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Question

# Find Karl Pearson's coefficient for the following data: Marks in English 45 70 65 30 90 40 50 75 85 60 Marks in Maths 35 90 70 40 95 40 60 80 80 50

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Solution

## Marks in English (X) Marks in Maths (Y) X2 Y2 XY 45 70 65 30 90 40 50 75 85 60 35 90 70 40 95 40 60 80 80 50 2025 4900 4225 900 8100 1600 2500 5625 7225 3600 1225 8100 4900 1600 9025 1600 3600 6400 6400 2500 1575 6300 4550 1200 8550 1600 3000 6000 6800 3000 $\Sigma X$=610 $\Sigma Y$=640 $\Sigma {X}^{2}$=40700 $\Sigma {Y}^{2}$=45350 $\Sigma XY$=42575 N = 10 $r=\frac{N\Sigma XY-\Sigma X\Sigma Y}{\sqrt{N\Sigma {X}^{2}-{\left(\Sigma X\right)}^{2}}\sqrt{N\Sigma {Y}^{2}-{\left(\Sigma Y\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{or},r=\frac{10×42575-610×640}{\sqrt{10×40700-{\left(610\right)}^{2}}\sqrt{10×45350-{\left(640\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{or},r=\frac{35350}{186.81×209.52}\phantom{\rule{0ex}{0ex}}⇒\mathrm{r}=0.9031$ Hence, Karl Pearson's coefficient is 0.9031.

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