Question

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units):	$65-85$	$85-105$	$105-125$	$125-145$	$145-165$	$165-185$	$185-205$
No. of consumers	$4$	$5$	$13$	$20$	$14$	$8$	$4$

Answer 1

$Classinterval$	$x_i$	$f_i$	$f_i{x}_i$	$Cumulativefrequency$
$65-85$	$75$	$4$	$300$	$4$
$58-105$	$95$	$5$	$475$	$9$
$105-125$	$115$	$13$	$1495$	$22$
$125-145$	$135$	$20$	$2700$	$42$
$145-165$	$155$	$14$	$2170$	$56$
$165-185$	$175$	$8$	$1400$	$64$
$185-205$	$195$	$4$	$780$	$68$
		$\sum f_i=68$	$\sum f_i{x}_i=9320$

$\Rightarrow$ $Mean=\frac{\sum f_i{x}_i}{f_i}=\frac{9320}{68}=137.05$

$l=$ lower limit of the modal class

$h=$ size of the class intervals

$f=$ frequency of the modal class

$f_1=$ frequency of the class preceding the modal class

$f_2=$ frequency of the class succeed in the modal class.

We have, $N=68$ and $\frac{N}{2}=34$

The cumulative frequency just greater than $\frac{N}{2}$ is $42$ then the median class is $125–145$ such that,

$l=125,h=145–125=20,f=20,cf=22$

$Median=l+\frac{\frac{N}{2}-cf}{f}\times h$

$\Rightarrow$ $Median=125+\frac{34-22}{20}\times 20$

$\Rightarrow$ $Median=125+12=137$

We know,

$Mode=l+\frac{f-f_1}{2f-f_1-f_2}\times h$

Here the maximum frequency is 20, then the corresponding class $125–145$ is the modal class

$l=125,h=145–125=20,f=20,f_1=13,f_2=14$

$\Rightarrow$ $Mode=125+\frac{140}{13}=135.77$

Now, after comparing all three values, we get following relation,

$Mean>Median>Mode$