Question

The upper limit of the median class of the following distribution is

Class	$0 - 5$	$6 - 11$	$12 - 17$	$18 - 23$	$24 - 29$
Frequency	$13$	$10$	$15$	$8$	$11$

(A) $17$
(B) $17.5$
(C) $18$
(D) $18.5$

Answer 1

Hint: Start with finding the continuous class for each class by adding and subtraction $0.5$ in upper and lower limits respectively. Now find the cumulative frequency for each class. The last cumulative frequency will be the value ‘N’. The class having the cumulative frequency just greater than the value $\dfrac{N}{2}$ will be called the median class.

Complete step-by-step answer:
Here in this problem, we are given with a distribution table divided into five classes of $0 - 5,6 - 11,12 - 17,18 - 23{\text{ and }}24 - 29$ with the respective frequencies as $13,10,15,8{\text{ and 11}}$ . With this information, we need to find the upper limit of the median class of the distribution.
And for finding out the upper limit of the median class, we need to first understand how to find the median class for distribution.
The median class can be found by finding the class which is having cumulative frequency just greater than the value of $\dfrac{N}{2}$ , where ‘N’ is the total number of observations or the sum of the frequencies. Cumulative frequency can be calculated for each class by adding up the frequency of the classes above it.
But first, we need to make the classes continuous by adding $0.5$ in the upper limit and subtracting $0.5$ from lower limits.

Class	$0 - 5$	$6 - 11$	$12 - 17$	$18 - 23$	$24 - 29$
Continuous Class	$0 - 5.5$	$5.5 - 11.5$	$11.5 - 17.5$	$17.5 - 23.5$	$23.5 - 29.5$
Frequency	$13$	$10$	$15$	$8$	$11$
Cumulative Frequency	$13$	$13 + 10 = 23$	$23 + 15 = 38$	$38 + 8 = 46$	$46 + 11 = 57$

So, we got the sum of the frequencies as the last value in cumulative frequency row, i.e. $N = 57$
$ \Rightarrow \dfrac{N}{2} = \dfrac{{57}}{2} = 28.5$
The number $28.5$ lies after the cumulative frequency of $23$ and is less than the cumulative frequency of $38$ .
Therefore the median class will be the class $\left( {11.5 - 17.5} \right)$
Thus the upper limit of the required median class will be $17.5$
Hence, the option (B) is the correct answer.
Note: Do not forget to find inclusive or continuous classes for the given exclusive class. Cumulative frequency is the addition of the frequencies of the above that class and represents the total observation till that class. Be careful while preparing the table with cumulative frequency. A small error in addition can completely change the result.