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)

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 $\frac{1}{2}(n+1)=\frac{1}{2}(159+1)=80th$ 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 $=l+\frac{\frac{N}{2}-C}{f}\times i=30+\frac{\frac{159}{2}-50}{45}\times 10$
$=30+\frac{295}{45}=36.55$