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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    If the data \[{{x}_{1}},{{x}_{2}},....,{{x}_{10}}\]is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000; then the standard deviation of this data is : [JEE Main Held on 12-4-2019 Morning]

    A) \[4\]                             

    B) \[2\]

    C) \[\sqrt{2}\]                              

    D) \[2\sqrt{2}\]

    Correct Answer: B

    Solution :

    \[{{x}_{1}}+...+{{x}_{4}}=44\]           \[{{x}_{5}}+...+{{x}_{10}}=96\]           \[\overline{x}=14,\Sigma {{x}_{i}}=140\] Variance\[=\frac{\Sigma x_{i}^{2}}{n}-{{\overline{x}}^{2}}=4\] Standard deviation = 2


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