Question

The height of $30$ boys of a class are given in the following table

Height in cm	frequency
$120-129$	$2$
$130-139$	$8$
$140-149$	$10$
$150-159$	$7$
$160-169$	$3$

If by joining of a boy of height $140$ cm, the median of the heights is changed from $M_1$ to $M_2$, then $M_1-M_2$ in cm is

• A. $0.1$
• B. $-0.1$
• C. $0$
• D. $0.2$

Answer 1

Answer: (C)

$0$

$Heightincm$	$Frequency$	$cf$
$119.5-129.5$	$2$	$2$
$129.5-139.5$	$8$	$10$
$139.5-149.5$	$10$	$20$
$149.5-159.5$	$7$	$27$
$159.5-169.5$	$3$	$30$

Here, $n=30$ then, $\frac{n}{2}=\frac{30}{2}=15$

$\therefore$ Median class $=139.5-149.5$

$l=139.5,c.f=10,f=10,h=10$

$Median=l+\frac{\frac{n}{2}-cf}{f}\times h$

$=139.5+\frac{15-10}{10}\times 10$

$=139.5+5$

$=144.5$

$\therefore$ $Median\left(M_1\right)=144.5$

$\Rightarrow$ After adding height of boy $140cm$, total number of boys becomes $31$

$\therefore$ $n=31$ and $\frac{n}{2}=\frac{31}{2}=15.5$

Median class $=139.5-149.5$

All the values remain same except $f=11$ and $\frac{n}{2}=15.5$

$Median\left(M_2\right)=139.5+\frac{15.5-10}{11}\times 10$

$=139.5+5$

$=144.5$

$\therefore$ $Median\left(M_2\right)=144.5$

$\Rightarrow$ $M_1-M_2=144.5-144.5=0$