Measures of
Central
TendencyFor Ungrouped Data
Recall:What are the three ways
on how we can define
“central tendency” or
“center of the
distribution”?
We can also say:
(1) the point on
which a distribution
would be balance;
We can also say:(2) the value whose
average absolute
deviation from all other
values is minimized;
and
We can also say:(3) the value whose
average squared
difference from all the
other values is
minimized.
Recall:
What are the
measures of central
tendency?
Do you know?
The mean is the
point on which the
distribution would
balance.
Do you know?The median is the
value that minimizes
the sum of absolute
deviations.
Do you know?The mean is the
value that minimizes
the sum of squared
deviations.
The
MeanFor Ungrouped Data
First
TypeArithmetic Mean
Arithmetic MeanIt is also called as “simple
mean” or “unweighted
mean”. It is the sum of a
collection of numbers
divided by the number of
numbers in the collection.
FormulaArithmetic Mean:
𝒙 =𝚺𝒙
𝒏where: x = items/scores
n=number of items/scores
Example:What is the mean age
of a group of children
whose ages are 9, 11, 7,
10, 9, 8, 8, 7, 12, 7 and
13?
Second
TypeWeighted Mean
Weighted MeanIt takes into
consideration the proper
weights assigned to the
observed values
according to importance.
FormulaWeighted Mean:
𝒙 =𝚺𝒘𝒙
𝒘where: x=items/scores
w=weight of each item/score
Example:A student took 3 exams in Math. He
finished the first exam in 45 minutes
and got a grade of 88; 60 minutes
on the second exam and got 92;
and 90 minutes on the third exam
and got 85. What was the student’s
mean score for the three exams?
Third
TypeMean for Simple Frequency
Distribution
FormulaMean for Simple Frequency
Distribution:
𝒙 =𝚺𝒇𝒙
𝒏Where: f=frequency; x=items/scores;
and n=number of items
Example:x f 𝚺fx
8 2
2 3
6 4
7 3
5 5
6 2
9 1
n=20 𝚺fx=
Example:x f 𝚺fx
8 2 16
2 3 6
6 4 24
7 3 21
5 5 25
6 2 12
9 1 9
n = 20 𝚺fx = 114
Let’s
PracticeMeasures of Central Tendency
for Ungrouped Data
Problem 1:A freshman college student for the
following final grades with the
number of units for each subject. (a)
Find the weighted mean grade and
(b) if all the subjects have uniform or
equal number of units, what would
be the mean grade?
Table 1. A freshman college student
final grades with the number of units
for each subject.Subjects Grades Units
Math 1 2.50 3
English 1 2.00 3
Filipino 1 2.50 3
History 1 1.75 3
PE 1 1.50 2
Chemistry Lecture 3.00 2
Chemistry Laboratory 2.50 1
Problem 2:Gottfried Wilhelm Leibniz answered
20 calculus problems. He spent 1 ½
hours for the first 6 problems; 45
minutes for 3 problems; and 3 hours
for 11 problems. What was the
average time he spent for each
problem?
Recall:What are the
symbols we use for
each of measure of
central tendency?
Something to think about…
If you are given a
dataset/frequency
distribution that is already
grouped, what will be the
most challenging part?
Why?
Measures of
Central
TendencyFor Grouped Data
The
MeanFor Grouped Data
Long Method/Midpoint
MethodFormula:
𝒙 =𝚺𝒇𝒙
𝒏Where: f=frequency; x=class
marks; and n=number of samples
Example:
Class Intervalsf
(frequency)
x
(class mark)
fx
(frequency x
class mark)
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Example:
Class Intervalsf
(frequency)
x
(class mark)
fx
(frequency x
class mark)
62-68 8 65 520
69-75 5 72 360
76-82 9 79 711
83-89 11 86 946
90-96 12 93 1,116
97-103 10 100 1,000
104-110 5 107 535
111-117 5 114 570
n=65 𝚺fx=5,758
Short Method or
Assumed Mean Method
It is the class mark of the class
interval near the center of the
distribution or the class mark of
the interval with the highest
frequency.
FormulaAssumed Mean
Method:
𝒙 = 𝒙𝟎 +𝚺𝒇𝒅
𝒏𝒘
Where:𝒙𝟎 = 𝒂𝒔𝒔𝒖𝒎𝒆𝒅𝒎𝒆𝒂𝒏
𝒇 = 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚𝒏 = 𝒔𝒊𝒎𝒑𝒍𝒆 𝒔𝒊𝒛𝒆𝒅 = 𝒄𝒐𝒅𝒆𝒅 𝒗𝒂𝒍𝒖𝒆𝒘 = 𝒄𝒍𝒂𝒔𝒔 𝒘𝒊𝒅𝒕𝒉
Example:Class
Intervals
f
(frequency)
x
(Class Mark)
fx
(frequency x
class mark)
fd
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Steps involved in
the calculation of
Mean by the
Short MethodFor Grouped Data
Step 1:
Tabulate the scores
in a frequency
distribution.
Example:Class
Intervals
f
(frequency)
x
(Class Mark)
fx
(frequency x
class mark)
fd
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Step 1:Take the midpoint of an interval
somewhere near the center of
the frequency distribution and,
if possible, the interval should
contain the largest frequency.
Example:Class
Intervals
f
(frequency)
x
(Class Mark)
fx
(frequency x
class mark)
fd
62-68 8 65
69-75 5 72
76-82 9 79
83-89 11 86
90-96 12 93
97-103 10 100
104-110 5 107
111-117 5 114
n=65
Note:
That will be our
assumed mean.
Therefore, 𝒙𝟎 = 𝟗𝟑.
Step 3:Fill in the column for the deviations
from the assumed mean in units of
class intervals (d). Starting with 0 for the class interval having the 𝒙𝟎
going up assign positive values and
going down assign negative values
for d.
Example:Class
Intervals
f
(frequency)
x
(Class Mark)
d
(coded
value)
fd
62-68 8 65 -4
69-75 5 72 -3
76-82 9 79 -2
83-89 11 86 -1
90-96 12 93 0
97-103 10 100 +1
104-110 5 107 +2
111-117 5 114 +3
n=65
Step 4:The fd column is the product
of f and d. The values greater than the 𝒙𝟎 are
positive and the values less than 𝒙𝟎are negative.
Example:Class
Intervals
f
(frequency)
x
(Class Mark)
d
(coded
value)
fd
62-68 8 65 -4 -32
69-75 5 72 -3 -15
76-82 9 79 -2 -18
83-89 11 86 -1 -11
90-96 12 93 0 0
97-103 10 100 +1 +10
104-110 5 107 +2 +10
111-117 5 114 +3 +15
n=65
Step 5:
Get the sum of the 𝒇𝒅 values.
Example:Class
Intervals
f
(frequency)
x
(Class Mark)
d
(coded
value)
fd
62-68 8 65 -4 -32
69-75 5 72 -3 -15
76-82 9 79 -2 -18
83-89 11 86 -1 -11
90-96 12 93 0 0
97-103 10 100 +1 +10
104-110 5 107 +2 +10
111-117 5 114 +3 +15
n=65 𝚺fd=-41
Step 6:
Determine the
class width
(w).
Therefore,
The class
width is 7.
Step 7:
Substitute the
values obtained
to the formula.
Therefore,
The𝒙 = 𝟖𝟖. 𝟔.
Something to think about…
What can you say to the
mean we obtained using
the Midpoint Method and
Assumed Mean Method?
The
MedianFor Grouped Data
Formula:
𝒙 = 𝑳𝑳𝑹 +
𝒏𝟐− 𝒇 ≤
𝒇𝒘
Where:
𝑳𝑳𝑹 = 𝒍𝒐𝒘𝒆𝒓 𝒓𝒆𝒂𝒍 𝒍𝒊𝒎𝒊𝒕𝒇 ≤= 𝒄𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆 𝒍𝒆𝒔𝒔 𝒕𝒉𝒂𝒏 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
𝒇 = 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚𝒏 = 𝒔𝒂𝒎𝒑𝒍𝒆
𝒘 = 𝒄𝒍𝒂𝒔𝒔 𝒔𝒊𝒛𝒆
Example:Class Intervals
f
(frequency)
f≤
(cumulative less than
frequency)
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Steps Involved in
the Calculation
of MedianFor Grouped Data
Step 1:
Record the
cumulative
frequencies.
Example:Class Intervals
f
(frequency)
f≤
(cumulative less than
frequency)
62-68 8 8
69-75 5 13
76-82 9 22
83-89 11 33
90-96 12 45
97-103 10 55
104-110 5 60
111-117 5 65
n=65
Step 2:
Determine 𝒏
𝟐.
Therefore,
𝒏
𝟐= 𝟑𝟐. 𝟓.
Step 3:
Identify the class
interval in which
the 32.5th case falls.
Example:Class Intervals
f
(frequency)
f≤
(cumulative less than
frequency)
62-68 8 8
69-75 5 13
76-82 9 f≤=22
83-89 f=11 33
90-96 12 45
97-103 10 55
104-110 5 60
111-117 5 65
n=65
Therefore,
it is the 4th class
interval with exact
limits 82.5-89.5. 𝑳𝑳𝑹is 82.5.
Step 4:
Determine the
class width
(w).
Therefore,
the class
width is 7.
Step 5:
Substitute the
values obtained
to the formula.
Answer:
The median
is 89.2
The
ModeFor Grouped Data
Formula:
𝒙 = 𝑳𝑳𝑹 +𝒅𝒖
𝒅𝒖 + 𝒅𝒍𝒘
Where:𝑳𝑳𝑹 = 𝒍𝒐𝒘𝒆𝒓 𝒓𝒆𝒂𝒍 𝒍𝒊𝒎𝒊𝒕
𝒅𝒖 = difference between the highest
frequency and the frequency of the
interval below it
𝒅𝒍 = difference between the highest
frequency and the frequency of the
interval above it
𝒘 =class size
Example:Class Intervals f
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Steps Involved in
the Calculation
of ModeFor Grouped Data
Step 1:
Tabulate the scores
in a frequency
distribution.
Example:Class Intervals f
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Step 1:
Determine the class
interval with the
highest frequency.
Example:Class Intervals f
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Then,
the 𝑳𝑳𝑹 will be the
lower real limit of the
class interval with the
highest frequency.
Thus,
𝑳𝑳𝑹 = 𝟖𝟗. 𝟓
Step 2:Determine 𝒅𝒖. 𝒅𝒖 is the
difference between the
highest frequency and
the frequency of the
interval below it.
Example:Class Intervals f
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Thus,
𝒅𝒖 = 𝟏𝟐 − 𝟏𝟏 = 𝟏
Step 3:Determine 𝒅𝒍. 𝒅𝒍 is the
difference between the
highest frequency and
the frequency of the
interval above it.
Example:Class Intervals f
62-68 8
69-75 5
76-82 9
83-89 11
90-96 12
97-103 10
104-110 5
111-117 5
n=65
Thus,
𝒅𝒍 = 𝟏𝟐 − 𝟏𝟎 = 𝟐
Step 4:
Determine the
class width.
Thus,
the class
width is 7.
Step 5:
Substitute the
values obtained
to the formula.
Thus,
the mode is
91.8.
Kinds of
ModeFor Grouped Data
The True
ModeKinds of Mode
Formula:
𝒙 = 𝟑𝑴𝒆𝒅𝒊𝒂𝒏 − 𝟐𝑴𝒆𝒂𝒏
From the example:
using the true
mode formula,
the mode is 90.4.
Crude
ModeKinds of Mode
Crude Mode
It is the midpoint of
the class interval
with the highest
frequency.
Formula:
𝒙 =𝒍𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝒖𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕
𝟐
From the example:
using crude
mode method,
the mode is 93.