JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    The outcome of each of 30 items was observed; 10 items gave an outcome \[\frac{1}{2}-d\]each, 10 items gave outcome \[\frac{1}{2}\]each and the remaining 10 items gave outcome\[\frac{1}{2}+d\] each. If the variance of this outcome data is \[\frac{4}{3},\]then \[\left| d \right|\]equals [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[\frac{2}{3}\]                                      

    B)               \[\frac{\sqrt{5}}{2}\]

    C)               2                                             

    D)               \[\sqrt{2}\]

    Correct Answer: D

    Solution :

    Since, variance is independent of origin, so we shift the data by \[\frac{1}{2}.\] Now, mean of the given items\[=\frac{-10d+0+10d}{30}=0\] Now, variance \[=\frac{10{{d}^{2}}+10\times {{0}^{2}}+10{{d}^{2}}}{30}-{{(0)}^{2}}=\frac{4}{3}\] \[\Rightarrow \frac{20}{30}{{d}^{2}}=\frac{4}{3}\,\,\,\,\,\,\,\,\,\Rightarrow d=\pm \sqrt{2}\] \[\therefore \]\[|d|=\sqrt{2}\]


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