Question

The median of the following data is $28$. Find the values of $x$and $y$, if the total frequency is $50$.

Marks	$0 - 7$	$7 - 14$	$14 - 21$	$21 - 28$	$28 - 35$	$35 - 42$	$42 - 49$
No. of students	$3$	$x$	$7$	$11$	$y$	$16$	$9$

Answer 1

Hint: First of all, we have to know what is median? So, the median is the middle number in a sorted, ascending or descending, list of numbers and in the case of a data system, the median is also based on the frequency of the data. We apply the appropriate formula for median for the given frequency table and compute the values of x and y.

Formula Used:
To find the median of the data, the formula used is,
Median, $M = I + \dfrac{{\dfrac{N}{2} - cf}}{f} \times h$
Where, $I$ is the lower limit of the median class, $N$ is the total frequency, $cf$ is the cumulative frequency of the previous class to the median class, $f$ is the frequency of the median class, $h$ is the width of the median class.Then, by using the given conditions, we can find the values of $x$ and $y$.

Complete step by step answer:
Given, total frequency $N = 50$.Median of the data is, $M = 28$.

Marks	No. of students (frequency)	Cumulative frequency
$0 - 7$	$3$	$3$
$7 - 14$	$x$	$3 + x$
$14 - 21$	$7$	$10 + x$
$21 - 28$	$11$	$21 + x$
$28 - 35$	$y$	$21 + x + y$
$35 - 42$	$16$	$37 + x + y$
$42 - 49$	$9$	$46 + x + y$

As given, total frequency, $N = 50$
So, from cumulative frequency, the total frequency can be observed is $46 + x + y$.
Therefore, we can write,
$ \Rightarrow 46 + x + y = 50$
$ \Rightarrow x + y = 4 - - - (1)$
Also, Median, $M = 28$
Therefore, from the above data table, we can say that, the median class is $28 - 35$.
Now, by using the formula, $M = I + \dfrac{{\dfrac{N}{2} - cf}}{f} \times h$, we get,
$ \Rightarrow 28 + \dfrac{{\dfrac{{50}}{2} - (21 + x)}}{y} \times 7 = 28$
$ \Rightarrow 7 \times \left( {\dfrac{{25 - 21 - x}}{y}} \right) = 28 - 28$
Subtracting $28$from both sides of equation, we get,
$ \Rightarrow 7\left( {\dfrac{{4 - x}}{y}} \right) = 0$
Dividing both sides of equation by $7$, we get,
$ \Rightarrow 4 - x = 0$
Multiplying y on both sides of equation, we get,
$ \Rightarrow x = 4$
Now, substituting the value of $x$in equation $(1)$:
$4 + y = 4$
$ \therefore y = 0$
Therefore, the values of $x$ and $y$ are, $x = 4$ and $y = 0$.

Note: In this question, the class intervals are joint or continuous. But in some questions, the class intervals may be disjoint or discontinuous. In such cases, the upper limit of the previous class must be increased by $0.5$ and by decreasing the lower limit of the class by $0.5$. For, example if a class interval is of the form,

Class Interval

$1 - 10$

$11 - 20$

$21 - 30$

 This class interval is disjoint or discontinuous. So, to make it continuous, we write it as,

Class Interval

$0.5 - 10.5$

$10.5 - 20.5$

$20.5 - 30.5$

Hence, the class intervals become continuous and can be easily solved.