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Measures of Central Tendency MODE (Grouped Data) Prepared by: Ryan L. Race Jenelyn A. Samsaman Rafaela M. Sarmiento Jorge O. Dela Cruz Maria Theresa S. Parajas Luningning B. Federizo
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Mode (Grouped Data)

Jun 03, 2018

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Purpose/Rationale of this Module

This module is designed to provide studentswith a step by step discussion on thecomputation of the mode for grouped

data.It is to let the students discover for

themselves how to compute for the mode(grouped data) through an easy visualpresentation of the subject.

Enjoy!

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What Will You Learn From ThisModule?

 After studying this module, you shouldbe able to:

1. Compute the mode for groupeddata.

2. Use mode for grouped data toanalyze and interpret data to solveproblems in daily life.

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Let’s See What You Already

Know

Before you start studying this

module, take the following testfirst to find out how prepared you are to solve for the mode of

grouped data.

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 Find the mode of each of thefollowing sets of numbers.

a. 2, 4, 5, 1, 4, 6

b. 77, 80, 90, 65, 77, 89, 80

c. 1299, 2580, 4098, 9100, 1100

 Answer

 Answer

 Answer

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 Answer: 4

Click meto answer

letter b.

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 Answer: 77 and 80

Click meto answerletter c.

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 Answer: No Mode

Go Back CONTINUE

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For the next set of questions, refer tothe given frequency distributiontable of the grades of a group of

students.

CONTINUE

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1. What is the modal class?

Grades Number of students

90 – 94 2

85 – 89 5

80 – 84 8

75 – 79 12

70 – 74 3

 ANSWER

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 ANSWER: 75 – 79

The modal class is the class with thehighest frequency.

CONTINUE

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2. What is the lower class boundaryof the modal class?

Grades Number of students

90 – 94 2

85 – 89 5

80 – 84 8

(Modal Class) 75 – 79  1270 – 74 3

 ANSWER

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 ANSWER: Lmo = 74.5

CONTINUE

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3. What is the frequency of themodal class?

Grades Number of students

90 – 94 2

85 – 89 5

80 – 84 8

(Modal Class) 75 – 79 12

70 – 74 3

 ANSWER

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 ANSWER: f mo = 12 

CONTINUE

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4. What is the frequency of the classpreceding the modal class?

Grades Number of students

90 – 94 2

85 – 89 5

80 – 84 8

(Modal Class)  75 – 79 12

70 – 74 3

 ANSWER

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 ANSWER: f 1 = 8

CONTINUE

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5. What is the frequency of the classafter the modal class?

Grades Number of students

90 – 94 2

85 – 89 5

80 – 84 8

(Modal Class) 75 – 79 12

70 – 74 3

 ANSWER

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 ANSWER: f 2 = 3

CONTINUE

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 ANSWER: i = 5

CONTINUE

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74.5 is the lower

class boundary ofthe modal class

(Lmo)

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12 is the frequency

of the modal class(f mo)

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3 is the frequency of

the class after themodal class (f 2).

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5 is the class size (i).

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T l f th d f d d t

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To solve for the mode of grouped data,use the formula

Mo = Lmo+{( f mo  – f 1 )/( 2f mo  – f 1  – f 2 )}iwhere,

Lmo is the lower class boundary of the modal

class,f mo is the frequency of the modal class,

f 1 is the frequency of the class preceding the

modal class,f 2 is the frequency of the class after the

modal class, and

i is the class size. CONTINUE

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Let’s Practice! 

The followingdistribution gives thenumber of hours

allotted by 50students to do theirassignments in aweek. Find the modal

hour.

Hours Number ofStudents

1 - 4 9

5 - 8 12

9 - 12 14

13 - 16 10

17 - 20 4SOLUTION

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Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Hours Number ofStudents

1 - 4 9

5 - 8 12

9 - 12 14

13 - 16 10

17 - 20 4

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  9 – 12Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Hours Numberof

Students1 - 4 9

5 - 8 12

9 - 12 14

13 - 16 10

17 - 20 4

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  9 – 12

8.5

14

Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Hours Numberof

Students1 - 4 9

5 - 8 12

9 - 12 14

13 - 16 10

17 - 20 4

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  9 – 12

8.5

14

12

Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Hours Numberof

Students1 - 4 9

5 - 8 12

9 - 12 14

13 - 16 10

17 - 20 4

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  9 – 12

8.5

14

12

10

Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Hours Numberof

Students1 - 4 9

5 - 8 12

9 - 12 14

13 - 16 10

17 - 20 4

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  9 – 12

8.5

14

12

10

4

Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Compute the mode

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More Practice?

Find the mode using thefrequency distributionof the heights of 40

students.

Height(inches) No. ofStudents

52 – 54 2

55 – 57 458 – 60 3

61- 63 8

64 – 66 10

67 – 69 9

70 - 72 4

SKIP

SOLUTION

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Mo = 63.5 + {[10 – 8] / [2(10) – 8 - 9]}3

= 63.5 + (2/3)3

= 63.5 + 2

= 65.5NEXT

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EVALUATE YOURSELF

Find the mode of 20students whosescores on a 15-point

test are given in thefollowing distribution:

Scores No. ofStudents

1 – 3 1

4 – 6 4

7 – 9 8

10 – 12 5

13 - 15 2

SOLUTION

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Mo = 6.5 + {[8 – 4] / [2(8) – 4 - 5]}3

= 6.5 + (4/7)3

= 6.5 + 1.714

= 8.21NEXT

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