Question

If the median of the following frequency distribution is $32.5$, find the missing frequencies.

Class interval	Frequency
0-10	$f_1$
10-20	5
20-30	9
30-40	12
40-50	$f_2$
50-60	3
60-70	2
Total	40

• A. $3,6$
• B. $4,9$
• C. $2,8$
• D. None of these

Answer 1

The correct option is A $3,6$

Frequency distribution table:

$ci$	$f_i$	$cf$
$0-10$	$f_1$	$f_1$
$10-20$	$5$	$5+f_1$
$20-30$	$9$	$14+f_1$
$30-40$	$12$	$26+f_1$
$40-50$	$f_2$	$26+f_1+f_2$
$50-60$	$3$	$29+f_1+f_2$
$60-70$	$2$	$31+f_1+f_2$

Median $=l+\frac{(\frac{N}{2}-cf)}{f_m}\times c$

Here,

Class Interval $c=10$

$N=\Sigma f_i=31+f_1+f_2$

$N=\Sigma f_i=40$ ....(given)

$\therefore 40=31+f_1+f_2$

$\therefore f_2=9-f_1$ ....(1)

Median $=32.5$

$\Rightarrow$ Median class $=30-40$

$\Rightarrow$Frequency of median class$f_m=12$

Lower boundary of median class $l=30$

previous cumulative frequency of median class $cf=14+f_1$

$\therefore 32.5$ $=30+\frac{(\frac{31+f_1+f_2}{2}-(14+f_1))}{12}\times 10$

$\therefore 32.5$ $=30+\frac{(3-f_1+f_2)}{12}\times 5$

$\therefore 32.5$ $=\frac{360+15-5f_1+5f_2}{12}$

$\therefore 390-375=-5f_1+5f_2$

$\therefore -5f_1+5f_2=15$

$\therefore -f_1+f_2=3$ ....(2)

substituting (1) in (2), we have

$-f_1+(9-f_1)=3$

$-2f_1=-6$

$\therefore f_1=3$

substituting in $f_1=3$ in (2), we get

$-3+f_2=3$

$\therefore f_2=6$

Hence, option $A$ is correct.