Question

The age distribution of $400$ persons in a colony having median age $32$ is given below

Age (in Years)	$20-25$	$25-30$	$30-35$	$35-40$	$40-45$	$45-50$
Frequency	$110$	$x$	$75$	$55$	$y$	$30$

Then, $(x-y)$ is :

• A. $10$
• B. $20$
• C. $-10$
• D. $-20$

Answer 1

The correct option is C $-10$

The cumulative frequency distribution table is

Age	No. of persons	Cumulative Frequency
$20-25$	$110$	$110$
$25-30$	$x$	$110+x$
$30-35$	$75$	$185+x$
$35-40$	$55$	$240+x$
$40-45$	$y$	$240+x+y$
$45-50$	$30$	$270+x+y$

$N=270+x+y$

Clearly, $270+x+y=400\Rightarrow x+y=130$
It is given that $32$ is the median. So $30-35$ is the median class such that
$l=30,h=5,f=75,N=400$ and $F=110+x$
$\therefore$ Median = $l+\frac{\frac{N}{2}-F}{f}\times h\Rightarrow 32=30+\frac{200-(110+x)}{75}\times 5$

$\Rightarrow 2=\frac{90-x}{15}\Rightarrow x=60$

$\because x+y=130\Rightarrow y=70$

Hence, $x-y=-10$