Question

Class Interval	$0-100$	$100-200$	$200-300$	$300-400$	$400-500$	$500-600$	$600-700$	$700-800$	$800-900$	$800-900$
Frequency	$2$	$5$	$x$	$12$	$17$	$20$	$y$	$9$	$7$	$4$

The median of the following data is $525$.

Find the value of $x$ and $y$, if the total frequency is $100$.

Answer 1

Find the value of $x$and $y$:

Given,

Class Interval	$0-100$	$100-200$	$200-300$	$300-400$	$400-500$	$500-600$	$600-700$	$700-800$	$800-900$	$900-1000$
Frequency	$2$	$5$	$x$	$12$	$17$	$20$	$y$	$9$	$7$	$4$

The value of the Median is $525$

The Median Class is$500–600$

Class Interval	Frequency(f)	Cumulative Frequency (cf)
$0-100$	$2$	$2$
$100-200$	$5$	$2+5$
$200-300$	$x$	$2+5+x$
$300-400$	$12$	$2+5+x+12$
$400-500$	$17$	$2+5+x+12+17$
$500-600$	$20$	$2+5+x+12+17+20$
$600-700$	$y$	$2+5+x+12+17+20+y$
$700-800$	$9$	$2+5+x+12+17+20+y+9$
$800-900$	$7$	$2+5+x+12+17+20+y+9+7$
$900-1000$	$4$	$2+5+x+12+17+20+y+9+7+4$

$$\begin{gathered} 76+x+y=100 \\ \Rightarrow x+y=24\dots \left(i\right) \end{gathered}$$

$\mathrm{Median}=l+\left[\frac{\left(\frac{N}{2}–cf\right)}{f}\right]\times h$

Since,$l=500,h=100,f=20,cf=36+x$ and$N=100$

Consequently, when using the Median formula with the value,

we obtain:

$$\begin{gathered} \mathrm{Median}=l+\left[\frac{\left(\frac{N}{2}–cf\right)}{f}\right]\times h \\ \Rightarrow 525=500+\left[\frac{\left(\frac{100}{2}–\left(36+x\right)\right)}{20}\right]\times 100 \\ \Rightarrow 525=500+\left[\frac{\left(50–36-x\right)}{20}\right]\times 100 \\ \Rightarrow 525-500=\left[\frac{\left(14-x\right)}{20}\right]\times 100 \\ \Rightarrow 25=\left(14-x\right)\times 5 \\ \Rightarrow 5=\left(14-x\right) \\ \Rightarrow x=14-5 \\ \therefore x=9 \end{gathered}$$

Put $x=9$ in (i),

$$\begin{gathered} 9+y=24 \\ \Rightarrow y=24-9 \\ \therefore y=15 \end{gathered}$$

Hence, the value of $x=9$ and $y=15$.