Question

Consider the following frequency distribution table:

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

The upper limit of the median class is
A. 17
B. 17.5
C. 18
D. 18.5

Answer 1

We find the values of class-limits and the class marks. We have been asked to find the upper-class limit of the median class. We put them in a table to find the cumulative frequency. From that value of cumulative frequency, we find the median class using the half value of total frequency. We use the formula of median to find the solution of the problem also.

Complete step by step answer:
We assume the frequencies as ${{f}_{i}}$ and the class marks as ${{x}_{i}}$.
We also need to find the cumulative frequencies.
We use class-limits and the class marks to find the value of the variables.
Total frequency is $n=13+10+15+8+11=57$. So, $\dfrac{n}{2}=\dfrac{57}{2}=28.5$.
From the cumulative frequency we can find the median class will be 11.5-17.5.

Class intervals	Class limits	Class marks (${{x}_{i}}$)	Frequency (${{f}_{i}}$)	Cumulative frequency (${{F}_{i}}$)
0-5	0-5.5	2.5	13	13
6-11	5.5-11.5	8.5	10	23
12-17	11.5-17.5	14.5	15	38
18-23	17.5-23.5	20.5	8	46
24-29	23.5-29.5	26.5	11	57
Total			$n=57$

We also the formula of median as $median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c$.
Here l is the lower limit of the median class. ${{F}_{l}}$ denotes the cumulative frequency of the previous class of that median class. ${{f}_{me}}$ denotes the frequency of the median class. Also, c is the class width of the frequency table. In our problem the value of c is 6.
So, we put the values in the equation and get
$median\left( {{x}_{i}} \right)=l+\dfrac{\dfrac{n}{2}-{{F}_{l}}}{{{f}_{me}}}\times c=11.5+\dfrac{\dfrac{57}{2}-23}{15}\times 6$
We solve this equation to get the value of median.
So, $median\left( {{x}_{i}} \right)=11.5+\dfrac{\dfrac{57}{2}-23}{15}\times 6=11.5+\dfrac{33}{15}=11.5+2.2=13.7$
The median value of the given frequency distribution table is 13.7.
The upper limit of the median class is 17.5.
The correct option is (B).

Note: We need to remember the median is the value of the frequency being at the most middle point. To find the median class upper bound we just need the cumulative frequency and half value of the frequency. So, instead of finding mean we use the formula of median as it considers the density of a cumulative grouped data. We need to be careful finding the median class.