JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    The mean and the standard deviation (s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes10, then their s.d. is?        [JEE Main 16-4-2018]

    A)  0                                

    B)  4

    C)  2                                

    D)  1

    Correct Answer: C

    Solution :

     \[\bar{x}=\frac{\sum{xi}}{n}=9\]                 \[s.d=\sqrt{\sum{\frac{xi-\bar{x}}{n}=0}}\]                 \[\Rightarrow \]\[{{X}_{i}}=\bar{x}\forall i\]                 \[\therefore \]each term in the original observation \[\Rightarrow \]observation\[1=\{9,9,9,9,9\}\] \[\bar{x}=\frac{\sum{xi}}{n}=\frac{9+9+9+9{{x}_{5}}}{5}\] \[\Rightarrow \]\[10=\frac{36+{{X}_{5}}}{5}\] \[\Rightarrow \]\[{{x}_{5}}=14\] \[s.d=\sqrt{\frac{\sum{{{({{x}_{i}}-{{x}^{-1}})}^{2}}}}{n}}\] \[=\frac{\sqrt{{{(9-10)}^{2}}+{{(9-10)}^{2}}+{{(9-10)}^{2}}+(10-10)}}{5}\] \[=\sqrt{\frac{4+{{4}^{2}}}{5}}\] \[=\sqrt{\frac{20}{5}}\]


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