Question

Find the missing frequencies $f_1$ and $f_2$ in the following frequency distribution table.$N=100$ and median is $32$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$Total$
Frequency	$10$	$f_1$	$25$	$30$	$f_2$	$10$	$100$

Answer 1

Given,

$N=100$ and median $=32$

also given, to find out the frequencies $f_1$ and $f_2$ for the given frequency table.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	Total
Frequency	$10$	$f_1$	$25$	$30$	$f_2$	$10$	$100$

Let us construct a table to find out frequencies $f_1$ and $f_2$.

Class	frequency	cumulative frequency
$0-10$	$10$	$10$
$10-20$	$f_1$	$10+f_1$
$20-30$	$25$	$25+f_1$ (C.F)
$30-40$	$30$	$65+f_1$
$40-50$	$f_2$	$65+f_1+f_2$
$50-60$	$10$	$75+f_1+f_2$
	$20+25+30+f_1+f_2=75+f_1+f_2$

as there $6$ frequency distribution's let us Consider the highest middle '$30$'

as the frequency. $\therefore f=30$

$N=100\Rightarrow \frac{n}{2}=50$

class interval $\left(C\right)=10;l=30$

let,

$75+f_1+f_2=100$

we get, $f_1+f_2=100-75=25$

$\Rightarrow f_1+f_2=25$

as we know that,

Median $\left(m\right)=L+\left(\frac{\frac{n}{2}-C.F}{f}\right)\times C$

$\Rightarrow 32=30+\left(\frac{50-(35+f_1)}{30}\right)\times C$

$\Rightarrow 32=30+\left(\frac{50-35-f_1}{30}\right)\times 10$

$\Rightarrow 32=30+\frac{15-f_1}{3}$

$\Rightarrow 96=90+15-f_1$

$\Rightarrow 96=105-f_1\Rightarrow f_1=105-96=9$

Substitute $f_1$ in $[f_1+f_2=25]$ we get,

$9+f_2=25\Rightarrow f_2=25-9=16$.

$\therefore f_1=9;f_2=16$