The median of the following data is 16. Find the missing frequencies a and b if the Total of frequencies is 70. [CBSE 2013]
Class |
0 - 5 |
5 - 10 |
10 - 15 |
15 - 20 |
20 - 25 |
25 - 30 |
30 - 35 |
35 - 40 |
Frequency |
12 |
A |
12 |
15 |
b |
6 |
6 |
4 |
Given: Median = 16 & N = 70
Assume
Σfi = N = Sum of frequencies,
h = length of median class,
l = lower boundary of the median class,
f = frequency of median class
and Cf = cumulative frequency
Lets form a table, where x is the unknown frequency.
CLASS |
FREQUENCY(fi) |
Cf |
0 - 5 |
12 |
12 |
5 - 10 |
a |
12 + a |
10 - 15 |
12 |
12 + a + 12 = 24 + a |
15 - 20 |
15 |
24 + a + 15 = 39 + a |
20 - 25 |
b |
39 + a + b |
25 - 30 |
6 |
39 + a + b + 6 = 45 + a + b |
30 - 35 |
6 |
45 + a + b + 6 = 51 + a + b |
35 - 40 |
4 |
51 + a + b + 4 = 55 + a + b |
TOTAL |
55 + a + b |
Median = 16 (as already mentioned in the question)
16 lies between 15 - 20 ⇒ Median class = 15 - 20
∴ l = 15, h = 5, f = 15, N/2 = (55 + a + b)/2 and Cf = 24 + a
Median is given by,

⇒ 
⇒ ![]()
⇒ 16 – 15 = (7 – a + b)/6
⇒ 6 = 7 – a + b
⇒ a – b = 1 …(i)
And given that N = 70
⇒ 55 + a + b = 70
⇒ a + b = 15 …(ii)
Solving equations (i) & (ii), we get
(a – b) + (a + b) = 1 + 15
⇒ 2a = 16
⇒ a = 8
Substituting a = 8 in eq.(i),
8 – b = 1
⇒ b = 7
Thus, the unknown frequencies are a = 8 and b = 7.
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