Question

The following frequency distribution gives the monthly consumption of electricity of $70$ consumers of a locality.

Monthly Consumption(Units)	No. of Consumers
$65-85$	$04$
$85-105$	$05$
$105-125$	$13$
$125-145$	$22$
$145-165$	$14$
$265-185$	$08$
$185-205$	$04$

Find its mean and median.

Answer 1

Mean:

Monthly consumption (units)\Class interval	Class mark $\left(x_i\right)$	No. of consumers Frequency ($f_i$)	$f_i{x}_i$
$65-85$	$75$	$04$	$300$
$85-105$	$95$	$05$	$475$
$105-125$	$115$	$13$	$1495$
$125-145$	$135$	$22$	$2970$
$145-165$	$155$	$14$	$2170$
$165-185$	$175$	$08$	$1400$
$185-205$	$195$	$04$	$780$
Total		$\sum f_i=70$	$\sum f_i{x}_i=9590$

$\therefore$ $Mean=\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{9590}{70}$
$=137$
$\therefore$ Mean of monthly consumption of electricity is $137$ units.

Monthly consumption (Units)	Frequency $\left(f_i\right)$	Cum. Frequency (less than type)
$65-85$	$04$	$04$
$85-105$	$05$	$09$
$105-125$	$13$	$22\to c.f$
$125-145$ Median class	$22\to f$	$44$
$145-165$	$14$	$58$
$165-185$	$08$	$66$
$185-205$	$04$	$70$

Here, total frequency $=\sum f_i=N=70$
$\therefore$ $\frac{N}{2}=\frac{70}{2}$
$=35$
Cumulative frequency is just greater than or equal to $35$ is $44$.
$\therefore$ Corresponding class, namely, $125-145$ is median class.
$\therefore$ Median $=L+(\frac{N}{-}c.f)\frac{h}{f}$
Here, $L=125,c.f=22,h=20,f=22$
$\therefore$ Median $=125+(35-22)\frac{20}{22}$
$=125+\left(13\right)\frac{10}{11}$
$=125+\frac{130}{11}$
$=125+11.8$
$=136.8$
$\therefore$ Median monthly consumption of electricity $=136.8$ units.