Additional Mathematics Project Work.docx

Additional Mathematics Project Work 2015

ContentsBil Title Page1 Title :”Probability distribution and

its use in solving problem in real life situations”

2

2 Appreciation 33 Part 1

Introductions4-6

4 Part 24.1Result of the survey 4.2Frequency distribution table: Representation Mean and standard

deviation for the weight

Ogive4.3Assuming the weight is normally distributed

7-14

5 Part 3:5.1BMI for each student5.2Mean and standard deviation for the BMI

Ogive5.3Assuming the BMI normally distributed

15-19

6 Further Exploration 207 Reflection 21

1


2

Title


Appreciation:

I would like to thank my Additional Mathematics teacher Puan Nor Liza Binti Mohammad for explaining to us and guide us throughout this project work.Without her , I wouldn’t have been able to do this project so smoothly.

I would also like to take this opportunity to thank my friends Teo Feng Tian ,Eileen Toh for all cooperating in searching for the information ,obtaining the data , sharing their ideas and helping out in writing this report . I would like to thank them for their full commitment in doing this project work .

Lastly I would like to thank my family members. Especially my parents for their moral and financial support.

3

Appreciation


Go to internet to explore “Probability distribution , binomial distribution , normal distribution “

Decisions made by government , business , education department , sporting bodies ,etc , are often made after careful consideration of statistical evidence . Statistics play a vital role in many areas of our society . Statistics are a tool for helping to made rational decisions about variables described by data sets.

Amongst other things , governments use statistics to help formulate future policies . Businesses often use statistics to aid decision making ,for example , whether or not to enter the market with an alternative to a product when there are already several of them on the market . Statistical information about sport has increased dramatically in recent years. We only need to watch a ‘one day international ‘ cricket match to observe the many statistical graph and tables used to help make the viewer more informed.

Following are some examples of the types of problems we may face , and where statistical methods may help to answer them:

A younger executive of a hotel chain claimed that lowering the room tariff by 10% would increase the patronage by 20%. Would this be true ?

A manufacturer of AAA batteries claimed that her batteries outlasted all other leading brands by at least 100 hour . Is she correct?

Does the unemployment rate affect the crime rate fir that city? An employer claims that younger employees (<30 years) have on

average twice as many sick days as older ones(≥30 years) Which drug for helping to quit smoking has the greatest chance of

success.

The binomial probability model:

4

Introduction


The binomial probability model is a sequence of trials, each of which consist of repetition of a single experiment .We assume the trials to be independent . Furthermore we assume that there are only two possible outcome for each trial that is success or failure . Many real-world situation have the characteristics of the binomial probability model.

Examples of binomial distribution being applied to real life problems:1. A machine produce light bulbs to meet certain specification ,and 80% of the

bulbs produced meet there specification . A sample of 6 bulbs is taken from the machine’s production .What is the probability that 3 or more of them fail to meet specification?

2. Suppose that 60% of the voters intend to vote for a conservative candidate . What is the probability that a survey poiling 8 people reveals that 3 or fewer intend to vote for a conservative candidate?

3. In a 15 item true-false examination , what is the probability that a student who guesses on each question will get at least 10 correct answer ? If another student has 0.8 probability of correctly answering each question , what is the probability that this student will answer at least 12 questions correctly ?

The normal distribution model :

Frequency distributions can assume almost any shape or form , depending on the data . However , the data obtained from many experiments often follow a common pattern which has been thoroughly investigated . For example , heights of people , weight of people , test scores and coin tossing all lead to data which have the same kind of frequency distribution is referred to as normal distribution or the Gaussian distribution. The graph of the normal distribution , called the normal distribution , called the normal curve , is the bell-shaped curve shown below:

Examples of normal distribution being applied to real life problems:

A company manufactures 60000 pencils each day . Quality control studies have shown that on the average, 4% of the pencils are defective . A random sample of 500 pencils is selected from each day’s production .What is the probability that in the sample there are at least 12 and no more 24 defective pencils ?

5


The average height of 2000 woman in a random sample is 158 cm . The standard deviation is 6 cm. The heights have a normal distribution . How many woman are between 155 cm and 162 cm?

Records shows that the average life expectancy of a pair of shoes is 2.2 years with a standard deviation of 1.7 years . A manufacturer guarantees that shoes lasting less than a year are replaced free. For every 1000 shoes sold , how many shoes should the manufacturer expect to replace free?

6

Part 2


1. A survey on the distribution of height and weight of 50 students in my school are shown in the table below:

Students Height (cm) Weight(kg)1 183 732 163 533 163 484 167 805 159 396 162 487 172 708 152 409 154 4510 155 3911 160 4712 155 4113 172 6314 171 6315 175 6516 156 4717 169 5918 178 5519 163 5120 160 4221 160 48

22 160 60

7


23 173 6724 175 4525 170 6026 158 5127 160 4828 151 3829 162 6430 154 5231 171 4932 159 7033 155 7434 165 5835 165 6936 158 6037 165 6338 158 5639 161 5440 161 5541 165 5942 164 7043 161 5844 150 6345 158 7846 156 5747 161 6448 154 5449 168 6150 173 63

2. Frequency distribution table for the weight obtained

8


Wight (kg) Tallies Frequency33-39 ∕∕∕ 340-46 ∕∕∕∕ 547-53 ∕∕∕∕ ∕∕∕∕

∕11

54-60 ∕∕∕∕ ∕∕∕∕ ∕∕∕

13

61-67 ∕∕∕∕ ∕∕∕∕ 1068-74 ∕∕∕∕ ∕ 675-81 ∕∕ 2

i. Represent the data obtained using 3 different statistical graph :

9


(a) Using Bar graph to represent the data:

33-39 40-46 47-53 54-60 61-67 68-74 75-81 Weight(kg)

0

2

4

6

8

10

12

14

(b) Using Line graph to represent the data

Weight(kg)

33-39 40-46 47-53 54-60 61-67 68-74 75-810

2

4

6

8

10

12

14

(c) Using pie chart to represent the data

10


33-39

40-46

47-53

54-60

61-67

68-74

75-81

11


2.2 Find the mean and standard deviation for the weight

Weight (kg) x f fx Fx2

33-39 36 3 108 388840-46 43 5 215 924547-53 50 11 550 2750054-60 46 13 728 4076861-67 64 10 640 4096068-74 71 6 426 3024675-81 78 2 156 12168

∑ f =50 ∑ fx=2823 ∑ f x2=164775

∑ fx

∑ f=2823

50 =56.46 σ 2= ∑ f x2

∑ f−x2=164775

50−(56.46)2=107.768

σ=√107.768=10.381

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ii. Use ogive to answer the problemsThe findings were also recorded in the following table:

Weight(kg) ≤32.5 ≤39.5 ≤46.5 ≤53.5 ≤60.5 ≤67.5 ≤74.5 ≤81.5No of pupil 0 3 8 19 32 42 48 50

32.5 39.5 46.5 53.5 60.5 67.5 74.5 81.50

10

20

30

40

50

60

From the ogive , percentage of students with weight more than 60 kg

=18.550

×100=37 %

13


3. The weight of the students are normally distributed with mean 56.46 and standard deviation =10.381 that is x-N( 56.46,10.381)

i. Percentage of students with weight more than 60 kg

P(x>60)=P(z>60−56.46

10.381 )=P(z>0.3410)=0.3666

36.66% of the students weight more than 60 kg ii. Percentage of students with weight less than 45 kg

P(x<45)=P(z<45−56.46

10.381 )=P(z<-1.1039)=0.1348

13.48% of students weight less than 45 kgiii. The value of m if 90% students has weight more than m kg

P(x>m)=0.9

P(z>m−56.46

10.381 )=0.1

P(z>-m−56.46

10.381 )=0.1

-m−56.46

10.381 =1.281

m=43.16kg

90%of the students weight more than 43.16kg

4. The conclusion we obtained regarding the percentage of students weighting more than 60 kg using ogive is 37% and using normal distribution is 36.67%.Therefore they are the the same using two different methods.

5. P(x=3)=10∁ 3 (0.3667 )(0.6333)=0.2418

If 10 students are picked, the probability that 3 students have weight more than 60 kg is 0.2418.

6 My school has 800 students . The number of students with weight more than 60 kg is estimated to be

14


0.3667×800=293 students

The body weight index

Students Height(cm) Weight (kg) BMI Category1 183 73 21.80 Normal2 163 53 19.95 Normal3 163 48 22.65 Normal4 157 80 28.69 Overweight5 159 39 15.43 Underweight6 162 48 18.29 Underweight7 172 70 23.66 Normal8 152 40 17.31 Underweight9 154 45 18.97 Normal10 155 39 16.23 Underweight11 160 47 18.36 Underweight12 155 41 17.07 Underweight13 172 63 21.30 Normal14 171 63 21.55 Normal15 175 65 21.22 Normal16 156 47 19.31 Normal17 169 59 20.66 Normal18 178 55 17.36 Underweight19 163 51 19.20 Normal20 160 42 26.25 Overweight

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Part 3


21 160 48 18.75 Normal22 160 60 23.44 Normal23 173 67 22.39 Normal24 175 45 14.69 Underweight25 170 60 20.76 Normal26 158 51 20.43 Normal27 160 48 18.75 Normal28 151 38 16.67 Underweight29 162 64 24.39 Normal30 154 52 21.93 Normal31 171 49 16.76 Underweight32 159 70 27.69 Overweight33 155 74 30.80 Obese34 165 58 21.30 Normal35 165 69 25.34 Overweight36 158 60 24.03 Normal37 165 63 23.14 Normal38 158 56 22.43 Normal39 161 54 20.83 Normal40 161 55 21.22 Normal41 165 59 21.67 Normal42 164 70 26.03 Overweight43 161 58 22.38 Normal44 150 63 28 Overweight45 158 78 31.24 Obese46 156 57 23.42 Normal47 161 64 24.69 Normal48 154 54 22.77 Normal49 168 61 21.61 Normal

16


50 173 63 21.05 Normal

2)

i. The mean and standard deviation for the BMI

Mean= 1083.86

50=21.677

Standard Deviation=√ 24178.9250

−21.6772=3.698

ii. The findings were also recorded in the following table:

BMI ≤13 ≤16 ≤19 ≤22 ≤25 ≤28 ≤31 ≤33No of pupils

0 3 14 30 42 48 49 50

13 16 19 22 25 28 31 330

10

20

30

40

50

60

iii. a) percentage of students who are underweight

1050×100=20 %

17


b)percentage of students with BMI more than 25

7

50×100=14 %

3. The BMI of the students are normally distributed with mean 21.667 and standard deviation =3.698 that is x-N(21.677,3.698 )

i. Percentage of students who are overweight

P(25<x<29.9)= P(25−21.677

3.698<z< 29.9−21.677

3.698 )

=P(0.8986<z<2.2236)=0.1713

17.13 % of the students are overweight.ii. Percentage of students with BMI less than 18.5

P(x<18.5)=P(z<18.5−21.677

3.698 )=P(z<-0.8591)=0.1951

19.51% of the students with BMI less than 18.5iii. The value of k if 5% of students have BMI more than k

P(x>k)=0.05

P(z>k−21.677

3.698 )=0.05

k−21.6773.698

=1.645

K=27.765% of the students have BMI more than 27.76

4. Estimate the number of students who are overweight .

50×17.13100

=8.565

9 students who are overweight.

18


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

Dansinger recommends eating a diet that minimizes starches, added sugars,

and animal fat from meat and dairy foods. For rapid weight loss, he

recommends focusing on fruits, veggies, egg whites, soy products, skinless

poultry breasts, fish, shellfish, nonfat dairy foods, and 95% lean meat.

Here are more tips from Dawn Jackson Blatner, RD, author of The Flexitarian

Diet : Eat vegetables to help you feel full. Drink plenty of water. Get tempting foods out of your home. Stay busy -- you don't want to eat just because you're bored. Eat only from a plate, while seated at a table. No grazing in front of the 'fridge. Don't skip meals.

Keeping a food journal -- writing down everything you eat -- can also help you

stay on track.

We can also adopt a healthier lifestyle by following these easy step:

1. Think positive and focus on gratitude

2. Eat your vegetables

3. Set a “5-meal ideal”

4. Exercise daily

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5. Get at good night's sleep

One famous mathematician who contributed to the development of normal distribution .

Kari Frederik Gauss examined errors in measurement and found an appropriate error measurement function . It was still is agreed that the characteristic of a measurement error function are :

The measurement errors of the same size but of opposite sign are equally likely

Small errors are more likely than larger ones , and extremely larger errors are very unlikely,

The total area under the function is 1 The most likely measurement is the mean of the distribution.

Gauss was able to prove that the normal distribution is the only error function that satisfies all four conditions.

The normal distribution is fundamental in the study of probability and statistics , partly because it is the model for errors in measurement .Because of the significance of Gauss’ fundamental discovery, this distribution is also called the Gaussian distribution.

20

Further Exploration


From doing this project , we have learn the following skills and values:

(a) Use what we learn in the class to solve real life problemWe learned from the chapter 6 and 7 and 8 ,it helps us to calculate the question by using the formulae.

(b) Importance of teamwork.We work like a team to search for the information. Our friendship is being strengthened

(c) Creative in presenting the resultsBy using the table, bar chart ,line graph to present the results is more innovating. The data is shown more clearly and easy to be understand .

(d) Using technology to help us made our work easier.For example , we calculate the answers by setting the formulae in the Microsoft Word. Then, the scientific calculator also helps us a lot in doing this project. So ,we are to get the answer more accurately.

21

Reflection

Documents

Additional Mathematics Project Work.docx