Q.Calculate the co-efficient of correlation from the following data : $\Sigma X = 50$, $\Sigma Y = -30$, $\Sigma X^{2} = 290$, $\Sigma Y^{2} = 300$, $\Sigma XY = -115$, $N = 10$.

By Karl Pearson's coefficient, $r=\dfrac{N\Sigma XY-\Sigma X\,\Sigma Y}{\sqrt{[N\Sigma X^{2}-(\Sigma X)^{2}][N\Sigma Y^{2}-(\Sigma Y)^{2}]}}\approx +0.38$.

This applies Karl Pearson's coefficient of correlation from the TN HSC Class-11 Business Mathematics syllabus.

Given: $\Sigma X=50,\ \Sigma Y=-30,\ \Sigma X^{2}=290,\ \Sigma Y^{2}=300,\ \Sigma XY=-115,\ N=10.$

Step 1 — Formula.

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

Step 2 — Numerator.

$N\Sigma XY-\Sigma X\,\Sigma Y=10(-115)-(50)(-30)=-1150+1500=350.$

Step 3 — Denominator brackets. …