JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    If both the mean and the standard deviation of 50 observations \[{{x}_{1}},{{x}_{2}}....,{{x}_{50}}\] are equal to 16, then the mean of \[{{({{x}_{1}}-4)}^{2}},{{({{x}_{2}}-4)}^{2}},.....\]\[{{({{x}_{50}}-4)}^{2}}\]is : [JEE Main 10-4-2019 Afternoon]

    A) 525                 

    B) 380

    C) 480                 

    D) 400

    Correct Answer: D

    Solution :

    Mean \[(\mu )=\frac{\sum\limits_{{}}^{{}}{{{x}_{i}}}}{50}=16\] standard deviation\[(\sigma )=\sqrt{\frac{\sum\limits_{{}}^{{}}{x_{i}^{2}}}{50}-{{(\mu )}^{2}}}=16\]           \[\Rightarrow (256)\times 2=\frac{\sum\limits_{{}}^{{}}{x_{i}^{2}}}{50}\] \[\Rightarrow \] New mean             \[=\frac{\sum\limits_{{}}^{{}}{{{\left( {{x}_{i}}-4 \right)}^{2}}}}{50}=\frac{\sum\limits_{{}}^{{}}{x_{i}^{2}+16\times 50-8\sum\limits_{{}}^{{}}{{{x}_{i}}}}}{50}\]             \[=(256)\times 2+16-8\times 16=400\]


You need to login to perform this action.
You will be redirected in 3 sec spinner