In our article Median for Frequency Type Data, we explained how to calculate the median for ungrouped frequency distribution of a discrete variable. In this article we shall discuss how to calculate the median for grouped frequency distribution of discrete variables as well as continuous variables. For both cases, the method for calculating the median is the same. We shall also discuss a few properties of Median.
Median for Discrete and Continuous Frequency Type Data (grouped data) :
For the grouped frequency distribution of a discrete variable or a continuous variable the calculation of the median involves identifying the median class, i.e. the class containing the median. This can be done by calculating the less than type cumulative frequencies. It should be recalled that less than type cumulative frequencies correspond to the upper class boundaries of the respective classes. First we calculate and find out the less than type cumulative frequency just greater or equal to . Let that less than type cumulative frequency be denoted by . Now, we find out the class boundary corresponding to which the less than type cumulative frequency is equal to . Let us denote that class boundary by . The median class will be that class for which the upper class boundary is .Now, that we know the median class, the median can be calculated using the following formula:
Median or
where, is the lower class boundary of the median class,
is the total frequency,
is the less than type cumulative frequency corresponding to ,
is the frequency of the median class
and is the class width of the median class.
Example: The following distribution represents the number of minutes spent per week by a group of teenagers in going to the movies. Find the median number of minutes spent per by the teenagers in going to the movies.
Number of minutes per week |
Number of teenagers |
0-99 |
26 |
100-199 |
32 |
200-299 |
65 |
300-399 |
75 |
400-499 |
60 |
500-599 |
42 |
Solution:
Let us convert the class intervals given, to class boundaries and construct the less than type cumulative frequency distribution.
Number of minutes per week | Class Boundaries | Number of teenagers (Frequency) |
Cumulative Frequency (less than type) |
0-99 |
0-99.5 | 26 |
26 |
100-199 |
99.5-199.5 | 32 |
58 |
200-299 |
199.5-299.5 | 65 |
123 |
300-399 |
299.5-399.5 | 75 |
198 |
400-499 |
399.5-499.5 | 60 |
258 |
500-599 |
499.5-599.5 | 42 |
300 |
Here,
Here, the cumulative frequency just greater than or equal to 150 is 198.
and
is the less than type cumulative frequency corresponding to the class boundary 399.5
the median class is the class for which upper class boundary is
In other words, 299.5-399.5 is the median class, i.e. the class containing the median value.
using the formula for median we have,
Median or
where, (lower class boundary of the median class),
(total frequency),
( less than type cumulative frequency corresponding to ),
(frequency of the median class),
and (class width of the median class).
or, Median (
So, the median number of minutes spent per week by this group of 300 teenagers in going to the movies is 335.5, i.e. there are 150 teenagers for whom the number of minutes spent per week in going to the movies is less than 335.5 and there are another 150 teenagers for whom the number of minutes spent per week in going to the movies is greater than 335.5.
Note:
It is quite clear that in calculating the median of any grouped frequency distribution using this method, the nature of the variable (i.e. discrete or continuous) is of little consequence. Whatever be the nature of the variable, for grouped frequency distributions, this method is exhaustive and will ensure correct calculation of the median.
Properties of Median:
1. If we have two sets of va;ues having medians and respectively, then the combine set median, say, , lies between and
2. If be a real-valued, monotonic function of , then median of is given by , being the median of .
Example: Values of two variables, and , are related as .If the median of be , find the median of , i.e. .
Solution:
Here, the median of is
Exercise:
1. Obtain the median for the following frequency distribution of house rent for a sample of 30 families in a certain locality:
Rent (Rs.) |
Number of Families |
1800-2000 |
4 |
2000-2200 |
7 |
2200-2400 |
10 |
2400-2600 |
5 |
2600-2800 |
2 |
2800-3000 |
2 |
2. Frequency distribution of I.Q. of 309 6-year old children is given below:
Class Intervals |
Frequency |
40-49 |
1 |
50-59 |
2 |
60-69 |
3 |
70-79 |
5 |
80-89 |
17 |
90-99 |
65 |
100-109 |
69 |
110-119 |
79 |
120-129 |
37 |
130-139 |
19 |
140-149 |
7 |
150-159 |
3 |
160-169 |
2 |
Find the median I.Q. of these children.
Raj says
Solve this x is 1-10, 1-20, 1-30, 1-40, 1-50 and f is 1, 3, 6, 8, 10 find median
sathish says
26.66
Deepanshu sharma says
Nice..
Santosh Kumar says
Hi all,
Suppose I wish to take the middle points of the ranges in the example in order to take them as weights to calculate a weighted mean, What can I do for the highest range if there is no upper limit?
For instance in the first example on this page for minutes spent in going to movies, if the upper range says ‘600 or more’ what can I safely take as the weight (middle point) here?