Question

Find the upper limit of the median class from the frequency distribution table.

Class	$0 - 5$	$6 - 11$	$12 - 17$	$18 - 23$	$24 - 29$
frequency	$3$	$10$	$15$	$8$	$11$

A. $17$
B. $17.5$
C. $18$
D. $18.5$

Answer 1

Hint:
For finding the median class we need to calculate the cumulative frequency and convert the data into the continuous format by adding $0.5$ to the upper limit and subtracting the same from the lower limit.

Complete step by step solution:
So here we are given the distribution as:

Class	$0 - 5$	$6 - 11$	$12 - 17$	$18 - 23$	$24 - 29$
frequency	$3$	$10$	$15$	$8$	$11$

Here the frequency of $5 - 6,11 - 12,17 - 18,23 - 24$ is missing. So here first of all we need to take the average of the upper limit of one class interval and lower limit of the other class interval which is $\dfrac{{5 + 6}}{2} = 5.5$
Similarly the average of $17,18{\text{ is 17}}{\text{.5}}$ and solving for all we will get the averages of all as $5.5,11.5,17.5,23.5$
So now we can write this in the continuous distribution as follows:

Class	$0 - 5.5$	$5.5 - 11.5$	$11.5 - 17.5$	$17.5 - 23.5$	$23.5 - 29.5$
frequency	$3$	$10$	$15$	$8$	$11$

Now in the continuous for now we can find the cumulative frequency and represent it as follows:

class	frequency	Cumulative Frequency
$0 - 5.5$	$13$	$13$
$5.5 - 11.5$	$10$	$23$
$11.5 - 17.5$	$15$	$38$
$17.5 - 23.5$	$8$	$46$
$23.5 - 29.5$	$11$	$57$

Cumulative frequency is the sum of the preceding all the frequencies to that class interval whose cumulative frequency is to be found.
For the first interval that is $0 - 5.5$ the cumulative frequency is the same as the frequency as there is no preceding interval to that of the interval. Hence in rest of all we keep on adding the preceding all the frequencies to get the cumulative frequencies.
Here $N = 57$
Median is $\dfrac{N}{2} = 28.5$
We need to find the cumulative frequency just after $28.5$ which is $38$ and the interval is here $11.5 - 17.5$

Hence upper limit is $17.5$

Note:
If median class is given as $(a - b)$ then the value of the median would lie between the two values $a{\text{ and }}b$which means between the lower and the higher limit of the median class and the formula is given as:
${\text{median}} = a + \left( {\dfrac{{\dfrac{n}{2} - CF}}{f}} \right)h$
Here $a$is the lower limit of the median class and $n$ is the number of observations, $CF$ is the cumulative frequency of the class preceding the median class, $f$ is the frequency of the median class and $h$ is the class size.