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JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    Variance of the data given below:
    Size of item 3.5 4.5 5.5 6.5 7.5 8.5 9.5
    Frequency 3 7 22 60 85 32 8
    is

    A)  1.29                      

    B)  2.19

    C)  1.32                                      

    D)  none of these

    Correct Answer: C

    Solution :

     Let the assumed mean be A=6.5 Calculation of variance
    Size of item \[({{x}_{i}})\] \[{{f}_{i}}\] \[{{d}_{i}}=\] \[{{x}_{i}}-6.5\] \[{{d}_{i}}^{2}\] \[{{f}_{i}}{{d}_{i}}\] \[{{f}_{i}}{{d}_{i}}^{2}\]
    3.5 3 -3 9 -9 27
    4.5 7 -2 4 -14 28
    5.5 22 -1 1 -22 22
    6.5 60 0 0 0 0
    7.5 85 1 1 85 85
    8.5 32 2 4 64 128
    9.5 8 3 9 24 72
    \[N=\sum \] \[{{f}_{i}}=217\] \[\sum {{f}_{i}}{{d}_{i}}\] \[=128,\] \[a\sum {{f}_{i}}{{d}_{i}}^{2}\] \[=362\]
    Here\[N=217,\sum {{f}_{i}}{{d}_{i}}=128,\]\[\sum {{f}_{i}}{{d}_{i}}^{2}=362\] \[\therefore \]Var\[(X)=\left( \frac{1}{N}\sum {{f}_{i}}d_{i}^{2} \right)-{{\left( \frac{1}{N}\sum {{f}_{i}}d_{i}^{{}} \right)}^{2}}\] \[=\frac{362}{217}{{\left( \frac{128}{217} \right)}^{2}}\] = 1.668 - 0.347 = 1.321


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