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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    For the data
    \[x\] 4 7 8 3 4
    \[y\] 5 8 6 3 5
    the Karl Pearson coefficient is:

    A)  \[\frac{63}{\sqrt{94\times 66}}\]             

    B)         \[63\]

    C)         \[\frac{63}{\sqrt{94}}\]                

    D)         \[\frac{63}{\sqrt{66}}\]

    E)         \[\frac{1}{\sqrt{94\times 66}}\]

    Correct Answer: A

    Solution :

                    Take A = 5 and B = 5
    \[{{x}_{i}}\] \[{{y}_{i}}\] \[{{u}_{i}}\]\[={{x}_{i}}-A\] \[{{v}_{i}}\] \[={{y}_{i}}-A\] \[{{({{u}_{i}})}^{2}}\] \[{{({{v}_{i}})}^{2}}\] \[{{u}_{i}}{{v}_{i}}\]
    4 5 -1 0 1 0 0
    7 8 2 3 4 9 6
    8 6 3 1 9 1 3
    3 3 -2 -2 4 4 4
    4 5 -1 0 1 0 0
    \[\Sigma {{u}_{i}}=1\] \[\Sigma {{v}_{i}}=2\] \[\Sigma u_{i}^{2}\] =19 \[\Sigma v_{i}^{2}\] =14 \[\Sigma {{u}_{i}}{{v}_{i}}\]=13
    \[\therefore \]Karl Pearson coefficient \[=\frac{\Sigma {{u}_{i}}{{v}_{i}}-\frac{1}{n}\Sigma {{u}_{i}}{{v}_{i}}}{\sqrt{\Sigma u_{i}^{2}-\frac{1}{n}{{(\Sigma {{u}_{i}})}^{2}}}\sqrt{\Sigma v_{i}^{2}-\frac{1}{n}{{(\Sigma {{v}_{i}})}^{2}}}}\] \[=\frac{13-\frac{2}{5}}{\sqrt{19-\frac{1}{5}}\sqrt{14-\frac{4}{5}}}=\frac{63}{\sqrt{94}\sqrt{66}}\]


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