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JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where \[p\ne 0\] and\[q\ne 0\]. If the new mean and new s.d. become half of their original values, then q is equal to     [JEE MAIN Held On 08-01-2020 Morning]

    A) -10                   

    B) -20

    C) -5        

    D) 10

    Correct Answer: B

    Solution :

    [b] \[\mu =20,{{\sigma }^{2}}=2\] and \[p\mu -q=\frac{20}{2}\Rightarrow 20p-q=10\] and also \[\left| p \right|{{\sigma }^{2}}=\left| p \right|\cdot 2=1\Rightarrow p=\pm \frac{1}{2}\] If \[p=\frac{1}{2}\Rightarrow q=0\] (rejected) & \[p=-\frac{1}{2}\Rightarrow q=-20\]


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