Question

The following table shows the frequency distribution of number of rooms occupied in a hotel per day.

No. of rooms occupied	0-10	10-20	20-30	30-40	40-50	50-60
No. of days	5	27	17	11	9	1

Find the median number of rooms occupied per day in a hotel.

Answer 1

For the calculation of median, we need to prepare the cumulative frequency distribution (less than type)
table as shown:

No. of rooms occupied	Frequency of no. of days $\left(f_i\right)$	Cumulative frequency (less than type)
0-10	5	5
10-20	27	32
20-30	17	49
30-40	11	60
40-50	9	69
50-60	1	70

$$\begin{gathered} \mathrm{Total}\mathrm{frequency}=\sum _{}^{}{f}_i=N=70 \\ \therefore \frac{N}{2}=\frac{70}{2}=35 \\ \mathrm{So},\mathrm{cumulative}\mathrm{frequency}\mathrm{should}\mathrm{be}\mathrm{just}\mathrm{greater}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}35,\mathrm{i}.\mathrm{e}.49. \\ \mathrm{Therefore},\mathrm{the}\mathrm{median}\mathrm{class}\mathrm{will}\mathrm{be}20-30. \\ \mathrm{Median}=L+\left(\frac{N}{2}-c.f.\right)\frac{h}{f} \\ \mathrm{Here},L=\mathrm{lower}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=20 \\ c.f.=\mathrm{cumulative}\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{class}\mathrm{preceeding}\mathrm{the}\mathrm{median}\mathrm{clas}s=32 \\ h=\mathrm{width}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=10 \\ f=\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=17 \\ \therefore \mathrm{Median}=20+\left(35-32\right)\frac{10}{17} \\ =20+3\times 0.58 \\ =20+1.76 \\ =21.76 \\ \therefore \mathrm{Median}\text{number}\mathrm{of}\mathrm{rooms}\mathrm{occupied}\mathrm{per}\mathrm{day}\mathrm{is}21.76. \end{gathered}$$