Find the median form the following:

Marks (More than) 0 10 20 30 40 50 60 70

No. of Students 100 92 78 44 32 18 15 13

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Solution

Class Interval Frequency
(f) Cumulative
Frequency
(c.f.₎

0 − 10
10 − 20 8
14 8
22 (c.f.)

20 − 30 34 (f) 56

30 − 40
40 − 50
50 − 60
60 − 70
70 − 80 12
14
3
2
13 68
82
85
87
100

$N = \sum_{}^{} f = 100$

Median class is given by the size of ${(\frac{N}{2})}^{t h}$ item, i.e. ${(\frac{100}{2})}^{th}$ item, which is 50^th item.
This corresponds to the class interval of 20 − 30, so this is the median class.
$Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{100}{2} - c . f .}{f} \times i or, Median = 20 + \frac{50 - 22}{34} \times 10 or, Median = 20 + 8.235 = 28.23 Thus, Median = 28.23$

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Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
No. of Students	2	18	30	45	35	20	6	3

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	2	18	30	45	35	20	6	3

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	5	7	15	25	20	15	8	5

Marks obtained	0−10	10−20	20−30	30−40	40−50	50−60	Total
No. of Students	10	?	25	30	?	10	100

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of students	4	6	10	10	25	22	18	5

Marks (More than)	0	10	20	30	40	50	60	70
No. of Students	100	92	78	44	32	18	15	13