Question

If the median of the following frequency distribution is $46$. find the absolute difference of missing frequencies

Variable	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	Total
Frequency	$12$	$30$	$f_1$	$65$	$f_2$	$25$	$18$	$229$

Answer 1

Class Interval	Frequency	C.F.
$10-20$	$12$	$12$
$20-30$	$30$	$42$
$30-40$	$f_1$	$42+f_1$
$40-50$	$65$	$107+f_1$
$50-60$	$f_2$	$107+f_1+f_2$
$60-70$	$25$	$132+f_1+f_2$
$70-80$	$18$	$150+f_1+f_2$

Let the frequency of the class $30-40$ be $f_1$ and that of the class $50-60$ be $f_2$. The total frequency is $229$.
$12+30+$ $f_1$ $+65+$ $f_2$ $+25+18=229$
$\Rightarrow f_1+f_2=79$
It is given that median is $46$. Clearly $46$ lies in the class $40-50$. So, $40-50$ is the median class
$\therefore L=40,h=10,f=65$ and $C=42+f_1,N=229$
Median $=$ $L+\frac{\frac{N}{2}-C}{f}\times h$
$\Rightarrow 46=40+\frac{\frac{229}{2}-(42+f_1)}{65}\times 10$
$\Rightarrow 46=40+\frac{145-2f_1}{13}$
$\Rightarrow 6=\frac{145-2f_1}{13}\Rightarrow 2f_1=67$
$\Rightarrow f_1=33.5$ or $34$
Since $f_1+f_2=79$
$\Rightarrow f_2=45$
Hence $f_1=34$ and $f_2=45$

Hence, the difference between frequenncies is $11$.