Add Math 2
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NAME : RAJA MUHAMMAD HASIF BIN RAJA HASSAN I/C
: 911115-03-6335
FORM : 4 SAINS HAYAT 1 SUBJECT’S TEACHER : PN. NARIZAN BT RAFFLES
a) THE TABULATION OF DATA THE HEIGHT AND WEIGHT OF 50 STUDENTS IN FORM 4 AND FORM 5 Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
Height ( m ) 1.45 1.60 1.59 1.47 1.59 1.65 1.52 1.74 1.63 1.54 1.71 1.72 1.67 1.36 1.46 1.57 1.63 1.52 1.55 1.40 1.42 1.43 1.49 1.67 1.73 1.56 1.60 1.48 1.77 1.75 1.47 1.71 1.48 1.60 1.54 1.52 1.56 1.50 1.54
Weight ( kg ) 53 45 48 47 47 47 49 63 67 59 72 54 53 40 45 43 60 44 59 50 35 32 52 52 65 42 47 53 61 66 58 91 44 46 48 47 54 50 45
40 41 42 43 44 45 46 47 48 49 50
1.78 1.48 1.41 1.49 1.48 1.45 1.56 1.63 1.51 1.52 1.49
b)
65 36 47 49 46 46 59 46 42 43 40
Construct a frequency distribution table for the weights
obtained using a suitable class interval
The frequency distribution table: Cumulative Class Interval Frequency,f Frequency 0 21-30 0 5 31 - 40 5 30 41 - 50 25 42 51 - 60 12 48 61 - 70 6 49 71 - 80 1 81 - 90 91 - 100
0 1
49 50
Upper Boundary
Midpoint
20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5
25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5
b) i) Represent your data from the frequency distribution table by using three different statistical graphs. On the graph paper
b) ii) Find the mean, median and mode for the weights of the 50 students. Which one of the three is the best measure of central tendency to represent the data? Explain your choice.
Mean, x =
∑ ∑
fx f
=
=
35.5(5) + 45.5(25) + 55.5(12) + 65.5(6) + 75.5(1) + 95.5(1) 50
2545 50
= 50.9
Median class = ( 41 − 50 ) N L + 2 − F ( c) Median, M = fm 50 50.5 + 2 − 5 (10 ) = 25 = 28.2
CONCLUSION From the data obtained, the best measure of central tendency that is needed to represent the data is mean. It is because all the values in a set of data are taken into consideration while calculating the mean. Besides that, mean is suitable to represent data which are evenly distributed. c) Calculate the standard deviation of the weights by using three different methods [Note: Include “entering raw data into a calculator” as a method]
Method 1 By using calculator (entering the raw data into a calculator).
Calculator ( version fx-570MS ) Step 1 : Press [ MODE ] >> Press [ MODE ] again >> Press [1] for [ SD ] Step 2 : Key in the relevant data (weights of students) >> Press [ M+ ] to store the data. Step 3 : Repeat step 2 until all of the data are stored Step 4 : Press [ SHIFT ] >> Press [ S-VAR ] >> Press [ 2 ] for σn or standard deviation.
Method 2 By using formula i. Class Interval 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 – 90 91 - 100
Mid Point ( x ) 35.5 45.5 55.5 65.5 75.5 85.5 95.5
∑ fx ∑f
Standard deviation, σ =
f 5 25 12 6 1 0 1 Σf=50
fx 177.5 1137.5 666 393 75.5 0 95.5 Σfx=2545
fx2 6301.25 51756.25 36963 25741.5 5700.25 0 9120.25 Σfx2= 135582.5
2
−x
2
=
6301.25 + 51756.25 + 36963 + 25741.5 + 5700.25 + 9120.25 − (50.9) 2 50
=
2711.65 − 2590.81
= =
120.84 10.99
Method 3 By using formula ii.
Class Interval 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 – 90 91 - 100
Midpoint ( x) 35.5 45.5 55.5 65.5 75.5 85.5 95.5
=
6039 50
=
120.78
=
)
f x−x
5 25 12 6 1 0 1
1185.8 729 253.92 1278.96 605.16 0 1986.16
∑ f =50 ∑ f ( x − x )
∑ f ( x − x) ∑f
Standard deviation, σ =
(
F
2
2
=6039
2
10.99
d) What conclusion can you draw from the value of the standard deviation obtained in part c? The result for the standard deviation is 10.99. The result shows that the standard deviation is far away from the mean of the weight, which is 50.9 kg. This shows that from all 50 students, there are a few students that are too fat ad too thin. So, we can conclude that this result may not be consistent.
a) The BMI for each student:
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Height ( m ) 1.45 1.60 1.59 1.47 1.59 1.65 1.52 1.74 1.63 1.54 1.71 1.72 1.67 1.36 1.46 1.57 1.63 1.52 1.55 1.40 1.42 1.43 1.49 1.67 1.73 1.56 1.60 1.48 1.77 1.75 1.47 1.71 1.48 1.60 1.54 1.52 1.56 1.50 1.54 1.78 1.48 1.41 1.49 1.48 1.45
Weight ( kg ) 53 45 48 47 47 47 49 63 67 59 72 54 53 40 45 43 60 44 59 50 35 32 52 52 65 42 47 53 61 66 58 91 44 46 48 47 54 50 45 65 36 47 49 46 46
BMI 25.21 17.58 18.99 21.75 18.59 17.26 21.21 20.81 25.21 24.88 24.62 18.25 19.60 21.63 21.11 17.44 22.58 19.04 24.55 25.1 17.36 15.65 23.42 18.65 21.72 17.26 18.34 24.20 19.47 21.55 26.84 31.12 20.09 17.97 20.24 20.34 22.19 22.22 18.97 20.52 16.44 23.64 22.07 21.00 21.88
Category Overweight Underweight Normal Normal Normal Underweight Normal Normal Overweight Normal Normal Underweight Normal Normal Normal Underweight Normal Normal Normal Overweight Underweight Underweight Normal Normal Normal Underweight Underweight Normal Normal Normal Overweight Obese Normal Underweight Normal Normal Normal Normal Overweight Normal Underweight Normal Normal Normal Normal
b) Based on the information given and the data collected; determine the percentage of students that are underweight and the percentage of students that are obese by drawing an ogive. Represent your findings using statistical graphs and give comments.
Percentage of students who are underweight
=
13 × 100 50
=
22 %
Percentage of students who are obese =
1 × 100 50
=
2%
Class Interval 21 - 30 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 – 90 91 - 100
Frequency 0 5 25 12 6 1 0 1
Cumulative Frequency 0 5 30 42 48 49 49 50
Upper Boundary 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5
COMMENTS Based on the findings, majority of the students have normal weight but a few of them have underweight problem. The percentage of the students who are underweight is 22%. Apart from that, 2% of the student that is one of the 50 students is obese. Lastly, we can conclude that most of the students in Form 4 and Form 5 at my school have normal weight. c) As a student, describe the role that you can play to make the “Obesity Awareness Campaign” a success.
As a responsible student, organizing an exhibition about the importance of maintaining the good health in our life is one of the great ways in ensuring the succession of the “Obesity Awareness Campaign”. Besides, by inviting an officer from a health department or any specialist from NGOs to deliver a speech session on how to prevent obesity and how to have a healthy life. Finally, through the society of active & healthy lifestyles, I would get a chance to organize an essay-writing competition on the risks of obesity and the importance of having a healthy lifestyles.
: 911115-03-6335
FORM : 4 SAINS HAYAT 1 SUBJECT’S TEACHER : PN. NARIZAN BT RAFFLES
a) THE TABULATION OF DATA THE HEIGHT AND WEIGHT OF 50 STUDENTS IN FORM 4 AND FORM 5 Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
Height ( m ) 1.45 1.60 1.59 1.47 1.59 1.65 1.52 1.74 1.63 1.54 1.71 1.72 1.67 1.36 1.46 1.57 1.63 1.52 1.55 1.40 1.42 1.43 1.49 1.67 1.73 1.56 1.60 1.48 1.77 1.75 1.47 1.71 1.48 1.60 1.54 1.52 1.56 1.50 1.54
Weight ( kg ) 53 45 48 47 47 47 49 63 67 59 72 54 53 40 45 43 60 44 59 50 35 32 52 52 65 42 47 53 61 66 58 91 44 46 48 47 54 50 45
40 41 42 43 44 45 46 47 48 49 50
1.78 1.48 1.41 1.49 1.48 1.45 1.56 1.63 1.51 1.52 1.49
b)
65 36 47 49 46 46 59 46 42 43 40
Construct a frequency distribution table for the weights
obtained using a suitable class interval
The frequency distribution table: Cumulative Class Interval Frequency,f Frequency 0 21-30 0 5 31 - 40 5 30 41 - 50 25 42 51 - 60 12 48 61 - 70 6 49 71 - 80 1 81 - 90 91 - 100
0 1
49 50
Upper Boundary
Midpoint
20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5
25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5
b) i) Represent your data from the frequency distribution table by using three different statistical graphs. On the graph paper
b) ii) Find the mean, median and mode for the weights of the 50 students. Which one of the three is the best measure of central tendency to represent the data? Explain your choice.
Mean, x =
∑ ∑
fx f
=
=
35.5(5) + 45.5(25) + 55.5(12) + 65.5(6) + 75.5(1) + 95.5(1) 50
2545 50
= 50.9
Median class = ( 41 − 50 ) N L + 2 − F ( c) Median, M = fm 50 50.5 + 2 − 5 (10 ) = 25 = 28.2
CONCLUSION From the data obtained, the best measure of central tendency that is needed to represent the data is mean. It is because all the values in a set of data are taken into consideration while calculating the mean. Besides that, mean is suitable to represent data which are evenly distributed. c) Calculate the standard deviation of the weights by using three different methods [Note: Include “entering raw data into a calculator” as a method]
Method 1 By using calculator (entering the raw data into a calculator).
Calculator ( version fx-570MS ) Step 1 : Press [ MODE ] >> Press [ MODE ] again >> Press [1] for [ SD ] Step 2 : Key in the relevant data (weights of students) >> Press [ M+ ] to store the data. Step 3 : Repeat step 2 until all of the data are stored Step 4 : Press [ SHIFT ] >> Press [ S-VAR ] >> Press [ 2 ] for σn or standard deviation.
Method 2 By using formula i. Class Interval 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 – 90 91 - 100
Mid Point ( x ) 35.5 45.5 55.5 65.5 75.5 85.5 95.5
∑ fx ∑f
Standard deviation, σ =
f 5 25 12 6 1 0 1 Σf=50
fx 177.5 1137.5 666 393 75.5 0 95.5 Σfx=2545
fx2 6301.25 51756.25 36963 25741.5 5700.25 0 9120.25 Σfx2= 135582.5
2
−x
2
=
6301.25 + 51756.25 + 36963 + 25741.5 + 5700.25 + 9120.25 − (50.9) 2 50
=
2711.65 − 2590.81
= =
120.84 10.99
Method 3 By using formula ii.
Class Interval 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 – 90 91 - 100
Midpoint ( x) 35.5 45.5 55.5 65.5 75.5 85.5 95.5
=
6039 50
=
120.78
=
)
f x−x
5 25 12 6 1 0 1
1185.8 729 253.92 1278.96 605.16 0 1986.16
∑ f =50 ∑ f ( x − x )
∑ f ( x − x) ∑f
Standard deviation, σ =
(
F
2
2
=6039
2
10.99
d) What conclusion can you draw from the value of the standard deviation obtained in part c? The result for the standard deviation is 10.99. The result shows that the standard deviation is far away from the mean of the weight, which is 50.9 kg. This shows that from all 50 students, there are a few students that are too fat ad too thin. So, we can conclude that this result may not be consistent.
a) The BMI for each student:
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Height ( m ) 1.45 1.60 1.59 1.47 1.59 1.65 1.52 1.74 1.63 1.54 1.71 1.72 1.67 1.36 1.46 1.57 1.63 1.52 1.55 1.40 1.42 1.43 1.49 1.67 1.73 1.56 1.60 1.48 1.77 1.75 1.47 1.71 1.48 1.60 1.54 1.52 1.56 1.50 1.54 1.78 1.48 1.41 1.49 1.48 1.45
Weight ( kg ) 53 45 48 47 47 47 49 63 67 59 72 54 53 40 45 43 60 44 59 50 35 32 52 52 65 42 47 53 61 66 58 91 44 46 48 47 54 50 45 65 36 47 49 46 46
BMI 25.21 17.58 18.99 21.75 18.59 17.26 21.21 20.81 25.21 24.88 24.62 18.25 19.60 21.63 21.11 17.44 22.58 19.04 24.55 25.1 17.36 15.65 23.42 18.65 21.72 17.26 18.34 24.20 19.47 21.55 26.84 31.12 20.09 17.97 20.24 20.34 22.19 22.22 18.97 20.52 16.44 23.64 22.07 21.00 21.88
Category Overweight Underweight Normal Normal Normal Underweight Normal Normal Overweight Normal Normal Underweight Normal Normal Normal Underweight Normal Normal Normal Overweight Underweight Underweight Normal Normal Normal Underweight Underweight Normal Normal Normal Overweight Obese Normal Underweight Normal Normal Normal Normal Overweight Normal Underweight Normal Normal Normal Normal
b) Based on the information given and the data collected; determine the percentage of students that are underweight and the percentage of students that are obese by drawing an ogive. Represent your findings using statistical graphs and give comments.
Percentage of students who are underweight
=
13 × 100 50
=
22 %
Percentage of students who are obese =
1 × 100 50
=
2%
Class Interval 21 - 30 31 - 40 41 – 50 51 - 60 61 - 70 71 - 80 81 – 90 91 - 100
Frequency 0 5 25 12 6 1 0 1
Cumulative Frequency 0 5 30 42 48 49 49 50
Upper Boundary 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5
COMMENTS Based on the findings, majority of the students have normal weight but a few of them have underweight problem. The percentage of the students who are underweight is 22%. Apart from that, 2% of the student that is one of the 50 students is obese. Lastly, we can conclude that most of the students in Form 4 and Form 5 at my school have normal weight. c) As a student, describe the role that you can play to make the “Obesity Awareness Campaign” a success.
As a responsible student, organizing an exhibition about the importance of maintaining the good health in our life is one of the great ways in ensuring the succession of the “Obesity Awareness Campaign”. Besides, by inviting an officer from a health department or any specialist from NGOs to deliver a speech session on how to prevent obesity and how to have a healthy life. Finally, through the society of active & healthy lifestyles, I would get a chance to organize an essay-writing competition on the risks of obesity and the importance of having a healthy lifestyles.
