Question

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequency x and y.$\begin{array}{ccccccc}RunsScored& 2500-3500& 3500-4500& 4500-5500& 5500-6500& 6500-7500& 7500-8500\\ No.ofBatsmen& 5& x& y& 12& 6& 2\end{array}$

Answer 1

Runs scored	Number of Bats men	Cumulative frequency
$2500-3500$	$5$	$5$
$3500-4500$	$x$	$5+x$
$4500-5500$	$y$	$5+x+y$
$5500-6500$	$12$	$17+x+y$
$6500-7500$	$6$	$23+x+y$
$7500-8500$	$2$	$25+x+y$

Now, $N=\sum f_i=60$

$\Rightarrow 60=25+x+y$

$\Rightarrow x+y=35$ .....$\left(1\right)$

Since the median is given to be $5000$, it lies in the class $4500-5500$

$\Rightarrow l=4500$,$h=1000$,$f=y$,$F=5+x$ and $N=60$

median$=l+\frac{\frac{N}{2}-F}{f}\times h$

$\Rightarrow 5000=4500+\frac{\frac{60}{2}-(5+x)}{y}\times 1000$

$\Rightarrow 500=\frac{30-5-x}{y}\times 1000$

$\Rightarrow 500y=(25-x)\times 1000$

$\Rightarrow \frac{y}{2}=25-x$

$\Rightarrow y=50-2x$

$\Rightarrow 2x+y=50$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get

$2x+(35-x)=50$

$\Rightarrow x=15$

Put $x=15$ in $\left(1\right)$, we get $y=20$

Hence, $x=15$ and $y=20$