ADD MATHS PROJECT 2009

NAME : Timothy Tai Ming Zheng

IC No : 921104-14-6589

CLASS / SET : 5C Set 2

TEACHER’S NAME : Ms.Leong

School Name : Wesley Methodist School

Content

No. Title Page

1 Introduction 2

2 Problem Solving strategies: (a) 3-4

1(a) (i) 5-6

(ii) 7

(iii) 8

1(b)(i) 9

(ii) 10

(iii) 11

1(c)(i) 12

(ii) 13

(iii) 14

3 Further Exploration: (a) 15-16

(b) 17-18

(c) 19

The reasons for carrying out this project

Pupils have been required to carry out this project so that they can apply correctly a variety of formulae’s and strategies to solve problems. Besides that, this project will also help to develop mathematical knowledge through problem solving in a way that increases students’ interest and confidence. Furthermore, students will also be able to improve their thinking skills. This project also aims to promote effective mathematical communication in the students. The students will be able to express mathematical ideas precisely and will be acquire and adapt a learning environment that stimulates and enhances effective learning. Last but not least, the reason for carrying out this project is also to develop a keen interest towards mathematics.

The moral values a student will learn from this project

There are many moral values that a student can benefit from this project. Among them, the students will learn to be meticulous at a task at hand. The students will have to learn how to get along with each other and maximize each others’ abilities. Doing this project in a group will also promote teamwork in each student as teamwork is prudent and indispensable. They will also be able to be more diligent and learn how to organize their time. Students will also learn the importance of responsibility to ensure that this project is completed

In this campaign, you are required to obtain the heights and weights of 50 students in Form 4 or Form 5.

a) Tabulate your data using the following format: I. Identifying the problem: I had to gather the weight and height of 50 different students from Form 4 or

Form 5.II. Strategy: I surveyed classes of Form 4 and Form 5 and got the weight and height of 50 different students.

III. Working:

Student Height(cm) Weight(kg)

Hooi Sin Yee 1.55 41

Yap Carl Mann 157 43

Diksha Dua 167 45

Lim Zheng Ern 161 45

Low Zi Yan 155 45

Chia Su-Yen 158 46

Elizabeth Tan 1.59 47

Grace Lim Yen-Tiung 160 47

Lo Chia Wei 164 47

Loong Joo Lee 162 47

Lee Ming Ling, Emelia 160 48

Tan Li Li 164 48

Catherine Kok 1.61 50

Fionna Tan Sze Fong 160 50

Goh Siau Wei 163 50

Lim Sin Yein 162 50

Pan Chern Li 165 50

Yong Joy Lene 154 50

Fiona Tan Sze Fong 1.60 50

Lim Khai Shing 173 52

Loh Chou Yun 168 52

Mark William Chew Houwei 182 52

Tan Wai Hwa 172 52

Stephanie Heng Shean Fei 153 52

Lionel Wong 1.76 52

Choong Sheue Li 1.58 54

John Tay Kwang Ming 165 55

Lee Wen Han 1.75 55

Choo Jin Wai 172 56

Jonathan Ong 1.77 58

Chua Yue Han 179 60

Tan Eng Yew 170 61

Eugene Yu Tze Ping 180 62

Melvyn Leong Jun Lam 171 62

Tan Ti Yoong, Alex 180 62

Lee Ki Yip 172 63

Lim Zhuo Wei 175 64

Daniel Wan Kok Jun 174 65

Gan Kok Jin 178 65

Ler Zhi Hui 181 66

Jonathan Yeoh Chong Yoong 188 67

Leong Wei Ian 178 69

Lim Kah Wai-Calvin 174 73

Timothy Tai Ming Zheng 170 73

Tan Yee Wen 168 74

Lim Xian Wei 173 77

Tang Chee Yan 181 80

Brian Naveen a/l Mariyarajan 175 84

David Ho 1.85 90

Joey Khoo Hwai Joo 176 90

IV. Conclusion: The table was completed. Weight and height of 50 students in a class

b)Construct a frequency distribution table for the weights obtained using a suitable class interval.I. Represent your data from the frequency distribution table by using three different statistical graphs

Identifying the problem: I needed to construct a frequency distribution table for the weights obtained using a suitable class interval.

1. Ogivea. Strategy:

i. A table of suitable class boundary, frequency, lower boundary and cumulative frequency is tabulated.

ii. An ogive graph is plotted.

b. Working:Weight (kg) Frequency Cumulative Frequency Upper Boundary

31-40 0 0 35.5

41-50 19 19 45.5

51-60 12 31 55.5

61-70 11 42 65.5

71-80 5 47 75.5

81-90 3 50 85.5

c. Conclusion :

The table is completely tabulated and the graph is plotted. The ogive is completed. The data can be represented in the form of an ogive.

2. Histogramc. Strategy:

i. A table of suitable class boundary, frequency, lower boundary and cumulative frequency is tabulated.

ii. A bar chart is plotted.

d. Working:Weight (kg) Frequency

31-40 0

41-50 19

51-60 12

61-70 11

71-80 5

81-90 3

c) Conclusion :

The table is completely tabulated and the graph is plotted. The histogram is completed. Data can be represented in the form of a histogram.

3. Frequency Polygon

a. Strategy:i. A table of suitable class boundary, frequency, lower boundary and cumulative

frequency is tabulated.ii. A frequency polygon is plotted.

b. Working:

Weight (kg) Frequency

31-40 0

41-50 19

51-60 12

61-70 11

71-80 5

81-90 3

c. Conclusion :

The table is completely tabulated and the graph is plotted. Data can be represented in the form of a frequency polygon.

Ogive

Histogram

Frequency Polygon

Weight,(kg)

Frequency,ƒ

Midpoint,χ

Midpoint x Frequency,

Cumulative Frequency

UpperBoundary

ƒχ41-50 19 45.5 864.5 19 50.551-60 12 55.5 666 31 60.561-70 11 65.5 720.5 42 70.571-80 5 75.5 377.5 47 80.581-90 3 85.5 256.5 50 90.5

I. Mean

a. Strategy :i. A table was constructed to display the product of frequency and midpoint.ii. The formula, Mean = ∑ f x , was applied. ∑f

iii. The calculation of the product of the frequency and midpoint over total frequency is needed to find the mean.

b. Working : Means, = sum of (frequency x class mark) Total frequency = Σƒχ Σƒ

= 864.5+666+720.5+377.5+256.5 50 = 2885 50 = 57.7 kg c. Conclusion :The mean calculated for the data is 59.10.

2. Median

a. Strategy: a table was constructed to display the lower boundary, cumulative frequency and frequency.

b. Working:Median Class = 51-60

Therefore , median, m = L + N/2-F x C

fм

= 50.5 + 50/2-19 x 10 12 = 50.5 + 5 = 55.5 kg

c. Conclusion : The median calculated for the data is 55.5kg

3. Mode

a. Identifying the problem: I need to find the mode.b. Strategy: The histogram is used to find the mode.

c. Conclusion :

The estimated value of the mode from the data is 47kg.

Therefore, Based on the three methods use, the best measure of central tendency to represent the data is the mean. The mean is the most suitable because it can detect the small changes in the data. The mode is not suitable and cannot be used because it only measures the highest frequency. The median is also unsuitable as small changes in the data cannot be detected and measured. Since central is the "middle" value or perhaps a typical value of the data used like an average, the median is the best measure of central tendency in this case.

Standard deviation

Method 1 : Using calculator

n Σχ Σχ² Standard deviation

50 57.7 28.85 832.32 12.60

Method 2 : Using formula 1

a. Identifying the problem : The standard deviation has to be calculated using the following formulas:

b. strategy:

i. With the help of a table showing various data, the standard

deviation was counted using the formula for grouped data.

c. Working:

Weight (kg) Frequency,ƒ Midpoint , χ ƒχ ƒχ²41-50 19 45.5 864.5 39334.7551-60 12 55.5 666 36963.0061-70 11 65.5 720.5 47192.7571-80 5 75.5 377.5 28501.2581-90 3 85.5 256.5 21930.75

Σƒ = 50 Σƒχ = 2885 Σƒχ² = 173922.50

Mean, = 55.3kgσ ² = Σƒχ² - (Σƒχ)² Σƒ = 173922.50/50-(2885/50)² = 158.81kgσ = 158.811/2

=12.60kgMethod 3 : Using formula 2

a. Identifying the problem : The standard deviation has to be calculated using the following formulas:

b. strategy:

i. With the help of a table showing various data, the standard

deviation was counted using the formula for grouped data.

c. Working:

Weight (kg) Frequency, ƒ Midpoint, χ ( χ - )² ƒ( χ - )²41-50 19 44.5 174.24 3310.5651-60 12 55.5 4.84 58.0861-70 11 65.5 60.84 669.2471-80 5 75.5 316.84 1584.281-90 3 85.5 772.84 2318.52

Σƒ = 50 Σ( χ - )²= 1329.6

Σƒ( χ - )²= 7940.6

= 55.3kg

σ² = Σƒ( χ - )² Σƒ = 7940.6 50 = 158.81

σ = 176.361/2

= 12.60kg Conclusion: The value of the standard deviation σ = 12.60kg indicates that there is a wide dispersion from the value of mean obtained.

Student Height (m) Weight (kg) BMI

Hooi Sin Yee 1.55 41 17.01

Yap Carl Mann 157 43 17.24

Diksha Dua 167 45 16.16

Lim Zheng Ern 161 45 18.03

Low Zi Yan 155 45 18.21

Chia Su-Yen 158 46 18.43

Elizabeth Tan 1.59 47 18.59

Grace Lim Yen-Tiung 160 47 19.95

Lo Chia Wei 164 47 15.59

Loong Joo Lee 162 47 17.91

Lee Ming Ling, Emelia 160 48 18.75

Tan Li Li 164 48 17.85

Catherine Kok 1.61 50 18.40

Fionna Tan Sze Fong 160 50 24.11

Goh Siau Wei 163 50 17.72

Lim Sin Yein 162 50 18.31

Pan Chern Li 165 50 18.37

Yong Joy Lene 154 50 21.08

Fiona Tan Sze Fong 1.60 50 19.53

Lim Khai Shing 173 52 17.37

Loh Chou Yun 168 52 18.42

Mark William Chew Houwei 182 52

15.70

Tan Wai Hwa 172 52 17.58

Stephanie Heng Shean Fei 153 52 22.21

Lionel Wong 1.76 52 16.78

Choong Sheue Li 1.58 54 21.68

John Tay Kwang Ming 165 55 18.42

Lee Wen Han 1.75 55 19.83

Choo Jin Wai 172 56 18.93

Jonathan Ong 1.77 58 18.51

Chua Yue Han 179 60 17.33

Tan Eng Yew 170 61 21.11

Eugene Yu Tze Ping 180 62 18.60

Melvyn Leong Jun Lam 171 62 21.22

Tan Ti Yoong, Alex 180 62 18.25

Lee Ki Yip 172 63 16.22

Lim Zhuo Wei 175 64 18.36

Daniel Wan Kok Jun 174 65 24.06

Gan Kok Jin 178 65 20.83

Ler Zhi Hui 181 66 20.10

Jonathan Yeoh Chong Yoong 188 67

18.96

Leong Wei Ian 178 69 21,78

Lim Kah Wai-Calvin 174 73 24.11

Timothy Tai Ming Zheng 170 73 25.26

Tan Yee Wen 168 74 26.22

Lim Xian Wei 173 77 25.73

Tang Chee Yan 181 80 24.42

Brian Naveen a/l Mariyarajan 175 84

27.43

David Ho 1.85 90 26.30

Joey Khoo Hwai Joo 176 90 29.05

BMI Frequency Cumulative frequency Upper boundary

10-14 0 0 14.5

15-19 29 29 19.5

20-24 11 40 24.5

25-29 10 50 29.5

Ogive

From the ogive above, it shows that

About 48 % of the student are underweight About 40% of the student are normal weight About 12% of the student are overweight 0% of the student are obese The mean of the BMI = Σƒχ

Σƒ = 1130 50 = 22.6

None of the student are obese but most of the students are underweight or normal weight.

From the investigation and the report made, the problems of obesity must be persevered since obesity cause not only physical change but also endanger the life of the person. In order to prevent obesity, diet control is very important. Have a balanced diet everyday for enough nutrition and energy especially for those who undergo puberty.Eat more fruit and vegetables to gain enough vitamins and fibre, taken enough amount protein and carbohydrate from meat and rice and also noodle. Prevent from drinking too much carbonate drink since it contains a lot of sugar that might cause obesity. We also should avoid from taking too much junk food since there are no nutrition.

Moreover, we must have enough exercise for our body. Exercise twice or three times in a week for a healthy body. Take a walk to digest food well after meals.