positive r: when the price of coffee rises, the demand for tea rises; better irrigation, higher yield. (1)
(4) Range −1 ≤ r ≤ 1: outside means an error in calculation. (1)
(5) Unaffected by origin and scale: r_XY = r_UV; basis of the step deviation method. (1)
(6) r = 0: no linear relation; r = ±1: exact linear relation; close to ±1 strong, close to 0 weak. (1)
Q. How is Spearman’s coefficient found when ranks are repeated? Explain with the book’s Example 5 (ΣD² = 198, n = 12; in Y, 75 twice, 90 twice and 50 three times).
Answer: Average rank: equal values get the mean of the ranks they would have taken if slightly different; e.g. 50 (9th, 10th, 11th) gets 10, and the next value gets 12. (1)
Formula: rs = 1 − 6[ΣD² + Σ(m³ − m)/12]/[n(n² − 1)] (1)
Correction: 90 (m = 2): 0.5; 75 (m = 2): 0.5; 50 (m = 3): 2; total 3 (1½)
Substituting: rs = 1 − 6(198 + 3)/(12 × 143) = 1 − 1206/1716 = 1 − 0.703 (1½)
Answer: rs = 0.297 ≈ 0.30: positive, but not strong. (1)
Q. Find r between X: 2, 4, 5, 6, 8 and Y: 3, 5, 4, 7, 9 by the direct method (the N ΣXY formula).
Answer: Sums: ΣX = 25, ΣY = 28, ΣX² = 145, ΣY² = 180, ΣXY = 6 + 20 + 20 + 42 + 72 = 160 (2)
Formula: r = [NΣXY − ΣXΣY]/[√(NΣX² − (ΣX)²) · √(NΣY² − (ΣY)²)] (1)
Numerator: 5 × 160 − 25 × 28 = 800 − 700 = 100 (1)
Denominator: √(725 − 625) · √(900 − 784) = √100 · √116 = 10 × 10.77 = 107.7 (1)
Answer: r = 100/107.7 = 0.93, a strong positive correlation. (1)
6 10-Mark Questions
Q. (UP Board 10 marks) Explain the meaning of correlation and the properties of Karl Pearson’s coefficient. Find r between X: 10, 20, 30, 40, 50 and Y: 8, 12, 15, 21, 24 and interpret the result.
Answer: 1. Meaning: correlation studies and measures the direction and intensity of relationship between variables; it measures covariation, not causation (ice-cream and drowning, temperature behind both). (2)
2. Properties: no unit; −1 ≤ r ≤ 1; unaffected by origin and scale; sign of covariance = sign of r; r = 0 no linear relation; r = ±1 perfect. (3)
3. Calculation: X̄ = 150/5 = 30, Ȳ = 80/5 = 16; x = −20, −10, 0, 10, 20; y = −8, −4, −1, 5, 8 (1)
Σxy = 160 + 40 + 0 + 50 + 160 = 410; Σx² = 1000; Σy² = 64 + 16 + 1 + 25 + 64 = 170 (2)
r = 410/√(1000 × 170) = 410/√170000 = 410/412.3 = 0.994 (1)
4. Interpretation: almost perfect positive correlation: as X rises, Y rises almost along a straight line. In a scatter diagram the points would lie very close to a rising line. (1)
7 Source-Based Questions
(A) 1 mark What type of correlation is there between the two variables in the passage?
Answer: Positive correlation: both rise together.
(B) 2 marks What ‘one and the same thing’ is the officer pointing to? Explain.
Answer: A third variable, temperature. (1) In hot weather more ice-cream is sold and more people go to swimming pools to beat the heat, so drowning incidents also rise. (1)
(C) 3 marks Why is the reader’s suggestion wrong statistically? Distinguish correlation from causation.
Answer: Correlation measures only covariation, moving together; it does not say that one variable causes the other. (1)
People do not drown because they eat ice-cream; banning ice-cream will not reduce drowning, because the cause is temperature. (1)
Correlation can exist without causation; it may come from coincidence or a third variable. Causation is decided by theory and reasoning. (1)
8 Data / Graph-Based Questions
(A) 2 marks Write the ranks of the students in both subjects (highest marks = 1) and D².
Answer: Rx = 4, 2, 5, 1, 3; Ry = 5, 2, 3, 1, 4 (1)
D = −1, 0, 2, 0, −1; D² = 1, 0, 4, 0, 1; ΣD² = 6 (1)
(B) 2 marks Find the rank correlation coefficient and interpret the result.
Answer: rs = 1 − 6 × 6/(125 − 5) = 1 − 36/120 = 0.7 (1)
The ranks in the two subjects are fairly strongly positively related: students ranked high in Mathematics are mostly ranked high in Economics too. (1)