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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Statistics

Given $ n=10, \: \Sigma x = 3,\: \Sigma x^2=8,\: \Sigma y^2=9,\: \Sigma xy=3$ Find correlation coefficient

$\begin {array} {1 1} (A)\;25 & \quad (B)\;.25 \\ (C)\;2.5 & \quad (D)\;0 \end {array}$

1 Answer

Toolbox:
  • Coeff. of correlation =$ r = \large\frac{\Sigma xy-\large\frac{1}{n} ( \Sigma x)( \Sigma y)}{\sqrt{ \Sigma x^2-\large\frac{1}{n} ( \Sigma x)^2} \sqrt{\Sigma y^2-\large\frac{1}{n} ( \Sigma y)^2}}$
Coeff. of correlation=$ r = \large\frac{\Sigma xy-\large\frac{1}{n} ( \Sigma x)( \Sigma y)}{\sqrt{ \Sigma x^2-\large\frac{1}{n} ( \Sigma x)^2} \sqrt{\Sigma y^2-\large\frac{1}{n} ( \Sigma y)^2}}$
$ = \large\frac{3-\large\frac{1}{\not{10}^5} \times \not{4}^2 \times 3}{\sqrt{8-\large\frac{1}{10} (4)^2} \sqrt{9-\large\frac{1}{10} (3)^2}}$
$ = \large\frac{3-\large\frac{6}{5}}{\sqrt{8-\large\frac{16}{10}} \sqrt{9-\large\frac{9}{10}}}$
$ = \large\frac{\large\frac{15-6}{5}}{\sqrt{\large\frac{80-16}{10}} \sqrt{\large\frac{90-9}{10}}}$
$ = \large\frac{\large\frac{9}{5}}{\sqrt{\large\frac{64}{10}} \sqrt{\large\frac{81}{10}}}$
$ = \large\frac{1.8}{\sqrt{6.4} \sqrt{8.1}}$
$ = \large\frac{18}{8 \times 9}$
$ = \large\frac{1}{4}$
$ = 0.25$
Ans : (B)
answered Jan 31, 2014 by thanvigandhi_1
edited Mar 26, 2014 by rvidyagovindarajan_1
 

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