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10th Class Mathematics Statistics Question Bank Statistics

  • question_answer
    Find the mean, median and mode of the following data.                       
    Classes 0-20 20-40 40-60 60-80 80-100 100-120 120-140
    Frequency 6 8 10 12 6 5 3
               

    A)
    Mean Median Mode
    50.67  62.67      64.5
                   

    B)
    Mean Median Mode
    60.67  52.42     64.5
                   

    C)
    Mean Median Mode
    62.4    61.67     65
                   

    D)
    Mean Median Mode
    62.4    62.42     65

    Correct Answer: C

    Solution :

    Class \[({{x}_{i}})\] Frequency \[({{f}_{i}})\] \[{{x}_{i}}{{f}_{i}}\] Cumulative frequency (cf)
    0-20 10 6 60 6
    20-40 30 8 240 14
    40-60 50 10 500 24
    60-80 70 12 840 36
    80-100 90 6 540 42
    100-120 110 5 550 47
    120-140 130 3 390 50
    Total \[\Sigma {{f}_{i}}=50\] \[\Sigma {{f}_{i}}{{x}_{i}}=3120\]
    Mean \[=\frac{\Sigma {{f}_{i}}{{x}_{i}}}{\Sigma {{f}_{i}}}=\frac{3120}{50}=62.4;\,n=50\,\Rightarrow \frac{n}{2}=25\] Median class is\[60-80\]. Median \[=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h\] \[=60+\left( \frac{25-24}{12} \right)\times 20=60+1.67=61.67\] Maximum frequency is 12, so modal class is\[60-80\]. Mode  \[=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h\] \[=60+\left( \frac{12-10}{2\times 12-10-6} \right)\times 20=60+5=65\]                


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