Q.
Compute the median from the following table
Marks obtained
No. of students
0-10
2
10-20
18
20-30
30
30-40
45
40-50
35
50-60
20
60-70
6
70-80
3
| Marks obtained | No. of students |
|---|---|
| 0-10 | 2 |
| 10-20 | 18 |
| 20-30 | 30 |
| 30-40 | 45 |
| 40-50 | 35 |
| 50-60 | 20 |
| 60-70 | 6 |
| 70-80 | 3 |
Statistics
Solution:
Marks obtained
No. of students
Cumulative frequency
0-10
2
2
10-20
18
20
20-30
30
50
30-40
45
95
40-50
35
130
50-60
20
150
60-70
6
156
70-80
3
159
$N=\Sigma f=159 $ (Odd number)
Median is $\frac{1}{2}(n+1)=\frac{1}{2}(159+1)=80 th$ value, which lies in the class $30-40$ (see the row of cumulative frequency 95 , which contains 80 ).
Hence median class is $30-40$.
$\therefore $ We have
$l=$ Lower limit of median class $=30$
$f=$ Frequency of median class $=45$
$C=$ Total of all frequencies preceding median class $=50$
$i=$ Width of class interval of median class $=10$
$ \therefore $ Required median $=l+\frac{\frac{N}{2}-C}{f} \times i=30+\frac{\frac{159}{2}-50}{45} \times 10$
$=30+\frac{295}{45}=36.55 $
| Marks obtained | No. of students | Cumulative frequency |
|---|---|---|
| 0-10 | 2 | 2 |
| 10-20 | 18 | 20 |
| 20-30 | 30 | 50 |
| 30-40 | 45 | 95 |
| 40-50 | 35 | 130 |
| 50-60 | 20 | 150 |
| 60-70 | 6 | 156 |
| 70-80 | 3 | 159 |