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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