Q.
The marks obtained by 60 students in a certain test are given below :
Marks
No. of students
Marks
No. of students
10 - 20
2
60 - 70
12
20 - 30
3
70 - 80
14
30 - 40
4
80 - 90
10
40 - 50
5
90 - 100
4
50 - 60
6
Mean and Median of the above data are respectively
| Marks | No. of students | Marks | No. of students |
| 10 - 20 | 2 | 60 - 70 | 12 |
| 20 - 30 | 3 | 70 - 80 | 14 |
| 30 - 40 | 4 | 80 - 90 | 10 |
| 40 - 50 | 5 | 90 - 100 | 4 |
| 50 - 60 | 6 |
Statistics
Solution:
We construct the following table taking assumed mean a = 55 (step deviation method).
Class
$x_i$
$f_i$
c.f.
$u_i - \frac{x_i - a}{10}$
$f_i u_i$
10 - 20
15
2
2
-4
-8
20 - 30
25
3
5
-3
-9
30 - 40
35
4
9
-2
-8
40 - 50
45
5
14
-1
-5
50 - 60
55
6
20
0
0
60 - 70
65
12
32
1
12
70 - 80
75
14
46
2
28
80 - 90
85
10
56
3
30
90 - 100
95
4
60
4
16
Total
60
56
$\therefore $ Mean = $a + \frac{\sum f_{i}u_{i}}{\sum f_{i}}\times c$
$= 55+\frac{56}{60}\times10 =55+\frac{56}{6}=64.333$
Here $n = 60 \, \Rightarrow \frac{n}{2} = 30 $ , therefore, 60-70 is the median class
Using the formula :
$M = l + \frac{\frac{n}{2}-C}{f}\times c = 60 + \frac{30 -20}{12}\times 10$
$ = 60 + \frac{100}{12}=60+8.333=68.333$
| Class | $x_i$ | $f_i$ | c.f. | $u_i - \frac{x_i - a}{10}$ | $f_i u_i$ |
| 10 - 20 | 15 | 2 | 2 | -4 | -8 |
| 20 - 30 | 25 | 3 | 5 | -3 | -9 |
| 30 - 40 | 35 | 4 | 9 | -2 | -8 |
| 40 - 50 | 45 | 5 | 14 | -1 | -5 |
| 50 - 60 | 55 | 6 | 20 | 0 | 0 |
| 60 - 70 | 65 | 12 | 32 | 1 | 12 |
| 70 - 80 | 75 | 14 | 46 | 2 | 28 |
| 80 - 90 | 85 | 10 | 56 | 3 | 30 |
| 90 - 100 | 95 | 4 | 60 | 4 | 16 |
| Total | 60 | 56 |