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Worked Examples · Example 44
Q.

Result of two simultaneous tests conducted in a class have been given in following table. Calculate the Karl Pearson's coefficient of correlation for scores obtained by students in these two different tests.

Name of studentPriyaRahulTanyaPriyankaSnehaNaveenNeerajSunil
Score in first test1346891114
Score in second test12445789
Puducherry CbseNCERTSubjective· 5mImportance★★★★★est
98% · 56/57 Questions
✓ Free question

Between the first-test and second-test scores of the 8 students, Karl Pearson's coefficient of correlation is r≈0.977r\approx0.977, indicating a very strong positive linear relationship.

r=∑(xi−xˉ)(yi−yˉ)∑(xi−xˉ)2  ∑(yi−yˉ)2r=\dfrac{\sum(x_i-\bar x)(y_i-\bar y)}{\sqrt{\sum(x_i-\bar x)^2\;\sum(y_i-\bar y)^2}}

where xi,yix_i,y_i are the paired scores and xˉ,yˉ\bar x,\bar y their means.

  1. Working table:
StudentPriyaRahulTanyaPriyankaSnehaNaveenNeerajSunilTotal
XX (test 1)134689111456
YY (test 2)1244578940
x−xˉx-\bar x-6-4-3-112470
y−yˉy-\bar y-4-3-1-102340
(x−xˉ)(y−yˉ)(x-\bar x)(y-\bar y)24123104122884
(x−xˉ)2(x-\bar x)^2361691141649132
(y−yˉ)2(y-\bar y)^2169110491656
  1. n=8n=8; xˉ=568=7\bar x=\dfrac{56}{8}=7; yˉ=408=5\bar y=\dfrac{40}{8}=5.
  2. From the table: ∑(x−xˉ)(y−yˉ)=84\sum(x-\bar x)(y-\bar y)=84, ∑(x−xˉ)2=132\sum(x-\bar x)^2=132, ∑(y−yˉ)2=56\sum(y-\bar y)^2=56.
  3. r=84132×56=847392r=\dfrac{84}{\sqrt{132\times56}}=\dfrac{84}{\sqrt{7392}}.
  4. 7392≈85.98\sqrt{7392}\approx85.98, so r≈8485.98≈0.977r\approx\dfrac{84}{85.98}\approx0.977.
  5. Self-check: 0.9772×7392≈70570.977^2\times7392\approx7057, and 842=705684^2=7056 — matches (rounding). Since −1≤r≤1-1\le r\le1 and rr is close to +1+1, this confirms a very strong, positive linear relationship. ✓
✓Final answer

Karl Pearson's coefficient of correlation r≈0.977r\approx0.977 (very strong positive correlation between the two tests)

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