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JEE Main & Advanced Mathematics Statistics Question Bank Critical Thinking

  • question_answer
    The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are

    A)                 14.98, 39.95           

    B)                 39.95, 14.98

    C)                 39.95, 224.5           

    D)                 None of these

    Correct Answer: B

    Solution :

                    Corrected \[\Sigma x=40\times 200-50+40=7990\]                                 \ Corrected \[\bar{x}=7990/200\]\[=39.95\]                                 Incorrect\[\Sigma {{x}^{2}}=n\,[{{\sigma }^{2}}+{{\bar{x}}^{2}}]=200[{{15}^{2}}+{{40}^{2}}]=365000\]                                 Correct \[\Sigma {{x}^{2}}=365000-2500+1600\]\[=364100\]                                 \ Corrected \[\sigma =\sqrt{\frac{364100}{200}-{{(39.95)}^{2}}}\]                                             \[=\sqrt{(1820.5-1596)}\]\[=\sqrt{224.5}=14.98\].


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