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CA Foundation · Quantitative Aptitude · Correlation and Regression

For two variables x and y, the standard deviations are σx = 3 and σy = 4, and the variance of the difference (x - y) is 13. What is Karl Pearson's coefficient of correlation between x and y?

The coefficient is 0.50. Since the variance of x - y equals σx² + σy² - 2 times covariance, 13 = 25 - 2Cov, so covariance is 6. Dividing by σxσy = 12 gives r = 0.50.

  1. A0.50Correct
  2. B-0.50
  3. C0.24
  4. D1.00

Explanation

Var(x - y) = σx² + σy² - 2Cov(x,y), so 13 = 9 + 16 - 2Cov, giving Cov = 6. Then r = 6/(3×4) = 0.50. Using a plus sign for the covariance term gives Cov = -6 and r = -0.50, which is wrong. Omitting the factor 2 gives Cov = 12 and r = 1.

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