Question

The median of $125$ observation for the given frequency distribution is $22.12$. Find missing frequencies $f_1$ and $f_2$.

Class	$0-4$	$5-9$	$10-14$	$15-19$	$20-24$	$25-29$	$30-34$	$35-39$	$40-44$
Frequency	$3$	$8$	$12$	$f_1$	$35$	$21$	$f_2$	$6$	$2$

Answer 1

Class	$0-4$	$5-9$	$10-14$	$15-19$	$20-24$	$25-29$	$30-34$	$35-39$	$40-44$
Frequency	$3$	$8$	$12$	$f_1$	$35$	$21$	$f_2$	$6$	$2$
Cumulative Frequency	$3$	$11$	$23$	$23+f_1$	$58+f_1$	$79+f_1$	$79+f_1+f_2$	$85+f_1+f_2$	$87+f_1+f_2$

Given: Median $=22.12$

Number of observations $n=125$

As median is 22.12 which is between $20-24$ so this class is the median class.

$l=$ lower limit of the median class $=20$

$h=$ size of class interval $=4$

$\frac{n}{2}=\frac{125}{2}$

$f=$ frequency of median class$=35$

$F=$ cumulative frequency of the class preceding the median class $=23+f_1$

Now $87+f_1+f_2=125$

$f_1+f_2=38$ ....... $\left(1\right)$

$Median=l+\left(\frac{\frac{n}{2}-F}{f}\right)\times h$

$22.12=20+\left(\frac{\frac{125}{2}-23-f_1}{35}\right)\times 4$

$2.12=\frac{39.5-f_1}{35}\times 4$

$0.53=\frac{39.5-f_1}{35}$

$18.55=39-f_1$

$f_1=20.95$ .... $\left(2\right)$

Using $\left(2\right)$ in $\left(1\right)$

$20.95+f_2=38$

$f_2=17.05$

Hence $f_1=20.95$ and $f_2=17.05$