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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    Let the observations \[{{x}_{i}}(1\le i\le 10)\] satisfy the equations, \[\sum\limits_{i=1}^{10}{({{x}_{i}}-5)=10}\]  and \[\sum\limits_{i=1}^{10}{{{({{x}_{i}}-5)}^{2}}=40}\] If \[\mu \] and \[\lambda \] are the mean and the variance of the observations, \[{{x}_{1}}-3,\] \[{{x}_{2}}-3,.....,{{x}_{10}}-3,\] then the ordered pair \[(\mu ,\lambda )\]is equal to [JEE MAIN Held on 09-01-2020 Morning]

    A) \[(6,\,3)\]           

    B) \[(3,\,6)\]

    C) \[(3,\,3)\]

    D) \[(6,\,6)\]

    Correct Answer: C

    Solution :

    [c] Let \[({{x}_{i}}-5)={{y}_{i}}\] So \[\overline{y}=\frac{\sum{{{y}_{i}}}}{10}=\frac{10}{10}=1\] and \[Var\,\,(y)=\frac{\sum{y_{i}^{2}}}{10}-{{(\bar{y})}^{2}}=3\] Now mean of \[({{x}_{i}}-3)=({{y}_{i}}+2)\]is \[\bar{y}+2,\] which is 3. And variance remains same.        


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