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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 8

  • question_answer
    If \[\sum\limits_{i=1}^{9}{({{x}_{i}}-5)=9}\] and \[\sum\limits_{i=1}^{9}{{{({{x}_{i}}-5)}^{2}}=45}\], then the standard deviation of the 9 items \[{{x}_{1}},{{x}_{2}},....,{{x}_{9}}\] is

    A) 9         

    B) 4     

    C) 3                     

    D) 2

    Correct Answer: D

    Solution :

    [d]: Clearly, \[\sum\limits_{i=1}^{9}{{{x}_{i}}-45=9}\Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}=45}\] Now,\[\sum\limits_{i=1}^{9}{x_{i}^{2}-10\times 54+25\times 9=45}\] \[\Rightarrow \]\[\sum\limits_{i=1}^{9}{x_{i}^{2}=360}\] \[\therefore \]\[\sigma =\sqrt{\frac{360}{9}-{{\left( \frac{54}{9} \right)}^{2}}}=2\]


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