251median 9/15/04
Example for the median of grouped data
The formulas we are using are position 
1
2
n  1 and
 .5n  F 
x.5  Lm  
 w . The given data was as
 fm 
below.
Profit Rate
9-10.99%
11-12.99%
13-14.99%
15-16.99%
17-18.99%
Total
f
3
3
5
3
1
15
a. Before we start, we must find the cumulative frequency F , from the frequencies under f .
The first of these is identical to the frequency of the 9-10.99% class. The next one represents the total
number of items under 13, and is 3 + 3 = 6, the third cumulative frequency is the sum of the frequencies
below 15 and is 3 + 3 + 5 = 11, or, using the second cumulative frequency, 6 + 5 = 11. If we continue this
way our data table reads.
Profit Rate f F
9-10.99% 3 3
11-12.99% 3 6
13-14.99% 5 11
15-16.99% 3 14
17-18.99% 1 15
Total
15
b. We now find the position of the median by using the first formula, position 
1
2
n  1
 .516   8. If the numbers are considered to be x1 through x15 , the number we want is x8 .
c. To locate the class that contains x8 , compare 8 with the cumulative distribution. Since 8 is
above 6 and below 11, the median group must be the 13 – 14.99 class.
 .5n  F 
d. Now that we have chosen the median group, we use the second formula, x.5  Lm  
w
 fm 
to find a specific number. In this formula F is the cumulative frequency up to but not including the
f  15 ; w  2 is the width of the
median, so F  6 ; f m  5 is the frequency of the median group; n 
group; Lm  13 is the lower limit of the median group.

 .515   6 
 7.5  6 
2  13  
Thus, we find that the median is x.5  13  

 2  13  0.32  13 .6 .
5
 5 


1