•
MEAN
•
MEDIAN
•
MODE
At the end of this lesson, you should be able to:
1. Determine the use of the three measures of central
tendency (mean, median, and mode)
2. Solve for mean, median, and mode for grouped and
ungrouped data
3. Solve problems involving measures of central tendency.
 The most appropriate measures found to be useful in
describing a distribution of observations.
 Measures of Central Tendency refers to the ‘center’ of a
distribution of observations.
 This is a value which gives a summary of the characteristics
of a given set of data.
1. Simple Mean
The MEAN, or average is one number that is
representative of the data. You can find the mean by adding
all the data and then dividing the sum by the number of
addends.
Formula: 𝑋 =
Where:
𝑥
𝑛
𝑋 = mean, read “x bar”
𝑥 = items/scores/data
𝑛 = number of items
Example: What is the mean score of 10 statistics students in
their midterm examination, if their scores are as follows:
46 44 40 41 45 32 38 40 49 36
Solution:
𝑋=
𝑋=
𝑋=
𝑥
𝑛
46+44+40+41+45+32+38+40+49+36
10
411
10
𝑋 = 41.1
The MEDIAN is the middle number/value when the
data are arranged in order.
Example:
46
44
Solution:
40
41
45
32
38
40
32, 36, 38, 40, 40, 41, 44, 45, 46, 49
𝑀𝑒𝑑𝑖𝑎𝑛 =
𝑀𝑒𝑑𝑖𝑎𝑛 =
40+41
2
81
2
𝑀𝑒𝑑𝑖𝑎𝑛 = 40.5
49
36
The MEDIAN is the middle number/value when the
data are arranged in order.
Example:
46
44
Solution:
40
41
45
32
38
40
32, 36, 38, 40, 40, 41, 44, 45, 46
𝑀𝑒𝑑𝑖𝑎𝑛 = 40
36
The MODE is the number that occurs most often in a
set of data. There may be one mode, more than one mode, or
no mode.
Example:
46
44
Solution:
40
41
45
32
38
40
32, 36, 38, 40, 40, 41, 44, 45, 46, 49
𝑀𝑜𝑑𝑒 = 40
UNIMODAL
49
36
The MODE is the number that occurs most often in a
set of data. There may be one mode, more than one mode, or
no mode.
Example: 32, 32, 36, 38, 40, 40, 41, 44, 45, 46, 49
𝑀𝑜𝑑𝑒 = 32 𝑎𝑛𝑑 40
BIMODAL
The MODE is the number that occurs most often in a
set of data. There may be one mode, more than one mode, or
no mode.
Example: 32, 32, 36, 38, 40, 40, 41, 44, 45, 46, 46, 49
𝑀𝑜𝑑𝑒 = 32, 40 𝑎𝑛𝑑 46
MULTIMODAL
The MODE is the number that occurs most often in a
set of data. There may be one mode, more than one mode, or
no mode.
Example: 32, 32, 36, 38, 40, 40, 41, 44, 45, 46, 46, 46, 49
𝑀𝑜𝑑𝑒 = 46
UNIMODAL
The MODE is the number that occurs most often in a
set of data. There may be one mode, more than one mode, or
no mode.
Example: 32, 34, 36, 38, 40, 41, 44, 45, 46, 49
𝑀𝑜𝑑𝑒 = 𝑁𝑜 𝑀𝑜𝑑𝑒
MEAN
Steps in Computing the Mean of Grouped Data
1. Calculate the midpoint or class marks of all class
intervals.
2. Multiply each class mark by their corresponding
frequency.
3. Add the products of each in #2.
4. Divide the sum by the total number of cases to obtain the
mean.
Formula:
Where:
𝑓𝑥
𝑋=
𝑁
𝑋 = Mean
𝑓 = Class Frequency
𝑥 = Class Mark/ Midpoint
𝑁 = Number of Cases
Class Interval
C.I.
Frequency
f
51-55
4
56-60
3
61-65
4
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
Class Mark
x
fx
Class Interval
C.I.
Frequency
f
51-55
4
56-60
3
61-65
4
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
𝑵 = 𝟓𝟎
Class Mark
x
fx
Class Interval
C.I.
Frequency
f
Class Mark
x
51-55
4
53
56-60
3
58
61-65
4
63
66-70
10
68
71-75
9
73
76-80
7
78
81-85
5
83
86-90
8
88
𝑵 = 𝟓𝟎
fx
Class Interval
C.I.
Frequency
f
Class Mark
x
fx
51-55
4
53
212
56-60
3
58
174
61-65
4
63
252
66-70
10
68
680
71-75
9
73
657
76-80
7
78
546
81-85
5
83
415
86-90
8
88
704
𝑵 = 𝟓𝟎
Class Interval
C.I.
Frequency
f
Class Mark
x
fx
51-55
4
53
212
56-60
3
58
174
61-65
4
63
252
66-70
10
68
680
71-75
9
73
657
76-80
7
78
546
81-85
5
83
415
86-90
8
88
704
𝑵 = 𝟓𝟎
𝒇𝒙 = 𝟑𝟔𝟒𝟎
Class Interval
C.I.
Frequency
f
Class Mark
x
fx
51-55
4
53
212
56-60
3
58
174
61-65
4
63
252
66-70
10
68
680
71-75
9
73
657
76-80
7
78
546
81-85
5
83
415
86-90
8
88
704
𝑵 = 𝟓𝟎
𝒇𝒙 = 𝟑𝟔𝟒𝟎
Solution:
𝑿=
𝒇𝒙
𝑵
𝑿=
𝟑𝟔𝟒𝟎
𝟓𝟎
𝑿 = 𝟕𝟐. 𝟖𝟎
1.
2.
3.
4.
5.
6.
7.
Steps in Computing the Median of Grouped Data
Determine the median
class by dividing n by 2
𝑛
𝑀𝑒𝑑𝑖𝑎𝑛 𝑐𝑙𝑎𝑠𝑠 = .
2
Construct the less than cumulative frequency column in the
table.
𝑛
Locate the in the cumulative frequency to determine the
2
median class.
Get the lower boundary of the median class.
𝑛
from the computed , subtract the < 𝑐𝑓.
2
Divide the difference by the frequency of the median class,
then multiply the quotient by the class size (𝑖).
Add the obtained value in #6 to the lower boundary of the
median class (𝑋𝐿𝐵 ).
Formula:
𝑛
− < 𝑐𝑓
2
𝑋 = 𝑋𝐿𝐵 +
𝑖
𝑓𝑚
Where:𝑋 = Median value, read as “tilde x”
𝑋𝐿𝐵 = Lower boundary of the median class
< 𝑐𝑓 = Cumulative frequency before the median class
𝑓𝑚 = Frequency of the median class
𝑛 = number of cases
𝑖 = class size
Class Interval
C.I.
Frequency
f
51-55
4
56-60
3
61-65
4
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
𝑵 = 𝟓𝟎
<cf
Class Interval
C.I.
Frequency
f
<cf
51-55
4
4
56-60
3
7
61-65
4
11
66-70
10
21
71-75
9
30
76-80
7
37
81-85
5
42
86-90
8
50
𝑵 = 𝟓𝟎
Class Interval
C.I.
Frequency
f
<cf
51-55
4
4
56-60
3
7
61-65
4
11
66-70
10
21
71-75
9
30
76-80
7
37
81-85
5
42
86-90
8
50
𝑵 = 𝟓𝟎
𝒏
𝑴𝒆𝒅𝒊𝒂𝒏 𝑪𝒍𝒂𝒔𝒔 =
𝟐
𝟓𝟎
𝑴𝒆𝒅𝒊𝒂𝒏 𝑪𝒍𝒂𝒔𝒔 =
𝟐
𝑴𝒆𝒅𝒊𝒂𝒏 𝑪𝒍𝒂𝒔𝒔 = 𝟐𝟓
Median Class
Class Interval
C.I.
Frequency
f
<cf
51-55
4
4
56-60
3
7
61-65
4
11
66-70
10
21
71-75
9
30
76-80
7
37
81-85
5
42
86-90
8
50
𝑵 = 𝟓𝟎
Lower Boundary of
the Median Class
𝑿𝑳𝑩 = 𝟕𝟏 − 𝟎. 𝟓
Median Class
𝑿𝑳𝑩 = 𝟕𝟎. 𝟓
Class Interval
C.I.
Frequency
f
<cf
51-55
4
4
56-60
3
7
61-65
4
11
66-70
10
21
71-75
9
30
76-80
7
37
81-85
5
42
86-90
8
50
𝑵 = 𝟓𝟎
Solution:
𝑿 = 𝑿𝑳𝑩 +
Median Class
𝑿 = 𝟕𝟎. 𝟓 +
𝑿 = 𝟕𝟐. 𝟕𝟐
𝒏
− <𝒄𝒇
𝟐
𝒇𝒎
𝟓𝟎
−𝟐𝟏
𝟐
𝟗
𝒊
𝟓
Steps in Computing the Mode of Grouped Data
1. Determine the modal class. The modal class has the
highest frequency in the distribution.
2. Get the lower boundary of the modal class.
3. Determine the ∆1 and ∆2 and divide ∆1 to the sum of ∆1
and ∆2.
4. Multiply the result obtained in #3 by the class size (𝑖).
5. Add the answer in number 4 by the lower boundary of
the modal class.
Formula:
∆1
𝑋 = 𝑋𝐿𝐵 +
𝑖
∆1 + ∆2
Where:
𝑋 = Mode, read as “caret x”
𝑋𝐿𝐵 = Lower Boundary of the modal class
∆1 = The difference of the highest frequency and the
frequency just below it.
∆2 = The difference of the highest frequency and the
frequency just above it.
Class Interval
C.I.
Frequency
f
51-55
4
56-60
3
61-65
4
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
𝑵 = 𝟓𝟎
Lower Boundary of
the Modal Class
𝑿𝑳𝑩 = 𝟔𝟔 − 𝟎. 𝟓
Modal Class
𝑿𝑳𝑩 = 𝟔𝟓. 𝟓
Class Interval
C.I.
Frequency
f
51-55
4
56-60
3
61-65
4
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
𝑵 = 𝟓𝟎
Delta 1 and Delta 2
=𝟔
Modal Class
=𝟏
∆𝟏 = 𝟏𝟎 − 𝟒
∆𝟏 = 𝟔
∆𝟐 = 𝟏𝟎 − 𝟗
∆𝟐 = 𝟏
Class Interval
C.I.
Frequency
f
51-55
4
Solution:
56-60
3
61-65
4
∆𝟏
𝑿 = 𝑿𝑳𝑩 +
𝒊
∆𝟏 + ∆𝟐
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
𝑵 = 𝟓𝟎
Modal Class
𝑿 = 𝟔𝟓. 𝟓 +
𝑿 = 𝟔𝟗. 𝟕𝟗
𝟔
𝟔+𝟏
𝟓
Class Interval
C.I.
Frequency
f
51-55
4
56-60
3
61-65
4
66-70
10
71-75
9
76-80
7
81-85
5
86-90
8
𝑵 = 𝟓𝟎
𝑿 = 𝟕𝟐. 𝟖𝟎
𝑿 = 𝟕𝟐. 𝟕𝟐
𝑿 = 𝟔𝟗. 𝟕𝟗
 Roc-Narag, Edlyn Castillo (2010). Basic Statistics with
Calculator and Computer Application. Rex Book
Store, Inc. 856 Nicanor Reyes, Sr., St., Sampaloc,
Manila. Pages 51-59