Question

The following table shows the distribution of monthly expenditure on electricity of households in a certain village.

Monthly Expenditure	150-225	225-300	300-375	375-450	450-525	525-600	600 and above
No. of households	65	171	196	75	53	26	14

Find the median expenditure on electricity of households per month.

Answer 1

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

Monthly expenditure	Frequency	Cumulative frequency (less than type)
150-225	65	65
225-300	171	$236\to c.f.$
300-375	$196\to f$	432
375-450	75	507
450-525	53	560
525-600	26	586
600 and above	14	600

$$\begin{gathered} \mathrm{Total}\mathrm{frequency}=\sum _{}^{}{f}_i=N=600 \\ \therefore \frac{N}{2}=\frac{600}{2}=300 \\ \mathrm{So},\text{the}\mathrm{cumulative}\mathrm{frequency}\mathrm{should}\mathrm{be}\mathrm{just}\mathrm{greater}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}300,\mathrm{i}.\mathrm{e}.432. \\ \mathrm{Therefore},\mathrm{the}\mathrm{median}\mathrm{class}\mathrm{will}\mathrm{be}300-375. \\ \mathrm{Median}=L+\left(\frac{N}{2}-c.f.\right)\frac{h}{f} \\ \mathrm{Here},L=\mathrm{lower}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=300 \\ c.f.=\mathrm{cumulative}\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{class}\mathrm{preceding}\mathrm{the}\mathrm{median}\mathrm{class}=236 \\ h=\mathrm{width}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=75 \\ f=\mathrm{frequency}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{class}=196 \\ \therefore \mathrm{Median}=300+(300-236)\frac{75}{196} \\ =300+64\times \frac{75}{196} \\ =300+16\times \frac{75}{49} \\ =300+24.49 \\ =324.49 \\ \therefore \mathrm{Median}\mathrm{expenditure}\text{of}\mathrm{households}\mathrm{on}\mathrm{electricity}\mathrm{per}\mathrm{month}\mathrm{is}\mathrm{Rs}324.49. \end{gathered}$$