• # 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

 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), $x=30-1=29$ $\therefore x=29,y=1$