Question

Calculate median from the following data-set:

Marks	Number of Students	Marks	Number of Students
1−5 6−10 11−15 16−20 21−25	7 10 16 32 24	26−30 31−35 36−40 41−45	18 10 5 1

Answer 1

Coverting the Inclusive series into exclusive series.

Marks	Frequency	Cumulative Frequency
0.5 − 5.5 5.5 − 10.5 10.5 − 15.5	7 10 16	7 7 + 10 = 17 17 + 16 = 33 (c.f.)
(l₁)15.5 − 20.5	32 (f)	33 + 32 = 65
20.5 − 25.5 25.5 − 30.5 30.5 − 35.5 35.5 − 40.5 40.5 − 45.5	24 18 10 5 1	65 + 24 = 89 89 + 18 = 107 107 + 10 = 117 117 + 5 = 122 122 + 1 = 123
	$\Sigma f=$N = 123

Median = Size of ${\left(\frac{\mathrm{N}}{2}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{123}{2}\right)}^{\mathrm{th}}$ item
= Size of 61.5^th item

Hence, the median lies in the class 15.5 − 20.5

$$\begin{gathered} \mathrm{Median}\mathrm{or}\mathrm{M}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f}{f}\times i \\ \mathrm{or},\mathrm{M}=15.5+\frac{61.5-33}{32}\times 5 \\ \mathrm{or},\mathrm{M}=15.5+\frac{28.5\times 5}{32} \\ \mathrm{or},\mathrm{M}=15.5+\frac{142.5}{32} \\ \mathrm{or},\mathrm{M}=15.5+4.45 \\ \Rightarrow \mathrm{M}=19.95\mathrm{marks} \end{gathered}$$

Hence, the median marks of the above data-set is 19.95