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JEE Main & Advanced Mathematics Statistics Question Bank Mock Test - Statistics

  • question_answer
    Computer the medium form the following table
       Marks obtained    No. of students
    0-10 2
    10-20 18
    20-30 30
    30-40 45
    40-50 35
    50-60 20
    60-70 6
    70-80 3

    A) 36.55   

    B) 35.55

    C) 6.85                 

    D) None of these

    Correct Answer: A

    Solution :

    [a]
    Marks obtained No. of students Cumulative frequency
    0-10 2 2
    10-20 18 20
    20-30 30 50
    30-40 45 95
    40-50 35 130
    50-60 20 150
    60-70 6 156
    70-80 3 159
    \[N=\Sigma f=159\](Odd number) Median is \[\frac{1}{2}(n+1)=\frac{1}{2}(159+1)=80th\]value, which lies in the class 30-40 (see the row of cumulative frequency 95, which contains 80). Hence median class is 30-40. \[\therefore \] We have l = Lower limit of median class =30 f = frequency of median class = 45 C = Total of all frequencies preceding median class = 50 i = width of class interval of median class=10 \[\therefore \]Required median\[=l+\frac{\frac{N}{2}-C}{f}\times i=30+\frac{\frac{159}{2}-50}{45}\times 10\] \[=3+\frac{295}{45}=36.55\]


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