JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    The mean of 5 observations is 5 and their variance is 124. If three of the observations are 1,2 and 6, then the mean deviation from the mean of the data is   JEE Main Online Paper (Held On 10 April 2016)

    A) 2.5                                                            

    B) 2.8                

    C) 2.6                                                            

    D) 2.4

    Correct Answer: D

    Solution :

                 (Bonus)                              This question is wrong              \[\overline{x}=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}}{5}=5\]              \[\sum\limits_{i=1}^{5}{{{x}_{i}}=25}\]                                      ????.. (i)              Also \[{{\sigma }^{2}}=124\] \[\Rightarrow \,\,\,\frac{\sum{x_{1}^{2}}}{5}-{{\left( \overline{x} \right)}^{2}}=124\] \[\Rightarrow \,\,\frac{\sum{x_{1}^{2}}}{5}=124+25=149\] \[\Rightarrow \,\,(x_{1}^{2}+x_{2}^{2}+\,......\,+x_{5}^{2})=745\] \[\Rightarrow \,\,x_{1}^{2}+x_{2}^{2}=704\]                          ???.....(ii) by (i) \[{{x}_{1}}+{{x}_{2}}=16\]                    ............. (iii) \[2{{x}_{1}}{{x}_{2}}+704=256\] \[{{x}_{1}}\,{{x}_{2}}=\frac{256-704}{2}\] \[{{x}_{1}}\,{{x}_{2}}=128-352=-224\]        ???.....(iv) Now \[\frac{\sum{\left| {{x}_{1}}-5 \right|}}{5}=\frac{\left| {{x}_{1}}-5 \right|+\left| {{x}_{2}}-5 \right|+4+3+1}{5}\] \[=\frac{8+\left| {{x}_{1}}-5 \right|+\left| 11-{{x}_{1}} \right|}{5}\] \[=\frac{8+6}{5}=2.8\]Ans.


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