Question

The following table gives the daily income of $50$ workers of a factory:

Daily income (in Rs.)	$100-120$	$120-140$	$140-160$	$160-180$	$180-200$
Number of workers:	$12$	$14$	$8$	$6$	$10$

Find the mean, mode and median of the above data

Answer 1

$Class$	$Frequency$ $f_i$	$Midvalue$ $x_i$	$f_i{x}_i$	$cf$
$100-200$	$12$	$110$	$1320$	$12$
$120-140$	$14$	$130$	$1820$	$26$
$140-160$	$8$	$150$	$1200$	$34$
$160-180$	$6$	$170$	$1020$	$40$
$180-200$	$10$	$190$	$1900$	$50$
	$\sum f_i=50$		$\sum f_i{x}_i=7260$

$\Rightarrow$ $Mean=\frac{\sum f_i{x}_i}{f_i}=\frac{7260}{50}=145.2$

$\Rightarrow$ We have, $N=50$. Then, $\frac{N}{2}=25$.

$\Rightarrow$ Median class is $120-140$

$l=$ 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.

$\Rightarrow$ So, $l=120,h=20,N=50cf=12,f=14$.

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

$\Rightarrow$ $Median=120+\frac{25-12}{14}\times 20$

$\Rightarrow$ $Median=120+\frac{130}{7}$

$\therefore$ $Median=120+18.6=138.6$

$\Rightarrow$ Modal class $=120-140$

$\Rightarrow$ $l=120,f=14f_1=12,f_2=8,h=20$

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

$\Rightarrow$ $Mode=120+\frac{14-12}{2\times 14-12-8}\times 20$

$\Rightarrow$ $Mode=120+\frac{40}{8}$

$\therefore$ $Mode=125$