Question

Find out median , first quartile and third quartile of the following series:

Height (in inches)	58	59	60	61	62	63	64	65	66
No of Persons	2	3	6	15	10	5	4	3	1

Answer 1

Height (X)	Persons (f)	c.f
58	2	2
59	3	5
60	6	11
61	15	26
62	10	36
63	5	41
64	4	45
65	3	48
66	1	49

(i)Median
$$\begin{gathered} \mathrm{Median}=\mathrm{Size}\mathrm{of}{\left(\frac{N+1}{2}\right)}^{\mathrm{th}}\mathrm{item} \\ =\mathrm{Size}\mathrm{of}{\left(\frac{49+1}{2}\right)}^{\mathrm{th}}={25}^{\mathrm{th}}\mathrm{item} \end{gathered}$$
This corresponds to height 61. Thus, the median is 61.

(ii) First Quartile
$$\begin{gathered} Q_1=\mathrm{Size}\mathrm{of}{\left(\frac{N+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ =\mathrm{Size}\mathrm{of}{\left(\frac{49+1}{4}\right)}^{\mathrm{th}}\mathrm{item}=12.5^{\mathrm{th}}\mathrm{item} \end{gathered}$$
This corresponds to height 61. Thus, first quartile is 61.

(iii) Third Quartile
$$\begin{gathered} Q_3=\mathrm{Size}\mathrm{of}3{\left(\frac{N+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ =\mathrm{Size}\mathrm{of}3{\left(\frac{49+1}{4}\right)}^{\mathrm{th}}=37.5^{\mathrm{th}}\mathrm{item} \end{gathered}$$
This corresponds to height 63. Thus, third quartile is 63.