JAMB 1993 · UME · Q47

Find the median of the observations in the table below.

Class	1–5	6–10	11–15	16–20	21–25	26–30	31–35	36–40
Frequency	2	4	5	2	3	2	1	1

Worked solution (try it first)

The cumulative frequencies are 2, 6, 11, 13, 16, 18, 19, 20, so $N = 20$ and the median is the $\frac{N}{2} = 10$ th value.
The 10th value lies in the class 11–15, where the running total goes from 6 to 11.
Its lower class boundary is 10.5, its frequency is 5 and its width is 5.
Median $= 10.5 + \frac{10 - 6}{5} \times 5$ , which is $10.5 + 4 = 14.5$ , option D.