Compute Karl Pearson's coefficient of correlation for the following data:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | 2 | 5 | 3 | 8 | 7 |
Concept understanding — Karl Pearson's Coefficient of Correlation
Pearson's r=∑(x−xˉ)(y−yˉ)/∑(x−xˉ)2∑(y−yˉ)2 (or the equivalent direct-method formula) is a unit-free number between −1 and +1 measuring the strength and direction of the linear relationship between two variables, independent of the units either is measured in.
The means are whole numbers, so use the deviation-from-mean formula.
xˉ=3, yˉ=5; ∑dxdy=13, ∑dx2=10, ∑dy2=26.
r=10×2613=26013=16.1213.
r≈+0.806 (high positive correlation).
Here n=5. First find the means:
xˉ=51+2+3+4+5=515=3,yˉ=52+5+3+8+7=525=5.
Take deviations dx=x−3 and dy=y−5:
| x | y | dx | dy | dxdy | dx2 | dy2 |
|---|---|---|---|---|---|---|
| 1 | 2 | −2 | −3 | 6 | 4 | 9 |
| 2 | 5 | −1 | 0 | 0 | 1 | 0 |
| 3 | 3 | 0 | −2 | 0 | 0 | 4 |
| 4 | 8 | 1 | 3 | 3 | 1 | 9 |
| 5 | 7 | 2 | 2 | 4 | 4 | 4 |
| Total | 0 | 0 | 13 | 10 | 26 |
(The check ∑dx=0 and ∑dy=0 confirms the means are correct.) Now
r=∑dx2 ∑dy2∑dxdy=102613=26013=16.124513=0.8062.
Independent check (raw-total formula). ∑x=15, ∑y=25, ∑xy=2+10+9+32+35=88, ∑x2=55, ∑y2=151.
r=n∑x2−(∑x)2n∑y2−(∑y)2n∑xy−∑x∑y=5(55)−2255(151)−6255(88)−(15)(25)=50130440−375=650065=80.6265=0.8062.
Both methods agree.
r≈+0.806 — a high degree of positive correlation.
The raw-total formula shown as the check can be used as the main method when the means are not whole numbers, since it avoids fractional deviations.
Forgetting to take the square root of each sum-of-squares separately, or writing ∑dx2×∑dy2 wrongly. Also, a value like r=1.6 is impossible — r can never exceed ±1, so recheck the arithmetic.
- CBSE 2026Set MARCH1 markQ.Define correlation coefficient.
›Reveal solutionSolution
r=σxσyCov(x,y) measures the degree and direction of linear relationship, −1≤r≤1.
Karl Pearson's coefficient of correlation between two variables X and Y is defined as the ratio of their covariance to the product of their standard deviations:
r=σxσyCov(x,y)=Σ(x−xˉ)2Σ(y−yˉ)2Σ(x−xˉ)(y−yˉ).
It is a pure number lying between −1 and +1; its sign shows the direction (positive/negative) and its magnitude shows the strength of the linear relationship.
✓Final answerr=σxσyCov(x,y), a unit-free measure of linear relationship with −1≤r≤1.
- CBSE 2024Set MARCH1 markMCQQ.Correlation co-efficient lies between :(a) −1 to 0(b) 0 to ∞(c) −1 to ∞(d) −1 to +1
›Reveal solutionSolution
The correlation coefficient always lies between −1 and +1.
Karl Pearson's coefficient of correlation r measures the strength and direction of the linear relationship between two variables and is a pure number with no unit. It is bounded:
−1≤r≤+1.
-
r=+1: perfect positive correlation,
-
r=−1: perfect negative correlation,
-
r=0: no linear correlation.
✓Final answerOption (d) −1 to +1.
-
- CBSE 2023Set MARCH1 markMCQQ.What does the numerator indicate in the formula for calculating correlation coefficient by Karl Pearson's method?(a) (A) Product of variance of X and Y(b) (B) Covariance of X and Y(c) (C) Variance of X(d) (D) Variance of Y
›Reveal solutionSolution
Karl Pearson's formula is r=SxSycov(x,y), so the numerator is the covariance of X and Y. Option (B).
The Karl Pearson coefficient of correlation is defined as
r=Sx⋅Sycov(x,y)=SxSyn1∑(x−xˉ)(y−yˉ).
Here the denominator is the product of the standard deviations Sx and Sy, while the numerator is the covariance cov(x,y) between the two variables.
✓Final answerCorrect option: (B) Covariance of X and Y.
- CBSE 2022Set MARCH1 markMCQQ.If the values of two variables move in opposite direction then the correlation is said to be :(a) Perfect positive(b) Negative(c) No correlation(d) Positive
›Reveal solutionSolution
Opposite-direction movement of two variables indicates negative correlation.
The direction of movement decides the sign of correlation:
- Same direction (both increase or both decrease together): positive correlation.
- Opposite directions (one increases as the other decreases): negative correlation.
Since the two variables here move in opposite directions, the correlation is negative.
✓Final answerOption (b) Negative.
- CBSE 2022Set MARCH1 markMCQQ.Example for positive correlation is :(a) Repayment period and EMI(b) Income and expenditure(c) Weight and Income(d) Price and demand
›Reveal solutionSolution
Income and expenditure move in the same direction, an example of positive correlation.
A positive correlation exists when both variables increase (or decrease) together. Examine the options:
- (a) Repayment period and EMI: a longer period lowers the EMI — opposite direction (negative).
- (b) Income and expenditure: as income rises, expenditure usually rises too — same direction (positive).
- (c) Weight and Income: no logical relationship — essentially no correlation.
- (d) Price and demand: as price rises, demand falls — opposite direction (negative).
Only income and expenditure move together, giving positive correlation.
✓Final answerOption (b) Income and expenditure.
- CBSE 2020Set MARCH1 markMCQQ.If two variables move in decreasing direction then the correlation is :(a) negative(b) positive(c) perfect negative(d) no correlation
›Reveal solutionSolution
If two variables move together in the same direction — both increasing or, as here, both decreasing — the correlation between them is positive.
The sign of correlation is decided by the direction in which the variables move together:
- Positive correlation: both variables change in the same direction (both increase together, or both decrease together).
- Negative correlation: the variables change in opposite directions (one increases while the other decreases).
Here both variables move in the decreasing direction together — i.e. the same direction — so the correlation is positive.
✓Final answerOption (b) positive.
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.