Calculate the correlation co-efficient from the following data. $N = 9$, $\Sigma X = 45$, $\Sigma Y = 108$, $\Sigma X^2 = 285$, $\Sigma Y^2 = 1356$, $\Sigma XY = 597$.

Calculate the correlation co-efficient from the following data. N = 9, Σ X = 45, Σ Y = 108, Σ X² = 285, Σ Y² =

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N = 9

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Substituting the summations into Karl Pearson's formula gives $r = \dfrac{513}{540} = 0.95$ , a strong positive correlation.

Step 1 — Formula.

$r = \frac{N\Sigma XY - \Sigma X\,\Sigma Y}{\sqrt{[N\Sigma X^2-(\Sigma X)^2]\,[N\Sigma Y^2-(\Sigma Y)^2]}}.$

Step 2 — Numerator.

$N\Sigma XY - \Sigma X\,\Sigma Y = 9(597) - (45)(108) = 5373 - 4860 = 513.$

Step 3 — Denominator terms.

$N\Sigma X^2-(\Sigma X)^2 = 9(285) - 45^2 = 2565 - 2025 = 540,$

$N\Sigma Y^2-(\Sigma Y)^2 = 9(1356) - 108^2 = 12204 - 11664 = 540.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set MARCH1 mark
$Q. Define correlation coefficient.$
›Reveal solutionSolution

$r=\dfrac{\text{Cov}(x,y)}{\sigma_x\sigma_y}$ measures the degree and direction of linear relationship, $-1\le r\le 1$ .

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:

$r=\frac{\text{Cov}(x,y)}{\sigma_x\,\sigma_y}=\frac{\Sigma(x-\bar x)(y-\bar y)}{\sqrt{\Sigma(x-\bar x)^2\,\Sigma(y-\bar y)^2}}.$ …

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CBSE 2024Set MARCH1 markMCQ
$Q. Correlation co-efficient lies between :$
(a) $-1$ to $0$
(b) $0$ to $\infty$
(c) $-1$ to $\infty$
(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:

$-1\le r\le +1.$ …

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CBSE 2023Set MARCH1 markMCQ
$Q. What does the numerator indicate in the formula for calculating correlation coefficient by Karl Pearson's method?$
(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 $r=\dfrac{\text{cov}(x,y)}{S_x\,S_y}$ , so the numerator is the covariance of $X$ and $Y$ . Option (B).

The Karl Pearson coefficient of correlation is defined as

$r=\frac{\text{cov}(x,y)}{S_x\cdot S_y}=\frac{\tfrac{1}{n}\sum (x-\bar x)(y-\bar y)}{S_x\,S_y}.$ …

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CBSE 2022Set MARCH1 markMCQ
$Q. If the values of two variables move in opposite direction then the correlation is said to be :$
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›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. …
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CBSE 2022Set MARCH1 markMCQ
$Q. Example for positive correlation is :$
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›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. …
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$Q. If two variables move in decreasing direction then the correlation is :$
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›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). …
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Q.Calculate the correlation co-efficient from the following data. $N = 9$ , $\Sigma X = 45$ , $\Sigma Y = 108$ , $\Sigma X^2 = 285$ , $\Sigma Y^2 = 1356$ , $\Sigma XY = 597$ .

Concept understanding — Karl Pearson's Coefficient of Correlation

Q.Calculate the correlation co-efficient from the following data. N=9N = 9N=9, ΣX=45\Sigma X = 45ΣX=45, ΣY=108\Sigma Y = 108ΣY=108, ΣX2=285\Sigma X^2 = 285ΣX2=285, ΣY2=1356\Sigma Y^2 = 1356ΣY2=1356, ΣXY=597\Sigma XY = 597ΣXY=597.

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Q.Calculate the correlation co-efficient from the following data. $N = 9$ , $\Sigma X = 45$ , $\Sigma Y = 108$ , $\Sigma X^2 = 285$ , $\Sigma Y^2 = 1356$ , $\Sigma XY = 597$ .

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