Question

The median of the following data in $525$. Find the missing frequency, if it is given that there are $100$ observations in the data:

Class interval	Frequency	Class interval	Frequency
0-100	$2$	500-600	$20$
100-200	$5$	600-700	$f_2$
200-300	$f_1$	700-800	$9$
300-400	$12$	800-900	$7$
400-500	$17$	900-1000	$4$

Answer 1

Class interval	Frequency	Cumulative frequency
$0-100$	$2$	$2$
$100-200$	$5$	$7$
$200-300$	$f_1$	$7+f_1$
$300-400$	$12$	$19+f_1$
$400-500$	$17$	$36+f_1$
$500-600$	$20$	$56+f_1$
$600-700$	$f_2$	$56+f_1+f_2$
$700-800$	$9$	$65+f_1+f_2$
$800-900$	$7$	$72+f_1+f_2$
$900-1000$	$4$	$76+f_1+f_2$
	$N=100$

$\Rightarrow$ $76+f_1+f_2=100$

$\Rightarrow$ $f_1+f_2=24$ -------- ( 1 )

$Median=525$ [ Given ]

Then, median class $=500-600$

Here, $N=100$ then, $\frac{N}{2}=50$.

$l=500,f=20,cf=36+f_1$ and $h=600-500=100$

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

$\Rightarrow$ $525=500+\frac{50-(36+f_1)}{20}\times 100$

$\Rightarrow$ $525-500=\frac{50-36-f_1}{20}\times 100$

$\Rightarrow$ $25=(14-f_1)\times 5$

$\Rightarrow$ $25=70-5f_1$

$\therefore$ $5f_1=45$

$\therefore$ $f_1=\frac{45}{5}=9$

Substituting in equation ( 1 ),

$\Rightarrow$ $9+f_2=24$

$\therefore$ $f_2=15$.

$\therefore$ $f_1=9,f_2=15$.