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JEE Main & Advanced Mathematics Statistics Question Bank Self Evaluation Test - Statistics

  • question_answer
    The marks obtained by 60 students in a certain test are given below:
    Marks No. of students Marks No. of students
    10-20 2 60-70 12
    20-30 3 70-80 14
    30-40 4 80-90 10
    40-50 5 90-100 4
    50-60 6
    Mean, median and mode of the above data are respectively

    A) \[64.33,68.33,76.33\]

    B) \[60,70,80\]

    C) \[66.11,71.11,79.11\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] We construct the following table taking assumed mean a = 55 (step deviation method).
    Class \[{{x}_{i}}\] \[{{f}_{i}}\] \[c.f.\] \[{{u}_{i}}=\frac{{{x}_{i}}-a}{10}{{f}_{i}}{{u}_{i}}\]
    10-20 15 2 2 -4-8
    20-30 25 3 5 -3-9
    30-40 35 4 9 -2-8
    40-50 45 5 14 -1-5
    50-60 55 6 20 0 0
    60-70 65 12 32 1 12
    70-80 75 14 46 2 28
    80-90 85 10 56 3 30
    90-100 95 4 60 4 16
    Total 60 56
    \[\therefore \]The mean \[=a+\frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}}\times c\] \[=55\times \frac{56}{60}\times 10=55+\frac{56}{6}=64.333\] Here \[n=60\Rightarrow \frac{n}{2}=30,\] therefore, 60-70 is the median class Using the formula: \[M=l+\frac{\frac{n}{2}-C}{f}\times c=60+\frac{30-20}{12}\times 10=68.333\] Using Empirical formula, we have Mode = 3 Median -2 Mean \[=(68.333)-2(64.333)=204.999-128.666=76.333\]


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