10th Class Mathematics Solved Paper - Mathematics-2015 Term-I

  • question_answer
    In a class test, marks obtained by 120 students are given in the following frequency distribution. If it is given that mean is 59, find the missing frequencies x and y.
    Marks No. of students
    0 ? 10 1
    10 ? 20 3
    20 ? 30 7
    30 ? 40 10
    40 ? 50 15
    50 ? 60 x
    60 ? 70 9
    70 ? 80 27
    80 ? 90 18
    90 ? 100 y

    Answer:

               

    Marks No. of Students \[{{f}_{i}}\] \[{{X}_{i}}\] \[{{d}_{i}}=\frac{{{X}_{i}}-55}{10}\] \[{{f}_{i}}{{d}_{i}}\]
    0 ? 10 1 5 \[5\] \[5\]
    10 ? 20 3 15 \[4\] \[-12\]
    20 ? 30 7 25 \[3\] \[-21\]
    30 ? 40 10 35 \[2\] \[-20\]
    40 ? 50 15 45 \[1\] \[-15\]
    50 ? 60 x A = 55 0 0
    60 ? 70 9 65 1 9
    70 ? 80 27 75 2 54
    80 ? 90 18 85 3 54
    90 ? 100 y 95 4 4y
      \[\sum{{{f}_{i}}=90+x+y}\]     \[\sum{{{f}_{i}}{{d}_{i}}=-73+117+4y=44+4y}\]
                \[\sum{{{f}_{i}}=90+x+y}\]
    But       \[\sum{{{f}_{i}}=120}\]                         [Given]
    \[\therefore 90+x+y=120\]
                        \[x=120-90-y=30-y\]                                  ...(i)
             Mean \[=A+\frac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}\times h\]
    \[\Rightarrow 59=55+\left( \frac{44+4y}{120}\times 10 \right)\]                   \[\left[ A=55,h=10,\sum{{{f}_{i}}=120} \right]\]
    \[\Rightarrow 59-55=\frac{4(11+y)}{12}\]
    \[\Rightarrow 4\times 3=11+y\]
    \[\Rightarrow y=12-11=1\]
    From eq. (i), \[x=30-1=29\]
    \[\therefore x=29,y=1\]


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