Question

Find the missing values where median and mode of the following distribution is given 33.5 and 34 , respectively, choose which of the following is a correct pair.

Class Internal		Frequency
	0-10		4
	10-20		16
	20-30		?
	30-40		?
	40-50		?
	50-60		6
	60-70		4

Column - I		Column - II
I	$I^{\text {st }}$ frequency	P	50
II	II $^{\text {nd }}$ frequency	Q	68
III	III $^{\text {rd }}$ frequency	R	70
		S	40
		T	100

• A. $I - Q , II - S$
• B. $I - Q , III - S$
• C. III-R, III-P
• D. III-T, III $- S$

Answer 1

Answer: (D)

Let missing frequency $(20-30)=x$
Missing frequency $(30-40)=y$
And missing frequency $(40-50)=230-(4+$ $16+(x+y+6+4)$ $=200-x-y$

$\text { Median }=l_1+\frac{\frac{N}{2}-f_0}{f} \times(l_2-l_1) $
$ 33.5=30+\frac{115-(20+x)}{y} \times(40-30) $
$y(33.5-30)=(11.5-20-x) 10 $
$ 3.5 y=1150-200-10 x $
$ 10 x+3.5 y=950 $
$ \text { Mode }=l_1+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times(l_2-l_1)$
$34=20+\frac{y-x}{2 y-2(-200-x-y)} \times(30-20) $
$4(3 y-200)=10(y-x)$
$10 x+2 y=800$
Subtracting equation (ii) from equation (i), we get
$1.5 y=150 $
$\Rightarrow y=100$
Equality value of $y$ in eq(l), we get
$10 x+3.5(100)=950 $
$10 x=950-350 $
$\Rightarrow x=60$
So, third frequency $=200-x-y=200-60$ $-100=40$