Question

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

• A. 36.55
• B. 35.55
• C. 40.05
• D. None of these

Answer 1

Answer: (A)

36.55

Answer 2

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

$n = \sum f = 159$

Here n = 159, which is odd.

Median number$= \frac{1}{2}(n + 1) = \frac{1}{2}(159 + 1) = 80$, which is in the class 30-40 (see the row of cumulative frequency 95, which contains 80).

Hence median class is 30-40.

∴ 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

∴ Required median

$= l + \frac{\frac{N}{2} - C}{f} \times i = 30 + \frac{\frac{159}{2} - 50}{45} \times 10 = 30 + \frac{295}{45} = 36.55$.