Median for Discrete and Continuous Frequency Type Data (grouped data)

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 $\dfrac{N}{2}$ and find out the less than type cumulative frequency just greater or equal to $\dfrac{N}{2}$ . Let that less than type cumulative frequency be denoted by $F_{k}$ . Now, we find out the class boundary corresponding to which the less than type cumulative frequency is equal to $F_{k}$ . Let us denote that class boundary by $x_{k}$ . The median class will be that class for which the upper class boundary is $x_{k}$ .Now, that we know the median class, the median can be calculated using the following formula:

Median or $\tilde{x}= x_{l} +\dfrac{\dfrac{N}{2}-F_{l}}{f_{m}} \times c$

where, $x_{l}$ is the lower class boundary of the median class,

$N$ is the total frequency,

$F_{l}$ is the less than type cumulative frequency corresponding to $x_{l}$ ,

$f_{m}$ is the frequency of the median class

and $c$ 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, $\dfrac{N}{2}=\dfrac{300}{2}=150$

Here, the cumulative frequency just greater than or equal to 150 is 198.

$\therefore F_{k}=198$ and

$\because F_{k}$ is the less than type cumulative frequency corresponding to the class boundary 399.5

$\therefore x_{k}=399.5$

$\therefore$ the median class is the class for which upper class boundary is $x_{k}=399.5$

In other words, 299.5-399.5 is the median class, i.e. the class containing the median value.

$\therefore$ using the formula for median we have,

Median or $\tilde{x} =299.5 +\dfrac{\dfrac{300}{2}-123}{75} \times 100$

where, $x_{l}=299.5$ (lower class boundary of the median class),

$N =300$ (total frequency),

$F_{l}=123$ ( less than type cumulative frequency corresponding to $x_{l} =299.5$ ),

$f_{m} =75$ (frequency of the median class),

and $c =100$ (class width of the median class).

$\therefore \tilde{x}=299.5+ \dfrac{150-123}{75} \times 100=299.5+\dfrac{27}{75} \times 100=299.5+3.6 \times 100=299.5+36=335.5$

or, Median ( $\tilde{x})=335.5$

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 $M_{1}$ and $M_{2}$ respectively, then the combine set median, say, $M$ , lies between $M_{1}$ and $M_{2}$

2. If $y=g(x)$ be a real-valued, monotonic function of $x$ , then median of $y$ is given by $\tilde{y}=g(\tilde{x})$ , $\tilde{x}$ being the median of $x$ .

Example: Values of two variables, $x$ and $y$ , are related as $y= 12x-311.5$ .If the median of $x$ be $\tilde{x}=130$ , find the median of $y$ , i.e. $\tilde{y}$ .

Solution:

Here, the median of $y$ is $\tilde{y} =12 \times 130 -311.5=124.5$

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.

Comments

Raj says

August 31, 2016 at 2:00 pm

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
  
  October 17, 2016 at 5:39 am
  
  26.66
  
- Sugam Rajbanshi says
  
  February 9, 2018 at 7:16 am
  
  X f cf
  less than 10 1 1
  10-20 2 3
  20-30 3 6
  30-40 2 8
  40-50 2 10
  N=10
  N
  class of median= —-
  2
  = 5 th terms
  
  N/2= 5 f =6 cf = 3 i = 10
  
  Mdn= 20.66
  
Deepanshu sharma says

October 11, 2016 at 5:49 am

Nice..

Santosh Kumar says

June 8, 2017 at 1:21 pm

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?

- Ayush says
  
  October 7, 2017 at 7:25 am
  
  I think u have to use ASSUME MEAN METHOD to find the answer…
  
Aishah says

October 13, 2017 at 1:35 pm

Plzzz solve the following guestion…
The no. Of runs scored by 11persons of a cricket team of a school are:
5,19,42,11,50,30,21,0,52,36,27
Compute the median for the given data.

Dan says

October 26, 2017 at 6:41 pm

I got 19.5, please kindly tell me if it is correct and tell me where I missed a move at.

prof.Nkoyoyo Johnpaul says

November 14, 2017 at 3:49 pm

its right

chaturveda reddy says

December 15, 2017 at 4:19 am

hi all this website is super and we can also gain huge informativeness in this

Sachin says

January 12, 2018 at 4:35 pm

Solve this compute the median
Class interval -20,30,40,50,60,70,80,90,100
Less than type

C. F-0,4,16,30,46,66,82,92,100

Sai Maharshi says

March 2, 2018 at 3:59 pm

why upper boundaries of a class are taken for calculating median when we use less than cumulative frequency why don t we take lower boundaries explain in detali

saurabh says

August 26, 2018 at 4:28 am

class interval- 0-adhik 10-adhik 20-adhik 30-adhik 40-adhik 50-adhik 60-adhik 70-adhik

frequency- 100 91 78 57 44 30 18 8

m-37.4
find median

saurabh says

August 26, 2018 at 4:33 am

class interval- 0-adhik 10-adhik 20-adhik 30-adhik 40-adhik 50-adhik 60-adhik 70-adhik
frequency- 100 91 78 57 44 30 18 8

Median for Discrete and Continuous Frequency Type Data (grouped data) :

Properties of Median:

Exercise:

Comments

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