If x = X- \(\bar{X}\) and y = Y - \(\bar{Y}\) and the number of pairs (X, Y) is n, then the Karl Pearson's coefficient of correlation is:

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SSC CGL JSO Tier-II, 2018 (Statistics) Official Paper-III (Held On: 14 Sept, 2019)
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  1. \(\dfrac{n\sum xy}{{\sqrt {\sum x^2\sum y^2}}}\)
  2. \(\dfrac{\sum xy}{(\sum x^2 \sum y^2)^\dfrac{1}{3}}\)
  3. \(\dfrac{\sum xy}{{\sqrt {\sum x^2\sum y^2}}}\)
  4. \(\dfrac{\sum xy}{n\sum x^2\sum y^2}\)

Answer (Detailed Solution Below)

Option 3 : \(\dfrac{\sum xy}{{\sqrt {\sum x^2\sum y^2}}}\)
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Detailed Solution

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Calculation

Here, correlation coefficient = r

⇒ r = Cov(x,y)/SxSy

Here, Cov(x,y) = ∑(X - Y)(Y - Y)/(n - 1)

⇒ Sx =  √∑(X - X̅) 2/(n - 1)

⇒ Sy = √∑(Y - Y̅ )2/(n - 1)

⇒ Cov(x,y)/SxSy = ∑(X - X̅)(Y - )/(n - 1)/√(∑(X - )2)√(∑(Y - Y)2)/(n - 1)

⇒ Cov(x,y)/SxSy = ∑(X - X̅ )(Y - Y̅ )/√(∑(X - X̅ )2)√(∑(Y - Y̅ )2)

⇒  Cov(x,y)/SxSy = ∑xy/√∑x2 × √∑y2 Here, x = (X - \(̅{X}\)) and y = (Y - \(̅{Y}\))

∴ The Karl Pearson's coefficient of correlation is ∑xy/√[∑x2 × ∑y2]

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