Question

The following frequency distribution gives the monthly consumption of electricity of $70$ customers of a certain locality:

Monthly consumptions of Electricity (in units)	Frequency No. of consumers
$65-85$	$4$
$85-105$	$5$
$105-125$	$13$
$125-145$	$22$
$145-165$	$14$
$165-185$	$8$
$185-205$	$4$

Hence find mean and mode

Answer 1

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

$\therefore$ Mean$\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}$
$=\frac{9590}{70}$
$=137$
$\therefore$ Mean of monthly consumption of electricity is $137$ units.
Mode:

Original Class	True Class Boundaries	Frequency
$65-85$	$64.5-84.5$	$4$
$85-105$	$84.5-104.5$	$5$
$105-125$	$104.5-124.5$	$13\to f_1$
$125-145$	$124.5-144.5$	$22\to f_m$
$145-165$	$144.5-164.5$	$14\to f_2$
$165-185$	$164.5-184.5$	$8$
$185-205$	$184.5-204.5$	$4$

Here, maximum frequency $f_m=22$.
The corresponding class namely, $124.5-144.5$ is 'modal' class.
Mode $=L+\left[\frac{f_m-f_1}{2f_m-f_1-f_2}\right]h$
Where $L:$ lower limit of modal class $=124.5$
$f_m:$ maximum frequency $=22$
$f_1:$ frequency of premodal class $=13$
$f_2:$ frequency of postmodal clas $=14$
$h:$ width of modal class $=20$
$\therefore$ Mode $=L+\left[\frac{f_m-f_1}{2f_m-f_1-f_2}\right]h$
$=124.5+\left[\frac{22-13}{2\left(22\right)-13-14}\right]20$
$=124.5+\left[\frac{9}{17}\right]20$

$=135.09$