Mode (Grouped Data)

8/12/2019 Mode (Grouped Data)

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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



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?


90 – 94 2

85 – 89 5

80 – 84 8

(Modal Class) 75 – 79 1270 – 74 3

ANSWER



ANSWER: Lmo = 74.5

CONTINUE

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3. What is the frequency of themodal class?


90 – 94 2

85 – 89 5

80 – 84 8

(Modal Class) 75 – 79 12

70 – 74 3

ANSWER



ANSWER: f mo = 12

CONTINUE

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


90 – 94 2

85 – 89 5

80 – 84 8


70 – 74 3

ANSWER



ANSWER: f 1 = 8

CONTINUE

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


90 – 94 2

85 – 89 5

80 – 84 8


70 – 74 3

ANSWER



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)



12 is the frequency

of the modal class(f mo)





3 is the frequency of

the class after themodal class (f 2).



5 is the class size (i).



T l f th d f d d t



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



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





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



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



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



9 – 12

8.5

14

12

10

4

Modal Class :

Lmo :

f mo :

f 1 :

f 2 :

i :

Compute the mode





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



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

= 63.5 + (2/3)3

= 63.5 + 2

= 65.5NEXT



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



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

= 6.5 + (4/7)3

= 6.5 + 1.714

= 8.21NEXT







Documents

Mode (Grouped Data)