Statistic Form 4

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STATISTICS

ADDITIONAL

MATHEMATICS

MODULE 13

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CHAPTER 7 : STATISTICS

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CONTENTS PAGE

7.0 STATISTICS

7.1 Concept Map 2

7.2 Mean, Mode and Median of Grouped Data 35

7.3 Test Your Self 1 6

7.4 Range and Interquartile Range of

Grouped Data 7 8

7.5 Test Your Self 2 89

7.6 Past Year Question 10

7.7 Assessment 11 13

Answers 14



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7.1 CONCEPT MAP

STATISTICS

Measures of

Central Tendency

Measures of

Dispersion

Mean

x

Mode &

ModalClass

Median

m=

Interquartile

Range

Q3Q1

Range

First Quartile

Q1 =Third Quartile

Q3 =Histogram



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7.2 MEAN, MODE AND MEDIAN OF GROUPED DATA

(Using a formula)

Activity 1: The number of vehicles that pass by a toll plaza from 1 p.m to 2 p.m. for 60

concecutive days is shown in the table below.

Number of vehicles Number of days

5950 4

6960 10

7970 24

8980 16

9990 6

(a) Calculate themeanof the number of cars.

(b) State themodal class.(c) Draw a histogram and estimate themodeof the number of cars from a

histogram .

(d) Calculate themedianof the number of cars using formula.

Solution :(a)

Number of

vehicles

Number of

days( f )

Midpoint

(x)

fx

5950 4 54.5 218

6960 10 ( ) ( )

7970 24 ( ) 1788

8980 16 84.5 13529990 6 94.5 567

f = ( ) )(fx

Mean,)(

)(

f

fxx = 76 . 17 vehicles.

(b) Modal class = 7970

(c)

Cl ass bou ndar y N umber of days (frequency)

49.5 59.5 4

10

24

16

6

This is the class with

the highest frequency



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(c) The histogram is shown below

49.5 59.5 69.5 79.5 89.5 99.5 Number of vehicles

Mode = 76 vehicles

Frequency

5

10

15

20

25



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(d)

Number of

vehicles

Number of days

(f )

Cumulative

frequency

5950 4 4

6960 10 (14)7970 (24) 38

8980 16 ( )

9990 6 ( )

Step1 : Median class is given by = 302

60

2

TTTn

Therefore, the median class is 7970 cm.

Step 2 : Median = cf

Fn

Lm

2

= (___)

24

142

60

( __ )

= 76.17 vehicles

L= lower boundary of the medianclass = 69.5

n = 60fF = cumulative fr equency before the

median class =14

fm= fr equency of the median class

=24

c = size of the median class

= upper boundarylower

boundar

m



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7.3 Test Yourself 1

For each set of data below, calculate the median by using a formula.

1. The table below shows the age of a group of students.

Age (years) 0 4 5 9 10 14 15 19 20 24 25 29

Number of students 10 20 25 32 28 25

Age F r equency Cumulative frequency

0 4

5 9

10 14

15 19

20 24

25 29

Median =

2. The table below shows the weight of a group of babies . 2. The table below shows theweight of a group of babies.

Weight

(kg)

3.0 3.4 3.5 3.9 4.0 4.4 4.5 4.9 5.0 5.4

Frequency 21 14 25 18 16

Weight (kg) F r equency Cumulative f requency

3.0 3.4

3.5 3.9

4.0 4.4

4.5 4.95.0 5.4

Median =



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7.4 RANGE AND INTERQUARTILE RANGE OF GROUPED DATA

( Using a formula)

Activity 2 : The table below shows the speed of 100 vehicles that pass by a small town

in a certain period of time.

Speed (km hour1

) Number of vehicles

40 44 8

45 49 18

50 54 16

55 59 26

60 64 22

65 69 10

Calculate(a) the range(b) the interquartile range using formula.

Solution :

(a) Range =2

)(

2

6965

= 67 42

= 25 km hour1

(b)

Speed N umber of vehicles(f)

Cumulativefrequency

40 44 8 (8)

45 49 (18) 26

50 54 16 ( )

55 59 26 (68)

60 64 (22) 90

65 69 10 ( )

Step 1 : The Q3class is given by = 254

100

4

TTTn

Q1class = . 45 49

Midpoint of thehighest class

Midpoint of the lowest class

1

3



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Step 2 : Q1 = c

Q

Fn

Q fL

1

1

4

= ( ) +

18

)(4

100

(5)

= 49.22 km hour1

Step 3 : The Q3class is given by = 75)100(

4

3

4

3 TTTn

Q3class = . 60

64

Step 4 : Q3 = c

Q

Fn

Q fL

3

3

4

3

= 59.5 +

)(

68(___)4

3

(5)

= 61.09 km hour1

Hence, the interquartile range = Q3 Q1

= ( ) ( )

= 11.87 km hour1

7.5 Test Yourself 2

For each set of data below, find (a) the range (b) the interquartile range by calculation.

L Q1 = lower boundary of th e class Q1class = 44.5

n = 100fF = cumulative frequency before theQ1class= 8

fQ1= frequency of theQ1 class= 18

c = class size

= upper boundarylower boundary

= 49.544.5

= 5

L Q3 = lower boundary of th e class

Q1 class = 59.5

n = 100fF = cumulative frequency before

th eQ3class= 68

fQ3= frequency of theQ1 class= 22

c = class size

= upper boundary

lowerboundary = 64.559.5

= 5



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1. The table below shows the number of chickens sold over a period of 100 days.

Number of chickens 10 14 15 19 20 24 25 29 30 34

Number of days 20 25 30 15 10

(a)

(b)

2. The table below shows the transport expenses of the workers of an electronic firm to their

working place.

Expenses

(RM)

0.10 1.00 1.102.00 2.103.00 3.10 4.00 4.10 5.00

Number of

workers

5 9 12 8 6

(a)

(b)



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7.6 Past Year Question

SPM 2005 : Paper 2 Question 4

The diagram below is a histogram which represents the distribution of the marks obtained

by 40 pupils in a test.

Number of Pupils

0.5 10.5 20.5 30.5 40.5 50.5 Marks

(a) Without using an ogive, calculate the median mark.

[3marks]

2

4

6

8

10

14

12

0



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7.7 Assesment

1. Table below shows the frequency distribution of the age of a group of people in atown.

Age 16

20 21

25 26

30 31

35 36

40 41

45Frequency 18 p 15 8 4 1

Given that the median of the age of a group of people is 24.25, find the value ofp.

[4marks]

2.Use graph paper to answer this question.

The following table shows the distance of 40 workers houses to their place of work.

D istan ce (km) N umber s of

workers

1 2 4

3

4 85 6 14

7 8 10

9 10 4



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(a) Draw a histogram and estimate the modal distance from your histogram

[4marks]

(c) Without drawing an ogive, find the median distance. [3marks]

(c) Calculate the mean distance. [3marks]



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3. The table below shows the distribution of the ages of 100 teachers in a secondary

school.

Age(years) Number of teachers

< 30 8< 35 22

< 40 42

< 45 68

< 50 88

< 55 98

< 60 100

(a) Based on the given table, complete the following :

A ges(year s) F r equ en cy 25 29

[2marks]

(b) Without drawing an ogive, calculate the interquartile range of the above

distribution. [6marks]



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Answers:

Test Yourself 11. Median = 16.84 years

2. Median = 4.19 kg

Test Yourself 2

1. ( a ) range = 20

( b ) Q1 = 15.5Q3 = 24.5

Interquartile Range = 9 chickens

2. ( a ) range = 4

( b ) Interquartile Range = RM 1. 94

SPM 2005

Median = 24.07

Assessment

1. p = 20

2. (a) Mode = 5.7 km

(b) 5.64 km

(c) x =

f

fx=

40

224= 5.6 km

2. (a)

Age F r equency Cumulative

frequency

25 29 18 8

30 34 14 22

35 39 20 42

40 44 26 68

45 49 20 88

50 56 10 98

55 59 2 100

(b) Q1 = 35.25 years

Q3 = 46.25 years

Interquartile Range = 11 years



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STATISTICS

ADDITIONAL

MATHEMATICS

MODULE 14



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CHAPTER 7 : STATISTICS

CONTENT PAGE

7.1. Concept Map 2

7.2 Standard Deviation And Variance Of Ungrouped Data 3

7.3. Activity 1 5

7.4. Standard Deviation And Variance Of Ungrouped Data

( In Frequency Table )

8

7.5. Standard Deviation And Variance Of Grouped Data

( With Class Interval )

9

7.6. Activity 2 11

7.7. Activity 3 : SPM Focus Practice 13

7.8. Answers 16



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7.1 CONCEPT MAP

STATISTICS

MEASURES OF CENTRALTENDENCY MEASURES OF DISPERSION

Mean

x =n

x

GROUPED

DATA

UNGROUPED

DATA

Standard Deviation

= 22

)(xn

x

Variance

= 2

2

)(xn

x

Standard Deviation

= 22

)(xf

fx

Variance

= 22

)(xf

fx

UNGROUPED

DATA

GROUPED

DATA

Mean

x =

f

fx



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7.2: STANDARD DEVIATION AND VARIANCE OF UNGROUPED DATA

Example 1: Calculate the mean, standard deviation and variance of the set of data 15,

17, 20, 23, 25.

Solution :

Step 1: Find the sum of squares of x

x 15 17 20 23 25 x = 100

x 225 400 625 x = 2068

Step 2: Substitute the values into the formulae

Mean, x = n

x=

100

= 20

Standard deviation, = 22

)(xn

x

= 2)(

5

= 3.688

Variance, = (standard deviation) = ( ) = 13.6

Standard deviation,

=

2

)(xn

where

2x = sum of squares of xn = number of data

x = mean =n

x

Variance,

= 22

)(xx



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Example 2: The set of six integersp, q,6, 10, 15 and 18 has a mean of 9 and a standarddeviation of 6. Find

(a) the value ofp + q,

(b) the value ofp and ofq ifq>p.

Solution :

(a) Given that, mean, x = 9,

n

x= 9

9

6

156

qp

p+ q + 49 = 9 x 6

p+ q = 54 ( )

p+ q = 5

(b) Given that, standard deviation, = 6

2

2

)(xn

x

= 6

2

2

)(xn

x

= 6

2

222222

96

1815106qp

36668522

qp

366

68522

qp+ 81

6

68522 qp

p +q + 685 = 117 x 6

p +q = ( ) 685

(5 q) + q = 1725 10q+ q +q - 17 = 0

2q - 10q+ ( ) = 0

2, q - 5q+ 4 = 0(q 1)(q ) = 0

(q 1) = 0 or (q 4) = 0

q= 1 or q= 4Whenq = 1,p = 5 1 = 4 (not accepted because it does not satisfy the condition q > p )

Whenq = 4,p = 5 4 =1 (accepted)

From (a),p + q =5

p= 5 -q



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7.3 ACTIVITY 1

1. Find the mean, variance and standard deviation of the set of data 3, 8, 5, 7, 9, 6.

2. Find the mean, variance and standard deviation of the set of data 2.3, 2.8, 3.0, 3.2,2.5, 3.9, 3.3.



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3. The sum of 10 numbers is 180. The sum of the squares of these numbers is 3360.Find the variance of these 10 numbers.

4. Given a set of data (k 2),k,k, (5k 2), wherek 0, find the value of kfor which the

mean equals the standard deviation.



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5. The mean and the standard deviation for the numbers 1, 3, 5, 6 and 8 are 4.6 and 2.42respectively. Find in terms ofk

(a) The mean for numbers 1 +k, 3 +k, 5 +k, 6 +kand 8 + k.

(b) The standard deviation for the numbers k+ 4, 3k+ 4, 5k+ 4, 6k+ 4 and 8k+ 4.



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7.4: STANDARD DEVIATION AND VARIANCE OF UNGROUPED DATA (in

Frequency

Table)

Example : A group of 90 typist took a typing test. They were given a document to type.The following table shows the times they took to type.

Time(minutes), x Number of typists, f

15 10

16 28

17 24

18 13

19 10

29 5

Calculate

(a) the mean,(b) the standard deviation

Solution :

Time,(x) f f x fx

15 10 150 2250

16 ( ) 448 7168

17 24 ( ) 6937

18 13 234 4212

19 10 190 ( )

20 5 100 2000

f = 90 fx 261762fx

(a) Mean, x =

171530

f

fxminutes

Variance,

= 22

)(

f

fx

Standard deviation,

=

22

)(xfx

wheref = frequency

x = mean =

f

fx



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(b) Standard deviation,

= 22

)(xf

fx

= 217

90

= 1.36 minutes

7.5: STANDARD DEVIATION AND VARIANCE OF GROUPED DATA (with class

interval)

Example : The following table shows the distribution of heights of a group of 40 students.

Height(cm) Nu mber of students

159 162 1

163 166 4

167 170 11

171 174 12

175 178 6179 182 4

183 186 2

Calculate

(a) the mean,

(b) the standard deviationof the distribution.

Standard deviation,

=

2)(xfx

where

f = frequency

x = class midpoint

x = mean =

f

fx

Variance,

=

2

2

)(xfx



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Solution:

Step 1:

Height(cm Mid-point,x f fx fx

159 162 160.5 1 160.5 25760.25

163 166 164.5 4 ( ) 108241.00

167 170 ( ) 11 1853.5 312314.75

171 174 172.5 ( ) 2070.0 357075.00

175 178 176.5 6 1059.0 ( )

179 182 ( ) 4 722.0 130321.00

183 186 184.5 2 369.0 68080.50

40f fx 11887062fx

Step 2: Substitute the values into the formulae

(a) Mean, x =

3.1726892

f

fxcm

(b) Standard deviation,

= 22

)(xf

fx

=

40

1188706

= 5.51 cm



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7.6 ACTIVITY 2:

1. Find the mean and variance of the set of data below.

Number of tickets 26 28 30 32Number of days 10 12 15 11

The table above shows the number of tickets sold over a period of 48 days.



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2. Find the mean and standard deviation of the set of data below.

Score 10 14 15 19 20 24 25 29 30 34

Frequency 6 10 14 12 8

The table above shows the scores of a group of students who fail in a test.



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7.7 ACTIVITY 3 : SPM FOCUS PRACTICE

1. SPM 2005, PAPER 1. QUESTION 23.

The mean of four numbers is m. The sum of squares of the numbers is 100 and thestandard deviation is 3k.

Expressmin terms ofk. [3marks]

2. SPM 2003, PAPER 2. QUESTION 5.

A set of examination marks 654321 ,,,,, xxxxxx has a mean of 5 and a standard

deviation of 1.5.

(a) Find

(i) the sum of the marks,x ,

(ii) the sum of the squares of the marks, 2

x . [3marks](b) Each mark is multiplied by 2 and then 3 is added to it.

Find, for the new set of marks,

(i) the mean,(ii) the variance. [4marks]



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3. SPM 2004, PAPER 2, QUESTION 4.

A set of data consists of 10 numbers. The sum of the numbers is 150 and the sum of

the squares of the numbers is 2472.

(a) Find the mean and variance of the 10 numbers. [3marks]

(b) Another number is added to the set of data and the mean is increased by 1.Find

(i) the value of this number,

(ii) the standard deviation of the set of the set of 11 numbers. [3marks]



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4. SPM 2005, PAPER 2. QUESTION 4(b)

The diagram below is a histogram which represents the distribution of the marks

obtained by 40 pupils in a test.

Number of Pupils

0.5 10.5 20.5 30.5 40.5 50.5

(b) Calculate the standard deviation of the distribution.

2

4

6

8

10

14

12

0



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7.8 ANSWERS

ACTIVITY 1:

1. x = 6.333

= 3.893

= 1.973

2. x = 3

= 0.2457

= 0.4957

3. = 12 4. k = 2

5. (a) 5

235 k

(b) 2.42k

ACTIVITY 2:

1. x = 29.13 = 4.193

2. x = 22.6 = 6.216

ACTIVITY 3:

1. m = 25 9k 2. (a) (i) 30x(ii) 5.1632 x

(b) (i) x = 13(ii) = 9

3. (a) x = 15

= 22.2

(b) (i) k = 26(ii) = 5.494

4. (b) = 11.74



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Statistic Form 4

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