Question

Find out median of the following data-set:

Class Interval	60−69	50−59	40−49	30−39	20−29	10−19
Frequency	13	15	21	20	19	12

Answer 1

This is an inclusive series given in the descending order. It should first be converted into an exclusive series and placed in the ascending order.

Class Interval	Frequency	Cumulative Frequency
9.5 − 19.5 19.5 − 29.5	12 19	12 12 + 19 = 31 (c.f.)
(l₁)29.5 − 39.5	20 (f)	31 + 20 = 51
39.5 − 49.5 49.5 − 59.5 59.5 − 69.5	21 15 13	51 + 21 = 72 72 + 15 = 87 87 + 13 = 100
	N = 100

Median = Size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item
= Size of 50^th item

Hence, median lies in the class interval 29.5 − 39.5
$$\begin{gathered} \mathrm{Median}\mathrm{or}\mathrm{M}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{or},\mathrm{M}=29.5+\frac{50-31}{20}\times 10 \\ \mathrm{or},\mathrm{M}=29.5+\frac{19}{20}\times 10 \\ \mathrm{or},\mathrm{M}=29.5+\frac{190}{20} \\ \mathrm{or},\mathrm{M}=29.5+9.5 \\ \Rightarrow \mathrm{M}=39 \end{gathered}$$

Hence, the median value of the above series is 39.