Question

If the median of the distribution given below is $28.5$, find the values of $x$ and $y$.

Class-interval	Frequency
$0 - 10$	$5$
$10 - 20$	$x$
$20 - 30$	$20$
$30 - 40$	$15$
$40 - 50$	$y$
$50 - 60$	$5$
$Total$	$60$

Answer 1

Hint: The formula used to find the median of a given data is given as follows: $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$ is the lower limit of median class, $n$ is the sum of all frequencies, $cf$ is the cumulative frequency before the median class, $f$ is the frequency of median class and $h$ is the size of median class.

Complete step-by-step answer:
The cumulative frequency for the given data is calculated as follows:

Class-interval	Frequency	Cumulative frequency$$
$0 - 10$	$5$	$5$
$10 - 20$	$x$	$5 + x$
$20 - 30$	$20$	$25 + x$
$30 - 40$	$15$	$40 + x$
$40 - 50$	$y$	$40 + x + y$
$50 - 60$	$5$	$45 + x + y$
	$n = \sum f $=$60$

Here, it is given that Median=$28.5$ and $n = \sum f $=$60$.
Here, $n = 60$$ \Rightarrow \dfrac{n}{2} = 30$
Since, the median of the given data is$28.5$, which lies in the interval $20 - 30$.
Therefore, median class=$20 - 30$
Lower limit of the median class, $l = 20$
Frequency of the median class, $f = 20$
Cumulative frequency of the class preceding the median class, $cf = 5 + x$
Class size, $h = 10$
Therefore, $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$28.5 = 20 + \left( {\dfrac{{30 - \left( {5 + x} \right)}}{{20}}} \right) \times 10$
$ \Rightarrow 28.5 - 20 = \left( {\dfrac{{30 - 5 - x}}{2}} \right)$
$ \Rightarrow 8.5 \times 2 = 25 - x$
$ \Rightarrow 17 = 25 - x$
$ \Rightarrow x = 25 - 17$
$ \Rightarrow x = 8$ ….. (1)
Also, we have $n = \sum f $=$60$
We know that the last term of cumulative frequency is always equal to the sum of all frequencies.
$\therefore 45 + x + y = 60$
$ \Rightarrow x + y = 60 - 45$
$ \Rightarrow x + y = 15$
$ \Rightarrow y = 15 - x$
$ \Rightarrow y = 15 - 8$ [from (1)]
$ \Rightarrow y = 7$
  Hence, $x = 8$ and $y = 7$.

Note: The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the sum of all frequencies, since all frequencies will already have been added to the previous total.