Correlation Class 11: Scatter Diagram, Karl Pearson and Spearman Rank Correlation

What Correlation is about

Weightage: This chapter is part of Unit 3, Statistical Tools and Interpretation, which carries 25 of the 80 theory marks in the school annual exam.Source: CBSE Economics (030) syllabus 2026-27, Class XI course structure, Unit 3 Statistical Tools and Interpretation

Key concepts

Meaning of correlation; correlation is not causation

NCERT section 1

Correlation studies and measures the direction and intensity of the relationship between two variables. When one variable changes, does the other also change, in which direction, and how strongly?

Correlation measures covariation, not causation. A high correlation may arise because:

• one variable really affects the other (rainfall and crop yield);
• both are affected by a third variable (ice cream sales and drowning deaths both rise with temperature);
• pure coincidence (arrival of migratory birds and the birth rate in a town; shoe size and pocket money).

So a correlation coefficient should never by itself be read as proof that X causes Y. Economic theory or common sense must explain the link.

Worked example

Suppose data show that towns with more fire engines have more fire damage. Should a town reduce its fire engines to reduce damage?

Try it first, then show the solutionSolution

1. Bigger towns have more buildings, so more fires and more damage, and also more fire engines
2. A third variable (size of town) drives both

Answer: No. The correlation reflects town size, not a cause-and-effect link.

Types of correlation

NCERT section 2

Positive correlation: both variables move in the same direction: income and consumption; temperature and ice cream sales; price of coffee and demand for tea (a substitute).

Negative correlation: they move in opposite directions: price of apples and quantity demanded; interest rate and demand for loans; hours of study and chances of failing.

Linear and non-linear: correlation is linear if the points lie around a straight line (a constant ratio of change), and non-linear (curvilinear) if they follow a curve. Karl Pearson's r measures only linear relationships.

Degree: perfect (r = ±1), high, moderate, low, or zero (no linear relationship).

Worked example

State the type of correlation: (a) price of petrol and demand for cars; (b) fertiliser used and crop output; (c) age of a car and its resale value.

Try it first, then show the solutionSolution

1. (a) Petrol price up, car demand down: negative
2. (b) More fertiliser (within limits), more output: positive
3. (c) Older car, lower value: negative

Answer: (a) Negative; (b) positive; (c) negative.

Scatter diagram

NCERT section 3

A scatter diagram plots each pair of values (X, Y) as a point on a graph. It shows the relationship visually, without calculating a number.

How to read it:

• Points rising from bottom-left to top-right: positive correlation.
• Points falling from top-left to bottom-right: negative correlation.
• Points lying exactly on a rising (falling) straight line: perfect positive (negative) correlation.
• Points close together around a line: high degree; widely spread: low degree.
• Points scattered with no pattern: no correlation.
• Points following a curve: non-linear relationship.

Merits: simple; shows the nature of the relationship and outliers. Limit: gives no exact measure of the degree.

Worked example

Suppose income and consumption (₹ thousand) of 7 families are (20, 18), (25, 21), (30, 25), (35, 28), (40, 31), (45, 34), (50, 39). What will the scatter diagram show?

Try it first, then show the solutionSolution

1. As income rises, consumption rises every time
2. Points rise from left to right and lie close to a straight line

Answer: A high positive, nearly linear correlation.

Karl Pearson's coefficient of correlation (actual mean method)

NCERT section 3

Karl Pearson's coefficient (product moment correlation) is

$r=\frac{\sum xy}{\sqrt{\sum x^2}\sqrt{\sum y^2}}$, where x = X − X̄ and y = Y − Ȳ.

It is used for two variables, ungrouped data, with a linear relationship. Steps: find the means; find deviations x and y; find x², y² and xy; apply the formula.

Income (X) and consumption (Y) of 7 families (₹ thousand), X̄ = 35, Ȳ = 28:

X	Y	x = X − X̄	y = Y − Ȳ	x²	y²	xy
20	18	-15	-10	225	100	150
25	21	-10	-7	100	49	70
30	25	-5	-3	25	9	15
35	28	0	0	0	0	0
40	31	5	3	25	9	15
45	34	10	6	100	36	60
50	39	15	11	225	121	165
Σ 245	196	0	0	700	324	475

r = 475 ÷ √(700 × 324) = 0.997.

Worked example

Interpret r = 0.997 between income and consumption of families.

Try it first, then show the solutionSolution

1. Positive: consumption rises with income
2. Very close to +1: a very strong linear relationship

Answer: Almost perfect positive correlation: families with higher income spend more, by a nearly constant amount per extra rupee, as the theory of consumption suggests.

Karl Pearson's r: direct formula and step deviation method

NCERT section 3

Direct (raw values) formula:

$r=\frac{N\sum XY-\sum X\sum Y}{\sqrt{N\sum X^2-(\sum X{)}^2}\sqrt{N\sum Y^2-(\sum Y{)}^2}}$

Step deviation method (change of origin and scale): take u = (X − A) ÷ B and v = (Y − C) ÷ D, then

$r=\frac{\sum uv-\frac{\sum u\sum v}{N}}{\sqrt{\sum u^2-\frac{(\sum u{)}^2}{N}}\sqrt{\sum v^2-\frac{(\sum v{)}^2}{N}}}$

This works because r is unaffected by change of origin and scale (B and D of the same sign). It saves effort with large numbers.

Price (P) and quantity demanded (Q) by the direct formula:

P	Q	P²	Q²	PQ
10	60	100	3600	600
12	55	144	3025	660
14	48	196	2304	672
16	42	256	1764	672
18	38	324	1444	684
20	30	400	900	600
90	273	1420	13037	3888

r = (6 × 3888 − 90 × 273) ÷ √[(6 × 1420 − 90²)(6 × 13037 − 273²)] = -0.997 (also -0.997 by the actual mean method).

Worked example

Interpret r = -0.997 between price and quantity demanded.

Try it first, then show the solutionSolution

1. Negative: quantity falls as price rises
2. Close to −1: very strong

Answer: A very strong inverse relationship, as the law of demand predicts.

Properties of the correlation coefficient

NCERT section 3

1. r has no unit; it is a pure number (r between height in cm and weight in kg is just a number like 0.7).
2. −1 ≤ r ≤ +1. A value outside this range means a calculation error.
3. Sign shows direction: positive r, same direction; negative r, opposite directions.
4. r = +1: perfect positive; r = −1: perfect negative; r = 0: no linear relationship (there may still be a non-linear one).
5. Unaffected by change of origin and scale: adding, subtracting, multiplying or dividing all X values (or all Y values) by a constant (positive) does not change r.
6. r is symmetric: r between X and Y = r between Y and X.
7. r measures only linear association and is affected by extreme values.

Rough guide to strength: |r| ≥ 0.75 high; 0.5-0.75 moderate; below 0.5 low (use judgement, not a rule).

Worked example

Rainfall X (mm) and yield Y (quintals/hectare): (100, 20), (120, 22), (140, 26), (160, 28), (180, 30), (200, 36). Use u = (X − 140) ÷ 20 and v = (Y − 28) ÷ 2 to find r.

Try it first, then show the solutionSolution

1. Rainfall X	Yield Y	u = (X − 140) ÷ 20	v = (Y − 28) ÷ 2	u²	v²	uv
   100	20	-2	-4	4	16	8
   120	22	-1	-3	1	9	3
   140	26	0	-1	0	1	0
   160	28	1	0	1	0	0
   180	30	2	1	4	1	2
   200	36	3	4	9	16	12
   		3	-3	19	43	25

2. r = [25 − 3 × (−3) ÷ 6] ÷ √[(19 − 3² ÷ 6)(43 − (−3)² ÷ 6)] = 26.5 ÷ √(17.5 × 41.5) = 0.983
3. Check with actual values: r = 0.983

Answer: r = 0.983: rainfall and yield are very strongly positively related in these fields.

Spearman's rank correlation

NCERT section 3

Spearman's rank correlation coefficient uses ranks instead of values:

$r_s=1-\frac{6\sum D^2}{n^3-n}$

where D = difference between the two ranks of each item and n = number of pairs.

When to use it: when variables can only be ranked (honesty, beauty, taste); when measurement is not possible but ranking is; when data have extreme values (ranks are not affected by them); when the relationship is non-linear but clearly in one direction. It lies between −1 and +1 and is read like r, but it is generally less precise because the actual values are not used.

Repeated (tied) ranks: give tied items the average of the ranks they would occupy (two items tied for 3rd and 4th get 3.5 each). Then add a correction for each tie group of m items:

$r_s=1-\frac{6\left[\sum D^2+\sum \frac{m^3-m}{12}\right]}{n^3-n}$

Two judges ranking 8 contestants:

Contestant	Rank by Judge 1	Rank by Judge 2	D	D²
A	1	2	-1	1
B	2	1	1	1
C	3	4	-1	1
D	4	3	1	1
E	5	6	-1	1
F	6	8	-2	4
G	7	5	2	4
H	8	7	1	1
Total			0	14

rₛ = 1 − 6 × 14 ÷ (512 − 8) = 0.833.

Worked example

Interpret rₛ = 0.833 between the rankings of two judges.

Try it first, then show the solutionSolution

1. Positive and high
2. The judges largely agree on who is better

Answer: High agreement between the judges' tastes.

Correlation in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Karl Pearson's Coefficient of Correlation
• Spearman Rank Correlation and Scatter Diagram

Correlation: key facts

Karl Pearson's r (actual means)

r = \frac{\sum xy}{\sqrt{\sum x^2 \cdot \sum y^2}},\ x = X - \bar{X},\ y = Y - \bar{Y}

Two variables, ungrouped data.

Karl Pearson's r (direct)

r = \frac{N\sum XY - \sum X\sum Y}{\sqrt{N\sum X^2 - (\sum X)^2}\sqrt{N\sum Y^2 - (\sum Y)^2}}

Uses raw values.

Karl Pearson's r (step deviation)

r = \frac{\sum uv - \frac{\sum u\sum v}{N}}{\sqrt{\sum u^2 - \frac{(\sum u)^2}{N}}\sqrt{\sum v^2 - \frac{(\sum v)^2}{N}}}

u = (X − A) ÷ B, v = (Y − C) ÷ D.

Spearman's rank correlation

r_s = 1 - \frac{6\sum D^2}{n^3 - n}

D = difference in ranks.

Spearman with repeated ranks

r_s = 1 - \frac{6\left[\sum D^2 + \sum \frac{m^3-m}{12}\right]}{n^3 - n}

m = number of items in each tie.

Range of r

-1 \le r \le +1

Unit-free; unaffected by change of origin and scale.

How the school exam asks Correlation

The question patterns to expect, and the method that works for each.

Tricks to remember Correlation

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Spearman formula

  "One minus six D-squared over n-cubed minus n."

  Use it when: Rank correlation.

• Tie correction

  "m-cube minus m by twelve" for each tie.

  Use it when: Repeated ranks.

• Properties of r

  "No unit, minus one to one, origin and scale don't matter."

  Use it when: Property questions.

Correlation: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Correlation

The slips examiners see most often, and what to write instead.

• ✗Mistake: Ranking the two series in different directions

  One series ranked lowest = 1, the other highest = 1.

  ✓Do this: Use the same rule (highest = 1) for both series, or the sign of rₛ flips.

  Usually costs: The whole answer; 2-3 marks.

• ✗Mistake: Forgetting the correction for tied ranks

  Students average the ranks but stop there.

  Wrong working

  rₛ = 1 − 6 × 7 ÷ 504

  Right working

  rₛ = 1 − 6 × (7 + 1) ÷ 504

  ✓Do this: Add (m³ − m)/12 for every tie group.

  Usually costs: 1 mark.

• ✗Mistake: Getting r outside −1 to +1 and not noticing

  Arithmetic slip under exam pressure.

  ✓Do this: Check that |r| ≤ 1; with actual means also check Σx = 0 and Σy = 0.

  Usually costs: Most of the marks.

• ✗Mistake: Interpreting r as cause and effect

  High r feels like proof.

  ✓Do this: Say "is associated with", not "causes"; mention possible third factors when relevant.

  Usually costs: 1 mark in interpretation questions.

• ✗Mistake: Using r = 0 to say the variables are unrelated

  Forgetting non-linear relations.

  ✓Do this: r = 0 means no linear correlation; a curved relation may still exist.

  Usually costs: 1 mark in assertion-reason.

NCERT exercise guide: Correlation

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Correlation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   If r = −1, the correlation is:
   (a) perfect positive
   (b) perfect negative
   (c) zero
   (d) low negative

   Hint

   Sign and size.

   Answer

   (b) perfect negative

2. Q2 · 1 mark · MCQExam level

   Spearman's rank correlation is most suitable when:
   (a) data are in class intervals
   (b) variables can only be ranked
   (c) r is zero
   (d) N is very large

   Hint

   Qualities.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   Name the diagram that shows the correlation between two variables visually.

   Hint

   Points.

   Answer

   Scatter diagram.

4. Q4 · 3 marks · Short answerExam level

   State any three properties of Karl Pearson's coefficient of correlation.

   Hint

   Unit, range, origin.

   Answer

   (1) It has no unit. (2) It lies between −1 and +1. (3) It is unaffected by change of origin and scale (also: sign shows direction; r = 0 means no linear relation).

5. Q5 · 4 marks · Short answerHarder exam level

   Find r: X = 2, 4, 6, 8, 10; Y = 3, 7, 8, 12, 15.

   Hint

   Means 6 and 9.

   Answer

   x: −4, −2, 0, 2, 4; y: −6, −2, −1, 3, 6; Σxy = 24 + 4 + 0 + 6 + 24 = 58; Σx² = 40; Σy² = 86; r = 58 ÷ √(40 × 86) = 58 ÷ 58.65 ≈ 0.989.

6. Q6 · 3 marks · Short answerHarder exam level

   Find the rank correlation: ranks in X: 1, 2, 3, 4, 5, 6; ranks in Y: 3, 1, 2, 6, 4, 5.

   Hint

   n = 6.

   Answer

   D: −2, 1, 1, −2, 1, 1; ΣD² = 12; rₛ = 1 − 6 × 12 ÷ 210 = 1 − 0.343 = 0.657.

7. Q7 · 4 marks · Short answerHarder exam level

   Find rₛ when values are tied: X: 50, 60, 60, 70, 80; Y: 30, 45, 40, 50, 55.

   Hint

   Rank highest = 1; 60 gets 3.5.

   Answer

   Ranks X: 5, 3.5, 3.5, 2, 1; ranks Y: 5, 3, 4, 2, 1; D: 0, 0.5, −0.5, 0, 0; ΣD² = 0.5; correction (m = 2) = 0.5; rₛ = 1 − 6(0.5 + 0.5) ÷ 120 = 0.95.

8. Q8 · 3 marks · Short answerHarder exam level

   Explain with examples why correlation does not imply causation.

   Hint

   Third variable, coincidence.

   Answer

   Two variables may move together because a third variable affects both (ice cream sales and drowning both rise with temperature) or by pure coincidence (migratory birds and birth rate). Correlation only measures how they move together; theory must explain why.

9. Q9 · 3 marks · Short answerHarder exam level

   Distinguish between Karl Pearson's coefficient and Spearman's rank correlation on three bases.

   Hint

   Data used, extremes, precision.

   Answer

   Data: actual values vs ranks. Extremes: affected vs not affected. Precision and use: more precise, for measurable variables with a linear relation vs less precise, for qualities or rankable data (and non-linear monotonic relations).

10. Q10 · 4 marks · Short answerHarder exam level

    Show that r is unaffected by change of scale: for X = 1, 2, 3 and Y = 2, 4, 5, find r; then multiply all X by 10 and find r again.

    Hint

    Deviations scale too.

    Answer

    X̄ = 2, Ȳ = 11/3; x = −1, 0, 1; y = −5/3, 1/3, 4/3; Σxy = 3; Σx² = 2; Σy² = 14/3; r = 3 ÷ √(28/3) ≈ 0.982. With X × 10: x = −10, 0, 10; Σxy = 30; Σx² = 200; r = 30 ÷ √(200 × 14/3) ≈ 0.982. Same.

11. Q11 · 3 marks · Short answerHarder exam level

    Suppose the rank correlation between the price of a share and the number of shares traded over 10 days is −0.6. Interpret it and say whether Karl Pearson's r would necessarily be the same.

    Hint

    Ranks vs values.

    Answer

    A moderate negative association: on days of higher price, fewer shares tend to be traded. Pearson's r uses actual values and may differ, especially if the relation is non-linear or there are extreme values.

12. Q12 · 4 marks · Long answerHOTS / competency

    Suppose a study finds r = 0.85 between the number of mobile towers in districts and literacy rates. A politician says "Build towers to raise literacy." Evaluate the claim.

    Hint

    Third variable.

    Answer

    The correlation is high, but both may depend on a third variable such as income, urbanisation or infrastructure: richer, urban districts have more towers and more schools. Correlation does not show that towers cause literacy. The policy should target the real causes (schools, teachers, incomes), and a causal study is needed.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What does correlation measure?Show answer

   The direction and strength of the linear relationship between two variables.

2. 2.What does r = 0 mean?Show answer

   No linear correlation.

3. 3.Range of r?Show answer

   −1 to +1.

4. 4.Is r affected by change of origin?Show answer

   No.

5. 5.Formula for Spearman's rₛ?Show answer

   1 − 6ΣD² ÷ (n³ − n).

6. 6.What rank do two items tied for 3rd and 4th get?Show answer

   3.5 each.

7. 7.Which correlation measure is not affected by extreme values?Show answer

   Spearman's rank correlation.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why correlation is not causation, with an example?
• ☐Can you identify positive, negative and no correlation from a scatter diagram?
• ☐Can you compute Karl Pearson's r by two methods and get the same answer?
• ☐Can you state five properties of r?
• ☐Can you rank values and handle ties with average ranks?
• ☐Can you apply Spearman's formula with the tie correction?
• ☐Can you correct rₛ when a wrong D was used?
• ☐Can you interpret r = 0.9, −0.4 and 0 in words?

Correlation: frequently asked questions