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Exercises · Q6
Q.

The number of hours studied (xx) and marks obtained out of 50 (yy) by 5 students are given below. Find Karl Pearson's coefficient of correlation.

Student12345
Hours studied (xx)510152025
Marks obtained (yy)1218202530
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✓ Free question

Step 1 — Find the means. xˉ=5+10+15+20+255=15\bar x = \dfrac{5+10+15+20+25}{5}=15. yˉ=12+18+20+25+305=1055=21\bar y = \dfrac{12+18+20+25+30}{5}=\dfrac{105}{5}=21.

Step 2 — Tabulate deviations, products and squares.

xxyyx−xˉx-\bar xy−yˉy-\bar y(x−xˉ)(y−yˉ)(x-\bar x)(y-\bar y)(x−xˉ)2(x-\bar x)^2(y−yˉ)2(y-\bar y)^2
512−10−99010081
1018−5−315259
15200−1001
202554202516
25301099010081
Total215250188

Step 3 — Apply the formula.

r=215250188=21547000≈215216.79≈0.992r = \dfrac{215}{\sqrt{250}\sqrt{188}} = \dfrac{215}{\sqrt{47000}} \approx \dfrac{215}{216.79} \approx 0.992

Independent check (direct method). ∑xy=5(12)+10(18)+15(20)+20(25)+25(30)=60+180+300+500+750=1790\sum xy = 5(12)+10(18)+15(20)+20(25)+25(30) = 60+180+300+500+750=1790. ∑x2=25+100+225+400+625=1375\sum x^2=25+100+225+400+625=1375, ∑y2=144+324+400+625+900=2393\sum y^2=144+324+400+625+900=2393. Numerator: 5(1790)−(75)(105)=8950−7875=10755(1790)-(75)(105)=8950-7875=1075. Denominator: 5(1375)−56255(2393)−11025=1250940≈35.36×30.66≈1084.3\sqrt{5(1375)-5625}\sqrt{5(2393)-11025} = \sqrt{1250}\sqrt{940} \approx 35.36\times30.66\approx1084.3. r≈1075/1084.3≈0.991r \approx 1075/1084.3 \approx 0.991 — matches the deviation-method result of ≈0.992\approx0.992 closely (the tiny gap is rounding), confirming a very strong positive correlation.

✓Final answer

r≈0.99r \approx 0.99, indicating a very strong positive correlation between hours studied and marks obtained.

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