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.