Question

The median for the following data is $1,100$. Find the value of P.

X	$100-200$	$200-300$	$300-400$
F	$10$	P	$16$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

X	Frequencies(f)	Cumulative frequency(cf)
$100-200$	$10$	$10$
$200-300$	P	$10+P$
$300-400$	$16$	$26+P$
	$n=\Sigma f=26+P$ $\frac{n}{2}=\frac{26+P}{2}$

Median $=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$
$n=\Sigma f=$ Total number of frequency $=26+P$
$\frac{n}{2}=\frac{26+P}{2}$
So, the median class is $200-300$
L$=$ Lower limit of the median class $=200$
cf$=$ Cumulative frequency of the class preceding the median class $=10$
f$=$ Frequency of the median class $=P$
h$=$ class width$=100$
Therefore, Median $=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$
$1,100=200+\left(\frac{\frac{26+P}{2}-10}{P}\right)\times 100$
$1,100=200+\left(\frac{6+p}{2p}\right)\times 100$
$1,100=400p+600+100p$
$1,100-600=500p$
$P=500/500$
$P=1$