Measures of Central Tendency for Grouped Data
Mean: The following formula is used to calculate the mean for grouped data:
n
n
fm
x   i i where N   f i and mi are the mid points of class intervals.
i 1 N
i 1
Example1. (Exclusive series) The following table gives the frequency distribution of ages of
persons living in a locality:
Age (in years)
Number of persons
0-10
15
10-20
20
20-30
32
30-40
19
40-50
14
Calculate the mean age of persons.
Solution:
Class intervals Mid Points ( mi ) Frequency ( f i )
f i mi
0-10
5
15
75
10-20
15
20
300
20-30
25
32
800
30-40
35
19
665
40-50
45
14
630
Total
Thus,
5
f
i 1
i
 100
5
fm
i 1
i
 2470
i
f i mi
i 1 N
2470

 24.70 years
100
Example2. (Inclusive series) The following table gives the frequency distribution of the number
of orders received each day during the past 50 days at the office of a mail-order company.
Calculate the mean.
No. of orders
Frequency
7-9
4
10-12
10
13-15
20
16-18
13
19-21
3
Solution:
Class intervals Mid Points ( mi ) Frequency ( f i )
f i mi
7-9
8
4
32
10-12
11
10
110
13-15
14
20
280
16-18
17
13
221
19-21
20
3
60
n
x 
Total
5

i 1
f i  50
5
fm
i 1
i
i
 703
Thus,
f i mi
i 1 N
703

 14.06 orders
50
5
x 
Median: The following formula is used to calculate the median for grouped data:
N

 C
2
 h
Median ( Me)  l0  
f
where l0 = lower boundary of the median class
C = cumulative frequency of the class preceding the median class
f = frequency of the median class
h = difference of class interval or magnitude of the median class
n
N   fi
i 1
The median class is determined by taking N 2 and identifying the cumulative frequency which
includes N 2 .
Example3. (Exclusive series) Find the median for the following frequency distribution table:
Class Intervals
Frequency
140-145
3
145-150
5
150-155
10
155-160
6
160-165
5
165-170
1
Solution:
Class intervals Frequency ( f i )
Cumulative
frequency (C.F.)
140-145
3
3
145-150
5
8
150-155
10
18
155-160
6
24
160-165
5
29
165-170
1
30
5
N   f i  30
i 1
We see that N 2 = 30/2 = 15, is included in 18 (C.F.). Thus the median class is 150-155.
Now, we have l0 = 150, C = 8, f = 10 and h =155 - 150 = 5
N

 C
2
 h
Median ( Me)  l0  
f
15  8  5  150  3.5  153.5
 150 
10
Example4. (Inclusive series) The following table gives the frequency distribution of the number
of orders received each day during the past 50 days at the office of a mail-order company.
Calculate the median.
No. of orders
7-9
10-12
13-15
16-18
19-21
Frequency
4
10
20
13
3
Solution: The given series is in the form inclusive series, we have to convert it into exclusive
series.
Class intervals Frequency ( f i ) Cumulative frequency
(C.F.)
6.5-9.5
4
4
9.5-12.5
10
14
12.5-15.5
20
34
15.5-18.5
13
47
18.5-21.5
3
50
Total
5
N   f i  50
i 1
We see that N 2 = 50/2 = 25, is included in 34 (C.F.). Thus the median class is 12.5-15.5.
Now, we have l0 = 12.5, C = 14, f = 20 and h =15.5 - 12.5 = 3
N

 C
2
 h
Median ( Me)  l0  
f
 25  14   3  12.5  1.65  14.15
 12.5 
20
Mode: The following formula is used to calculate the mode for grouped data:
 f m  f1   h
Mode ( Mo)  l0 
 2 f m  f1  f 2 
where l0 = lower boundary of the modal class
f m = frequency of the modal class (maximum frequency)
f1 = frequency of the class preceding the modal class
f 2 = frequency of the class succeeding the modal class
h = difference of class interval or magnitude of the modal class
We determine modal class by identifying the class interval with maximum frequency.
Note: 1. Class intervals must be exclusive, equal, in ascending order, not cumulative.
2. If the first class is the modal class then f1 will be zero and if the last class is the modal class
then f 2 will be zero.
Example5. (Exclusive series) The following table gives the frequency distribution of ages of
persons living in a locality:
Age (in years)
Number of persons
0-10
15
10-20
20
20-30
32
30-40
19
40-50
14
Calculate the mode age of persons.
Solution:
Since the maximum frequency 32 occurs for the class interval 20-30, so the modal class is 20-30.
Age (in years)
Number of persons
0-10
15
10-20
20 (f1)
32 (fm)
20-30
30-40
19 (f2)
40-50
14
Now, we have l0 = 20, f m = 32, f1 = 20, f 2 = 19 and h = 30 - 20 = 10
Mode ( Mo)  l0 
 f m  f1 
h
 2 f m  f1  f 2 
 32  20  10
 20 
 2  32  20  19 
12
 10  20  4.8  24.8
 64  39 
Example6. (Inclusive series) The following table gives the frequency distribution of the number
of computers sold during each of the past 25 weeks at an electronics store. Calculate the mode.
 20 
No. of orders
4-9
10-15
16-21
22-27
28-33
Frequency
2
4
10
6
3
Solution: The given series is in the form inclusive series, we have to convert it into exclusive
series.
No. of orders
Frequency
3.5-9.5
2
9.5-15.5
4 (f1)
10 (fm)
15.5-21.5
21.5-27.5
6 (f2)
27.5-33.5
3
Since the maximum frequency 10 occurs for the class interval 15.5-21.5, so the modal class is
15.5-21.5
Now, we have l0 = 15.5, f m = 10, f1 = 4, f 2 = 6 and h = 21.5 - 15.5 = 6
Mode ( Mo)  l0 
 f m  f1 
h
 2 f m  f1  f 2 
10  4   6
 15.5 
 2  10  4  6
 15.5 
6
 6  15.5  3.6  19.1
 20  10
Note:
Inclusive Class Interval:
When the lower and the upper class limit is included, then it is an inclusive class interval. For
example: 1 -10, 11-20, 21-30,….,etc. are inclusive type of class intervals. Usually in the case of
discrete variate, inclusive types of class intervals are used.
Exclusive Class Interval:
When the lower limit is included, but the upper limit is excluded, then it is an exclusive class
interval. For example – 0-10, 10-20, 20-30,…,etc are exclusive type of class intervals. Usually in
the case of continuous variate, exclusive types of class intervals are used.
Questions for Practice (Ungrouped and Grouped Data)
Q1: Calculate the mean of the following observations:
30, 33, 34, 35, 37, 39, 41, 45, 48.
Q2: Calculate the mean of the following data:
x
4
6
8
f
1
3
4
10
1
Ans. 38
11
3
Ans. 8.083
Q3: The marks obtained by 30 students in an examination in Mathematics are given in the
following distribution table. Calculate the mean.
Marks
0-10
10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency
2
2
4
10
6
3
1
2
Ans. 38
Q4: Find the median of the observations:
23, 22, 21, 30, 38, 25, 28, 23, 18, 8, 13, 43.
Ans. 23
Q5: Find the median for the following frequency distribution:
x
3
5
6
9
10
15
f
5
9
18
13
8
7
Ans. 6
Q6: Find the median for the following frequency distribution:
x
60
50
52
61
56
f
5
8
6
3
7
Ans. 56
Q7: Find the median for the following frequency distribution:
Class (Marks)
0-10
10-20 20-30 30-40 40-50
No. of candidates
5
8
20
10
7
Ans. 26
Q8. Find the mean and median for the following frequency distribution:
Class Interval
1-10 11-20 21-30 31-40 41-50
Frequency
3
6
9
5
2
Ans. 24.3, 24.389
Q9: Find the mode from the following data:
38, 30, 28, 23, 25, 30, 36, 25, 30, 28.
Ans. 30
Q10: Find the mode from the following frequency distribution:
x
8
10
12
14
16
18
20
f
3
5
9
10
20
9
4
Ans. 16
Q11. Find the mode for the following frequency distribution:
Class Interval
0-10 10-20 20-30 30-40 40-50
Frequency
14
23
27
21
15
Ans. 24
Q12. Find the mode for the following frequency distribution of marks of 100 students in Statistics
examination:
Marks
1-10 11-20 21-30 31-40 41-50
Number of students
12
20
35
18
15
Ans. 25.1875