Fundamentals of Business Mathematics and Statistics · Correlation and Regression
Karl Pearson's Coefficient of Correlation: Formula and Method
Updated 10 October 2026 · Fact-checked
Karl Pearson's coefficient of correlation (r) measures the strength and direction of a linear relationship between two variables. It lies between −1 and +1. To solve a problem, find n, ΣX, ΣY, ΣXY, ΣX² and ΣY², then put them in r = [nΣXY − ΣXΣY] ÷ √{[nΣX² − (ΣX)²][nΣY² − (ΣY)²]}.
Understand Karl Pearson's Coefficient of Correlation
Correlation tells you whether two variables move together. When advertising spend rises and sales also rise, the variables are positively correlated. When the price of a product rises and its demand falls, they are negatively correlated.
Karl Pearson's coefficient of correlation, written r, puts a number on this. It measures only linear relationship. It is the covariance of X and Y divided by the product of their standard deviations. Dividing by the standard deviations removes the units, so r has no unit and can be compared across problems.
The value of r always lies from −1 to +1. A value of +1 means perfect positive linear correlation and −1 means perfect negative linear correlation. A value of 0 means no linear correlation. Values near ±1 show a strong relationship and values near 0 show a weak one.
Two properties matter a lot in MCQs. First, r is independent of change of origin and scale: if you subtract a constant from every value, or divide every value by a positive constant, r does not change. This is why the assumed mean and step-deviation methods work. Second, r is symmetric: r between X and Y equals r between Y and X.
Remember that correlation is not causation. A high r only shows that the variables move together. It does not prove that one causes the other. Also, r = 0 means no linear relationship. A curved relationship can still exist.
Key formulas to remember
- Definition of r
- r = Cov(X, Y) ÷ (σx × σy)
- Cov(X, Y) = Σ(X − X̄)(Y − Ȳ) ÷ n = ΣXY ÷ n − X̄Ȳ.
- Deviation-from-mean form
- r = Σ(X − X̄)(Y − Ȳ) ÷ √[Σ(X − X̄)² × Σ(Y − Ȳ)²]
- Use when the means are whole numbers. The factor n cancels out.
- Direct method
- r = [nΣXY − ΣXΣY] ÷ √{[nΣX² − (ΣX)²][nΣY² − (ΣY)²]}
- Best when the values are small. n is the number of pairs.
- Assumed mean method
- r = [nΣdxdy − ΣdxΣdy] ÷ √{[nΣdx² − (Σdx)²][nΣdy² − (Σdy)²]}
- dx = X − A and dy = Y − B, where A and B are assumed means. The same formula as direct, with d replacing X and Y.
- Step-deviation method
- u = (X − A) ÷ h, v = (Y − B) ÷ k; r = [nΣuv − ΣuΣv] ÷ √{[nΣu² − (Σu)²][nΣv² − (Σv)²]}
- If h and k are both positive, r is the same as for X and Y. If exactly one of h or k is negative, r changes sign.
- Limits of r
- −1 ≤ r ≤ +1
- r = ±1 means perfect linear correlation. r = 0 means no linear correlation.
- Change of origin and scale
- If u = (X − a) ÷ h and v = (Y − b) ÷ k with h, k > 0, then r(u, v) = r(X, Y)
- r does not depend on origin or on positive scale.
How to solve Karl Pearson's Coefficient of Correlation questions
Use this method for any Karl Pearson problem. Make a table and keep every column in it.
- 1Count the pairs of observations and write n.
- 2If the values are large, pick an assumed mean for each variable (near the middle) and, if all values are multiples of a common number, a step size. Convert to dx, dy or u, v.
- 3Make columns for X, Y, XY, X² and Y² (or the same for the deviations). Fill every row.
- 4Add each column and write the totals: ΣX, ΣY, ΣXY, ΣX², ΣY².
- 5Calculate the numerator nΣXY − ΣXΣY, and the two terms under the root, nΣX² − (ΣX)² and nΣY² − (ΣY)².
- 6Divide the numerator by the square root of the product of the two terms. Check that the sign matches the numerator.
- 7Check that r lies between −1 and +1, then state the meaning: strong or weak, positive or negative.
Quickest way: Shift and shrink, then use the sums formula
When to use it: Use it when the data have 5 to 10 pairs with large values, or when the question already gives the sums.
- Check first whether the question gives ΣX, ΣY, ΣXY, ΣX², ΣY². If so, go straight to the direct formula.
- Subtract a middle value from each variable. Divide by a common factor if possible. This keeps all numbers small.
- If one deviation column adds to 0, then the term ΣuΣv is 0. This saves a lot of work.
- Compute the numerator first. Its sign decides the sign of r, so you can eliminate options of the wrong sign immediately.
- Estimate the root. If the answer is a clean value like 0.8 or 0.9, the numbers were designed for it. Then check the answer against the options and mark it.
Common mistakes in Karl Pearson's Coefficient of Correlation
Writing (ΣX)² as ΣX² (or the reverse).
The two look alike, but one is the square of the total and the other is the total of the squares.
Fix: Keep separate columns. ΣX² comes from squaring each X and then adding. (ΣX)² comes from adding first and then squaring.
Forgetting to multiply ΣXY by n in the numerator.
Students remember the covariance form ΣXY ÷ n and mix it with the sums form.
Fix: In the sums formula, every first term has n: nΣXY, nΣX² and nΣY². Write the formula before you start.
Getting r outside the range −1 to +1 and still marking it.
A calculation slip is not checked at the end.
Fix: If |r| is more than 1, there is an error. Recheck the sums and the squares.
Thinking a change of origin or scale changes r.
Students confuse r with covariance or standard deviation, which do change with scale.
Fix: Remember that r has no unit. Positive shifts and divisions leave r unchanged. Only a negative divisor on one variable flips the sign.
Concluding that r = 0 means the variables are unrelated.
The word 'no correlation' sounds like 'no relationship'.
Fix: r = 0 means no linear relationship. A non-linear relationship can still exist.
Using the assumed mean method with mistakes in the signs of deviations.
Values below the assumed mean give negative deviations that students write as positive.
Fix: Always compute X − A with its sign. Check that Σdx matches the gap between the actual mean and A.
Worked examples
Example 1
The advertising spend X (₹ lakh) and sales Y (₹ crore) of a company in five months are: X = 1, 2, 3, 4, 5 and Y = 2, 1, 4, 3, 5. Find Karl Pearson's coefficient of correlation.
Show the solution
- n = 5.
- ΣX = 1 + 2 + 3 + 4 + 5 = 15. ΣY = 2 + 1 + 4 + 3 + 5 = 15.
- ΣX² = 1 + 4 + 9 + 16 + 25 = 55. ΣY² = 4 + 1 + 16 + 9 + 25 = 55.
- ΣXY = 1×2 + 2×1 + 3×4 + 4×3 + 5×5 = 2 + 2 + 12 + 12 + 25 = 53.
- Numerator = nΣXY − ΣXΣY = 5×53 − 15×15 = 265 − 225 = 40.
- nΣX² − (ΣX)² = 275 − 225 = 50. nΣY² − (ΣY)² = 275 − 225 = 50.
- r = 40 ÷ √(50 × 50) = 40 ÷ 50 = 0.8.
Answer: r = +0.8, a strong positive correlation.
Example 2
Find r for X = 10, 20, 30, 40, 50 and Y = 20, 25, 35, 30, 40, using the step-deviation method.
Show the solution
- Take A = 30, h = 10 for X, so u = (X − 30) ÷ 10 = −2, −1, 0, 1, 2.
- Take B = 30, k = 5 for Y, so v = (Y − 30) ÷ 5 = −2, −1, 1, 0, 2.
- n = 5. Σu = 0 and Σv = −2 − 1 + 1 + 0 + 2 = 0.
- Σu² = 4 + 1 + 0 + 1 + 4 = 10. Σv² = 4 + 1 + 1 + 0 + 4 = 10.
- Σuv = (−2)(−2) + (−1)(−1) + 0×1 + 1×0 + 2×2 = 4 + 1 + 0 + 0 + 4 = 9.
- Numerator = nΣuv − ΣuΣv = 5×9 − 0 = 45.
- Terms under the root: 5×10 − 0 = 50 and 5×10 − 0 = 50. r = 45 ÷ √(50 × 50) = 45 ÷ 50 = 0.9.
- Since h and k are both positive, r for X and Y is the same as r for u and v.
Answer: r = +0.9, a very strong positive correlation.
Exam tips
- Check the sign of the numerator first. If it is negative, r is negative, and you can drop all positive options at once.
- Learn the property questions: r lies between −1 and +1, has no unit, and is unchanged by positive change of origin and scale.
- If the question gives Cov(X, Y), σx and σy, just compute Cov ÷ (σx × σy). No table is needed.
- Remember that r = 0 means no linear relationship only, and that correlation does not prove cause and effect.
- With no negative marking, always attempt the question. If time is short, use the sign and the range to narrow the options and then guess.
Practice questions from Correlation and Regression
- Each value of x is transformed to u = (x − 50)/10 and each value of y to v = (y − 200)/25. If the correlation coefficient between u and v is…
- For a set of paired data, the covariance between X and Y is 12, the standard deviation of X is 4 and the standard deviation of Y is 5. What …
- The correlation coefficient between variables X and Y is 0.6. New variables are defined as U = 2X + 3 and V = 5 - 3Y. What is the correlatio…
- The regression lines of Y on X and X on Y are Y = 0.8X + 10 and X = 0.45Y + 5 respectively. What is the value of the correlation coefficient…
- The total variation in the sales (Y) of a Chennai retailer is 200 (in squared units). If the correlation between Y and the number of custome…
Karl Pearson's Coefficient of Correlation in other exams
The same ground in other exams, if you are preparing for more than one or want another angle on it.
Karl Pearson's Coefficient of Correlation: frequently asked questions
What is the formula for Karl Pearson's coefficient of correlation?
The basic formula is r = Cov(X, Y) ÷ (σx × σy). For calculation, use r = [nΣXY − ΣXΣY] ÷ √{[nΣX² − (ΣX)²][nΣY² − (ΣY)²]}. Both give the same value.
Does a change of origin or scale affect the correlation coefficient?
No, as long as you subtract constants and divide by positive constants. That is why the assumed mean and step-deviation methods give the same r as the direct method. If you divide one variable by a negative number, the sign of r flips.
What is the range of Karl Pearson's coefficient of correlation?
It lies from −1 to +1. A value of +1 is perfect positive linear correlation, −1 is perfect negative, and 0 is no linear correlation. A value outside this range means a calculation error.
Which method should I use in the exam, direct or step-deviation?
Use the direct method when the values are small. Use the assumed mean or step-deviation method when the values are large or are multiples of a common number. Both give the same answer, so choose the one with less arithmetic.