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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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The correlation coefficient between $x$ and $y$ is from the following data is \[\] $ \Sigma x = 40\: \: \Sigma y=50\: \: \Sigma xy = 220\: \: \Sigma x^2=200\: \: \Sigma y^2 = 262\: \: n=10$

$\begin {array} {1 1} (A)\;0.89 & \quad (B)\;0.76 \\ (C)\;0.91 & \quad (D)\;0.98 \end {array}$

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Correlation coefficient
$ r = \large\frac{n \Sigma xy - \Sigma x \Sigma y}{\sqrt{n \Sigma x^2- ( \Sigma x)^2} \sqrt{ n \Sigma y^2-( \Sigma y)^2}}$
= $ \large\frac{ 10 \times 220 - 40 \times 50}{\sqrt{10 \times 200 - ( 40)^2}{\sqrt{10 \times 262 - (50)^2}}}$
$ = \large\frac{200}{20 \times 10.945}$
$ = \large\frac{200}{218.9}$
$ = 0.91$
Ans : (C)
answered Jan 29, 2014 by thanvigandhi_1
 

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