Question

The median of the distribution below is $14.4$ . Find the values of $x$ and $y$ , if the total frequency is $20$

Class interval	$0 - 6$	$6 - 12$	$12 - 18$	$18 - 24$	$24 - 30$
Frequency	$4$	$x$	$5$	$y$	$1$

Answer 1

Here we need to find the values of $x$ and $y$ for the given grouped data. First, we shall create a frequency distribution table containing cumulative frequency. Then, we shall find the values that are needed to be applied in the formula. Also, it is given that the median of the distribution below is $14.4$ and the total frequency is $20$ . Using these given values, we can find the unknown values.
Formula to be used:
$Median = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right] \times h$

Complete step-by-step answer:
Here we are asked to calculate the values of $x$ and $y$ for the given grouped data. First, we need to prepare a frequency distribution table containing the cumulative frequency.
We shall add the frequencies of all the classes preceding the given class to obtain the cumulative frequency of a class.

Class interval	Frequency	Cumulative frequency
$0 - 6$	$4$	$4$
$6 - 12$	$x$	$4 + x$
$12 - 18$	$5$	$9 + x$
$18 - 24$	$y$	$9 + x + y$
$24 - 30$	$1$	$10 + x + y$
TOTAL	10+x+y=20

Here, the total frequency is $N = 20$
The given median is $14.4$ and it lies in the class interval $12 - 18$
Hence the required median class is $12 - 18$
Now, we shall calculate the value of $\dfrac{N}{2}$
Thus, we get $\dfrac{N}{2} = \dfrac{{20}}{2} = 10$
Let $l$ be the lower limit of the median class.
Hence, we have $l = 12$ .
Let $h$be the size of the class.
Here, for the above frequency distribution, $h = 6$
Let us assume that $cf$ be the cumulative frequency of class preceding the median class.
Then, $cf = 4 + x$ is the required cumulative frequency of class preceding $12 - 18$
Let $f$ be the frequency of the median class.
Therefore, we have $f = 5$
Now, we shall apply all the values in the formula $Median = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right] \times h$
Thus, $14.4 = 12 + \left[ {\dfrac{{10 - \left( {4 + x} \right)}}{5}} \right] \times 6$
$\Rightarrow 14.4 = 12 + \dfrac{{\left( {10 - 4 - x} \right) \times 6}}{5}$
$\Rightarrow 14.4 - 12 = \dfrac{{\left( {6 - x} \right) \times 6}}{5}$
$\Rightarrow 2.4 = \dfrac{{\left( {6 - x} \right) \times 6}}{5}$
$\Rightarrow 2.4 \times 5 = 36 - 6x$
$\Rightarrow 12 = 36 - 6x$
$\Rightarrow 6x = 36 - 12$
$\Rightarrow 6x = 24$
$\Rightarrow x = \dfrac{{24}}{6}$
Hence, we got $x = 4$
Now, we have $10 + x + y = 20$ (We have calculated $10 + x + y = 20$ in the frequency distribution table)
$x + y = 20 - 10$
$\Rightarrow x + y = 10$
We shall apply $x = 4$ in the above equation.
Thus, we have $4 + y = 10$
$\Rightarrow y = 10 - 4$
Hence, we got $y = 6$
Therefore, we found the unknown values $x = 4$ and $y = 6$

Note: Here, we have another method to find the median class. Since we are given the value of median, we have used the above method. If not given, we need to look at the value of cumulative frequency whose frequency is just greater than $\dfrac{N}{2}$ and $N$ is the total sum of the frequency.