Question

Find the missing frequency $f_1$ and $f_2$ in the table given below, it is being given that the mean of the given frequency distribution is $50$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	Total
Frequency	$17$	$f_1$	$32$	$f_2$	$19$	$120$

Answer 1

$Class$	$Class-Mark$ $x_i$	$Frequency$ $f_i$	$f_i{x}_i$
$0-20$	$10$	$17$	$170$
$20-40$	$30$	$f_1$	$30f_1$
$40-60$	$50$	$32$	$1600$
$60-80$	$70$	$f_2$	$70f_2$
$80-100$	$90$	$19$	$1710$
		$\sum f_i=f_1+f_2+68$	$\sum f_i{x}_i=3480+30f_1+70f_2$

$\sum f_i=120$ [ Given ]

$\sum f_i=f_1+f_2+68$

$120=f_1+f_2+68$

$\therefore$ $f_1+f_2=52$

$\Rightarrow$ $f_1=52-f_2$ ---- ( 1 )

$\Rightarrow$ $Mean\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}$

$\Rightarrow$ $50=\frac{3480+30f_1+70f_2}{120}$

$\Rightarrow$ $6000=3480+30(52-f_2)+70f_2$ [ From ( 1 ) ]

$\Rightarrow$ $2520=1560-30f_2+70f_2$

$\Rightarrow$ $40f_2=960$

$\therefore$ $f_2=24$

Substituting value of $f_2$ in ( 1 )

$f_1=52-24=28$

$\therefore$ $f_1=28,f_2=24$