Question

The distribution of profit per day of some shops in a certain town is shown below.

Profit (in Rs)	500-900	1000-1400	1500-1900	2000-2400	2500-2900	3000-3400	3500-3900
No. of shops	8	18	27	21	20	18	8

Calculate the median profit of the shops.

Answer 1

Here, the class intervals are inclusive type. For the calculation of median, we have to convert
the class intervals to continuous form by subtracting 50 from the lower limit and adding 50 to the upper limit
of each interval.Then, we have to prepare the cumulative frequency distribution (less than type) table
as shown below.

Original Class	Class Boundaries	Frequency $\left(f_i\right)$	Cumulative Frequency (less than type)
500-900	450-950	8	8
1000-1400	950-1450	18	26
1500-1900	1450-1950	27	$53\to c.f.$
2000-2400	1950-2450	$21\to f$	74
2500-2900	2450-2950	20	94
3000-3400	2950-3450	18	112
3500-3900	3450-3950	8	120

$$\begin{gathered} \mathrm{Total}\mathrm{frequency}=\sum _{}^{}{f}_{i_{}}=N=120 \\ \therefore \frac{N}{2}=\frac{120}{2}=60 \\ \mathrm{So},\mathrm{cumulative}\mathrm{frequency}\mathrm{should}\mathrm{be}\mathrm{just}\mathrm{greater}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}60,\mathrm{i}.\mathrm{e}.74. \\ \mathrm{Therefore},\mathrm{the}\mathrm{median}\mathrm{class}\mathrm{will}\mathrm{be}1950-2450. \\ \mathrm{Median}=L+\left(\frac{N}{2}-c.f.\right)\frac{h}{f} \\ \mathrm{Here},\mathrm{L}=\mathrm{lower}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=1950 \\ c.f.=\mathrm{cumulative}\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{class}\mathrm{preceding}\mathrm{the}\mathrm{median}\mathrm{class}=53 \\ h=\mathrm{width}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=500 \\ f=\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=21 \\ \therefore \mathrm{Median}=1950+(60-53)\frac{500}{21} \\ =1950+7\times \frac{500}{21} \\ =1950+\frac{500}{3} \\ =1950+166.67 \\ =2116.67\approx 2117 \\ \therefore \mathrm{Median}\mathrm{profit}\mathrm{of}\text{the}\mathrm{shops}\mathrm{is}\mathrm{Rs}2117. \end{gathered}$$