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10th Class Mathematics Statistics Question Bank Statistics and Probability

  • question_answer
    The height of 30 boys of a class are given in the following table :
    Height in cm Frequency
    120 - 129 2
    130 - 139 8
    140 - 149 10
    150 - 159 7
    160 - 169 3
    If by joining of a boy of height 140 cm, the median of the heights is changed from \[{{M}_{1}}\] to \[{{M}_{2}}\], then\[{{M}_{1}}-{{M}_{2}}\]in cm is

    A)  0.1                                        

    B)  -0.1

    C)  0                                              

    D)  0.2

    Correct Answer: C

    Solution :

     
    Height In cms Frequency Cumulative Frequency Actual Class limit
    120 - 129 2 2 \[119.5-129.5\]
    130 - 139 8 10 \[129.5-139.5\]
    140 - 149 10 20 \[139.5-149.5\]
    150 - 159 7 27 \[149.5-159.5\]
    160 - 169 3 30 \[159.5-169.5\]
    n = 30
    Here \[n=30\] \[\therefore \]  \[\frac{n}{2}+1=15+1=16\] \[\therefore \]  16 is under cumulative frequency 20. So median class be \[140-149\] \[{{L}_{1}}=139.5,\] \[{{L}_{2}}=149.5,\] \[f=10,\] \[n=30,\] \[c=10.\] Median \[{{M}_{1}}={{L}_{1}}+\frac{{{L}_{2}}-{{L}_{1}}}{f}\left( \frac{n}{2}-c \right)\]                 \[=139.5+\frac{10}{10}(15-10)\]                 \[=139.5+\frac{10}{10}\times 5=144.5\] If by joining f a boy of height 140 cms, the                 \[n=31,\,f=11\]. \[\therefore \] Median \[{{M}_{2}}=139.5+\frac{149.5-139.5}{11}(15.5-10)\]?                 \[=139.5+\frac{10}{11}\times 5.5\]                 \[=144.5\,cms\] then \[{{M}_{1}}-{{M}_{2}}=144.5-144.5=0\]


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