17/6/2015 12:11:00 AM

SMK DATO’ PENGGAWA BARAT

82000 PONTIAN,JOHOR

ADDITIONAL MATHEMATICS PROJECT WORK 2015

“HEALTHY CITIZEN-THE EARTHDREAM”

NAME : TY SHEE YANG

I/C NO : 980824-01-7371

CLASS : 5 DAHLIA

TEACHER : PN.AZNITA BINTI SAIRI

OBJECTIVES

The aims of carrying out this project are :

• To apply and adapt a variety of problem-solving strategies

• To improve critical thinking skills

• To promote effective mathematical communication

• To solve problems in real-life situation

• To use the language of mathematics to express mathematical

statements precisely

• To develop positive attitude towards mathematics

• To produce a systematic presentation

 

CONTENT

CONTENT

PAGE

Part 1 : Introduction

Binomial distribution

Normal distribution

Differences between Binomial Distribution&Normal Distribution

Part 2 : Distribution of Height and Weight

Part 3 : Body Mass Index (BMI)

Further Exploration

Reflection

Part1:Introduction

What is a Probability Distribution?

A probability distribution is a table or an equation that links each outcome of a statistical

experiment with its probability of occurrence.

Probability Distribution Prerequisites

To understand probability distributions, it is important to understand variables. random variables,

and some notation.

A variable is a symbol (A, B, x, y, etc.) that can take on any of a specified set of values.

When the value of a variable is the outcome of a statistical experiment, that variable is a random

variable.

Generally, statisticians use a capital letter to represent a random variable and a lower-case letter,

to represent one of its values. For example,

X represents the random variable X.

P(X) represents the probability of X.

P(X = x) refers to the probability that the random variable X is equal to a particular value,

denoted by x. As an example, P(X = 1) refers to the probability that the random variable X is

equal to 1.

Probability Distributions

An example will make clear the relationship between random variables and probability

distributions. Suppose you flip a coin two times. This simple statistical experiment can have four

possible outcomes: HH, HT, TH, and TT. Now, let the variable X represent the number of Heads

that result from this experiment. The variable X can take on the values 0, 1, or 2. In this example,

X is a random variable; because its value is determined by the outcome of a statistical

experiment.

A probability distribution is a table or an equation that links each outcome of a statistical

experiment with its probability of occurrence. Consider the coin flip experiment described above.

The table below, which associates each outcome with its probability, is an example of a

probability distribution.

Number of heads Probability

0 0.25

1 0.50

2 0.25

The above table represents the probability distribution of the random variable X.

Cumulative Probability Distributions

A cumulative probability refers to the probability that the value of a random variable falls within

a specified range.

Let us return to the coin flip experiment. If we flip a coin two times, we might ask: What is the

probability that the coin flips would result in one or fewer heads? The answer would be a

cumulative probability. It would be the probability that the coin flip experiment results in zero

heads plus the probability that the experiment results in one head.

P(X < 1) = P(X = 0) + P(X = 1) = 0.25 + 0.50 = 0.75

Like a probability distribution, a cumulative probability distribution can be represented by a table

or an equation. In the table below, the cumulative probability refers to the probability than the

random variable X is less than or equal to x.

Number of heads:

x Probability:

P(X = x) Cumulative Probability:

P(X < x)

0 0.25 0.25

1 0.50 0.75

2 0.25 1.00

Uniform Probability Distribution

The simplest probability distribution occurs when all of the values of a random variable occur

with equal probability. This probability distribution is called the uniform distribution.

Uniform Distribution. Suppose the random variable X can assume k different values. Suppose

also that the P(X = xk) is constant. Then,

P(X = xk) = 1/k

Example 1

Suppose a die is tossed. What is the probability that the die will land on 5 ?

Solution: When a die is tossed, there are 6 possible outcomes represented by: S = { 1, 2, 3, 4, 5, 6

}. Each possible outcome is a random variable (X), and each outcome is equally likely to occur.

Thus, we have a uniform distribution. Therefore, the P(X = 5) = 1/6.

What is the Normal Distribution?

In probability theory, the normal distribution is a continuous probability distribution. Normal

distributions are important in statistics and are often used in the natural and social sciences to

represent real-valued random variables whose distributions are not known. The normal

distribution is called the bell curve.

For standard normal distribution graph, any normal distribution curve can be converted to a

standard normal curve with mean = 0 and standard deviation 1. The standard normal distribution

is represented by N(0, 1). The continuous random variable for standard normal distribution is

called the standardized variable and is represented by Z.

From normal random variable X , to standardized variable Z ( Z-score) , a random variable X

that has a normal distribution with mean, and standard deviation, can be converted to Z-score

using the formula :

Z = X-µ

σ

In real-life, we can use normal distribution in the application of quality control. We can calculate

the probability and estimate the amount for each category or grade.  

WHAT IS BINOMIAL DISTRIBUTION?

A binomial distribution summarizes the number of trials, or observations, when each trial has the

same probability of attaining one particular value.

For example, flipping a coin would create a binomial distribution. This is because each trial can

only take one of two values (heads or tails), each success has the same probability (i.e. the

probability of flipping a head is 0.50) and the results of one trial will not influence the results of

another.

P(X = x) = nCr=p, where x = 0, 1, 2…, n and p + q = 1

Bernoulli has also shown that mean, variance and standard deviation of a binomial distribution

can be

calculated using :

If X ~ B(n , p),then,

Mean, μ = np

Variance, σ = npq

Standard deviation, σ =

Examples of Binomial Distribution

Suppose a biased coin comes up heads with probability 0.3 when tossed. What is

the probability of achieving 0, 1,..., 6 heads after six tosses?

\Pr(0\text{ heads}) = f(0) = \Pr(X = 0) = {6\choose 0}0.3^0 (1-0.3)^{6-0} \approx

0.1176

\Pr(1\text{ heads}) = f(1) = \Pr(X = 1) = {6\choose 1}0.3^1 (1-0.3)^{6-1} \approx

0.3025

\Pr(2\text{ heads}) = f(2) = \Pr(X = 2) = {6\choose 2}0.3^2 (1-0.3)^{6-2} \approx

0.3241

\Pr(3\text{ heads}) = f(3) = \Pr(X = 3) = {6\choose 3}0.3^3 (1-0.3)^{6-3} \approx

0.1852

\Pr(4\text{ heads}) = f(4) = \Pr(X = 4) = {6\choose 4}0.3^4 (1-0.3)^{6-4} \approx

0.0595

\Pr(5\text{ heads}) = f(5) = \Pr(X = 5) = {6\choose 5}0.3^5 (1-0.3)^{6-5} \approx

0.0102

\Pr(6\text{ heads}) = f(6) = \Pr(X = 6) = {6\choose 6}0.3^6 (1-0.3)^{6-6} \approx

0.0007[2]

 

DIFFERENCES BETWEEN NORMAL

DISTRIBUTION&BINOMIAL DISTRIBUTION

Normal distributions are continuous and have a special bell shape.

Binomial distributions are discrete ("stairsteppy"); they are close to normal only if the sample

size satisfies np ³ 10 and nq ³ 10.

Normal distributions arise in three general areas:

1) Natural processes where the data value (e.g., height) is the result of many small random

inputs.

2) Sampling distribution of xbar, where either the underlying distribution is normal or (more

commonly) where the sample size is large enough for the CLT to take effect. Rules of thumb are

on p.606 of textbook.

3) Repeated measurement of a fixed phenomenon (e.g., the orbital period of Mars, the mass of a

moon rock, or the height of a mountain). Most phenomena cannot be measured precisely—even

if we have an accurate pan balance or laser range finder or whatever, there will always be some

uncertainty or error in our measurement. For this reason, the normal distribution is sometimes

called the "error function." However, #3 is really just a special case of #1.

Binomial distributions arise whenever the r.v. of interest is the count of successes in a fixed

number (n) of independent trials. The four rules are listed near the beginning of the “binomial

distribution” section, before the second set of example problems.

PART 2

1. Distribution of height and weight of 50 students in school

Bil. Nama Height(cm) weight(kg) BMI

1 ABLE LEE SIOW YEE 160 42 16.4

2 AHMAD AIMAN BIN NOR LAHA 170 70 24.2

3 AHMAD ZULWAIE BIN SHAFIEE 174 76 25.1

4 CHENG FARN LIANG 182 75 22.6

5 CHIA JOO HIANG 180 65 20.1

6 CHUANG GAO JIE 182 56 16.9

7 ENG TING YIN 174 70 23.1

8 FATIN AQILAH BINTI SURATMAN 160 50 19.5

9 GRACE TAN JIA XIAN 155 49 20.4

10 HAFRINA SYAZWANI BINTI MOHD HUZAIMI 160 50 19.5

11 LIM JEI CHIN 164 50 18.6

12 LIM JING AN 151 69 30.3

13 MIMIE HATIRAH BINTI MOHD HISHAM 160 72 28.1

14 MUHAMMAD IDZHAM BIN HAMDAN 175 63 20.6

15 MUHAMMAD SYAFRIL AZMIN BIN ABD MALEK 175 61 19.9

16 NADIRA BINTI AZMI 159 41 16.2

17 NUR ZAFIRAH BINTI AHMAD 160 48 18.8

18 NUR DIANAH BINTI BACHIK 163 51 19.2

19 NUR HAYATI BINTI RAHMAT 173 65 21.7

20 NUR SYAMILAH BINTI MOHD YAZID 163 48 18.1

21 PUA HAN YING 165 58 21.3

22 PUA QIU HUI 160 47 18.4

23 ROZISYAHANIS BINTI SAHARUDDIN 163 75 28.2

24 SITI NORATIFAH NASHA BINTI MOHAMAD NORIZAN 160 63 24.6

25 SITI NUR SOLEHA BINTI AZIMAN 161 45 17.4

26 SYAHIDAH NUR RAIHAN BINTI SHAHROM NIZAM 160 52 20.3

27 TAN YI JUN 169 60 21.0

28 TOH MEY XUAN 158 45 18.0

29 UMMI AEISYA BINTI AZIZ 152 51 22.1

30 WAN NURUL NABILAH BINTI WAN SHUHAIMI 153 45 19.2

31 WONG YONG QIN 169 64 22.4

32 ZAHARAH BINTI MOHD KHAINI 160 63 24.6

33 AINA NATASHA BINTI RASHID 163 53 19.9

34 ALI IMRAN HAKIM BIN MOHAMMAD HOOD 174 60 19.8

35 CHING WEI YEE 163 63 23.7

36 LEE KAI SHERN 170 52 18.0

37 LEE WEN YEE 165 52 19.1

38 LEE YI XUAN 157 51 20.7

39 LIAN SIN JIE 165 53 19.5

40 MOHAMAD YAQZAN BIN MOHD ZAIDI 172 62 21.0

41 PUA CHIA MIN 164 58 21.6

42 SAIRAM A/L KRISHNAMUTTY 176 68 22.0

43 SEE ZHAI WEI 173 57 19.0

44 SHARIFAH ALYSSA BINTI SYED OMAR 168 42 14.9

45 SITI NADIA BINTI SHIFUDIN 150 50 22.2

46 SYAZA AMIRA BINTI ZULKEPLEE 159 48 19.0

47 TAN YONGXIU 156 51 21.0

48 THEY SIN SIAW 168 49 17.4

49 TONG YONG LIN 159 47 18.6

50 TY SHEE YANG 170 57 19.7

2. Draw frequency distribution table for the mass of 50 students using a suitable class

interval.

Weight Midpoints No. of students Cumulative Frequency X2 fx Fx2

41-50 45.5 17 17 2070.25 773.50 35194.25

51-60 55.5 16 33 3080.25 888.00 49284.00

61-70 65.5 13 46 4290.25 851.50 55773.25

71-80 75.5 4 50 5700.25 302.00 22801.00

Total 50 2815.00 163052.50

2. (i) Representation of data for weight

a. Histogram

b. Line Graph

c. Pie Chart

3.

(i)Use mean =

Such that f=number of students

Use standard deviation =

Such that f = number of students

x = mid point of class interval

Weight Midpoints No. of students Cumulative Frequency X2 fx Fx2

41-50 45.5 17 17 2070.25 773.50 35194.25

51-60 55.5 16 33 3080.25 888.00 49284.00

61-70 65.5 13 46 4290.25 851.50 55773.25

71-80 75.5 4 50 5700.25 302.00 22801.00

Total 50 2815.00 163052.50

(ii) Find the mean and standard deviation for the weight.

Mean

=281550

Standard deviation ,

∑∑

f

xf

2∑

∑2

)(∑∑

f

xf-

f

xf

=56.3 = 91.36

(iii)Find the percentage of the students with weight more than 60kg by drawing an ogive.

Percentage = total frequency – 25 X 100%

total frequency

= 25 X 100%

50

= 0.5 X 100%

= 50%

3. With assumption that the weight of the students are normally distributed, find the

(i)Percentage of students with weight more than 60kg.

(ii)The percentage of students with weight less than 45kg.

P( X<45 )=P(Z<45−56 .39.558 )

=P (Z<−1 .1823 )=P (Z>1.1823 )=0.1186=11. 86 %

(iii)The value of m if 90% students have weight more than m kg.

P( X>m)=0 .9

P(Z<m−56 . 39 .558 )=0 . 9

m−56 .39 .558

=−1.281

m=44 . 06kg

4. What conclusion can you draw from the answer obtained in 2(iii) and 3(i)?

- The actual number of students who are more than 60 kg is more than the assumption.

5. 10 students are picked at random, find the probability that exactly 3 students have weight more than 60kg?

P( X=3)=10 C3 p3q7

=10 C3 (0. 3494 )3 (0. 6506 )7

¿0 . 2526

6. Estimate the number of students with weight more than 60kg in your school. Compare your answer with 2

(iii) and give your comments.

Number of students =np

=50(0.2526)

=12.63

=13

2 22095.4.50

( 20.77 ) 2

PART 3

1. Using the data in Part 2, calculate the BMI for each student.

BMI Midpoints Frequency Cumulative Frequency x^2 fx fx^2

14.1-16.0 15.05 1 1 226.50 15.05 226.50

16.1-18.0 17.05 7 8 290.70 119.35 2034.92

18.1-20.0 19.05 17 25 362.90 323.85 6169.34

20.1-22.0 21.05 12 37 443.10 252.60 5317.23

22.1-24.0 23.05 6 43 531.30 138.30 3187.82

24.1-26.0 25.05 4 47 627.50 100.20 2510.01

26.1-28.0 27.05 0 47 731.70 0.00 0.00

28.1-30.0 29.03 2 49 842.74 58.06 1685.48

30.1-32.0 31.05 1 50 964.10 31.05 964.10

Total 50 1038.46 22095.40

2. (i)Based on the information given and data collected, find the mean and standard deviation for the BMI.

Mean

=1038 .550

=20 . 77

Standard deviation ,

= 10.52

(ii)Determine by drawing an ogive,

(a)The percentage of students who are underweight.

1250

X 100 %

=0.24 X 100 %=24 %

(b)The percentage of students with BMI more than 25.

550

X 100 %

=0.1 X 100 %=10 %

3. With assumption that the BMI of the students are normally distributed, find

(i)The percentage of students who are overweight

P(25< X<29 . 9)

=P(25−20 . 773. 249

<Z<29 . 9−20 .773 .249 )

=P (1 . 302<Z<2 . 81)

P( X>K )=0. 05

P(Z>k−20 .773 .249 )=0 .05

k−20 .773 .249

=−1 . 645

k=15 . 425

(ii)The percentage of students with BMI less than 18.5

P( X<18 .5 )

=P(Z<18 .5−20 . 773 . 349 )

=P (Z<−0 . 6778)=P (Z>0 . 6778)=0.2491=24 .91 %

(iii)The value of k if5% students have BMI more than k

P( X>K )=0. 05

P(Z>k−20 .773 .249 )=0 .05

k−20 .773 .249

=1 . 645

k=26 .115

4. Estimate the number of students who are overweight in your school.

Number of students =np

=50(0.9398)

=4.699

=5

5. Suggest ways and strategies that a person can adopt to reduce weight and live a healthier life

Quit smoking

It's no secret that nicotine use has been linked to a variety of life-threatening illnesses,

from cancer to heart disease and stroke. If you're a smoker, make quitting your top

priority.

Get regular checkups

Think of your doctor as an ally who helps keep you well, not just the person who treats

you when you're sick. Regular checkups and screenings are vital, especially if you or

your family are predisposed to certain medical conditions.

Watch your weight

Those extra pounds can cause big trouble. They strain your heart, raise your blood

pressure and significantly increase your risk of a heart attack. Eat more high-fiber,

nutrient-rich fruits and vegetables, and fewer high-fat foods.

Avoid excessive drinking

While drinking in moderation is usually fine, heavy drinking can lead to liver damage

and other serious health risks.

Get regular exercise

A healthy life requires periodic physical activity. To prevent heart disease, cancer, high

blood pressure and obesity, the American Heart Association recommends 30-60 minutes

of exercise at least four times a week.

FURTHER EXPLORATION

Carl Friedrich Gauss (1777-1855) discovered the normal

distribution in 1809. Gauss was the first to suggest the normal distribution

law. In 1809, Gauss published his monograph "Theoria motus corporum

coelestium in sectionibus conicis solem ambientium" to the method of least

squares. Gauss also used the normal curve to analyze astronomical data. The

normal distribution is often called the Gaussian distribution. The term bell

shaped curve is often used in everyday usage as the normal curve.

Although Gauss was the first to suggest the normal distribution law, Laplace made

significant contributions. It was Laplace who first posed the problem of aggregating several

observations in 1774, although his own solution led to the Laplacian distribution. It was Laplace

who first calculated the value of the integral ∫ e − t ² dt = √ π in 1782, providing the normalization

constant for the normal distribution. Finally, it was Laplace who in 1810 proved and presented to

the Academy the fundamental central limit theorem, which emphasized the theoretical

importance of the normal distribution

Conclusion

There are two types of probability distribution, binomial distribution and normal

distribution. Both distributions are very useful as they help us to solve problems in real

life. For binomial distribution, it can be described as the probability of a SUCCESS or

FAILURE outcome in an experiment, that is repeated multiple time. For a real-life

example, if a new drug is introduced to cure a disease, it either cures the disease

(successful) or it doesn’t cure the disease (failure). If we purchase a lottery ticket, we’re

either going to win money, or aren’t. In contrast, the normal distribution is very

important in statistic as it approximately fits into actual, observed probability of many

real-life measurements such as height, weight and blood pressure.

From the survey done, we can conclude that majority of the students are in the

normal BMI category. However, there are still many students that do not achieve an

ideal BMI, which is from 18.5 to 24.9. They were in the underweight, overweight and

obese categories. Proper diet arrangement and regular exercises are needed for them to

return to the normal BMI category, so that they live a healthier life.

Reference

1. http://en.wikipedia.org/wiki/Normal_distribution

2. http://en.wikipedia.org/wiki/Binomial_distribution

3. http://en.wikipedia.org/wiki/Pierre-Simon_Laplace

4. http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss

5. http://en.wikipedia.org/wiki/Abraham_de_Moivre

REFLECTION

While conducting this project, I have learnt the importance of using binomial distribution and

normal distribution to solve mathematical and statistics problems. Apart from that, this project

encourages students to gather information from the Internet, making surveys, improve thinking

skills and promote effective mathematical communication.

After conducting this project, I found out that most of the students have an ideal BMI index.

This means that they always maintain a healthy lifestyle. Besides, they always keep their body

fit.

On the other hand, from this project, I have given the chance to promote some moral values

while conducting this project. Firstly, I learn to be thankful to my additional mathematics

teacher, Pn.Aznita binti Sairi who has been so caring who has given guidance to me while

conducting this project. Besides, I also want to thank to my friends who have been very co-

operative with me while conducting this project. Furthermore, I am appreciate those

mathematician who invented normal distribution and binomial distribution so that our current life

is easy in solving questions concerning in this aspect.