Question

Calculate the median, lower quartile and upper quartile from the following data:

Marks	58	59	60	61	62	63	64	65	66
No. of Students	2	3	6	15	10	5	4	3	1

Answer 1

Marks	Number of students (f)	Cumulative Frequency (c.f.)
58 59 60 61 62 63 64 65 66	2 3 6 15 10 5 4 3 1	2 5 11 26 36 41 45 48 49
	N=49

Median

Median = size of ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, Median $=\frac{49+1}{2}={25}^{\mathrm{th}}\mathrm{item}$
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25^th is 26 (in the c.f. column), which is corresponding to 61. ​
Hence, median is 61.

Lower Quartile

$$\begin{gathered} Q_1=\mathrm{Size}\mathrm{of}{\left(\frac{\mathrm{N}+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},{\mathrm{Q}}_1=\mathrm{Size}\mathrm{of}{\left(\frac{49+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \Rightarrow {\mathrm{Q}}_1=12.5^{\mathrm{th}}\mathrm{item} \end{gathered}$$
This corresponds to 26 in the cumulative frequency.
Hence, lower quartile is 61 marks.

Upper Quartile

$$\begin{gathered} {Q}_3=\mathrm{Size}\mathrm{of}3{\left(\frac{\mathrm{N}+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \mathrm{or},{\mathrm{Q}}_3=\mathrm{Size}\mathrm{of}3{\left(\frac{49+1}{4}\right)}^{\mathrm{th}}\mathrm{item} \\ \Rightarrow {\mathrm{Q}}_3=37.5^{\mathrm{th}}\mathrm{item} \end{gathered}$$

This corresponds to 41 in the cumulative frequency.
Hence, upper quartile is 63 marks.