20111101151326 final hbmt4403_jan2010_updated
TRANSCRIPT
CONFIDENTIAL / SULIT
JANUARY SEMESTER 2010 / SEMESTER JANUARI 2010
FINAL EXAMINATION / PEPERIKSAAN AKHIR
COURSE / KURSUS : TEACHING MATHEMATICS IN FORM SIX
CODE / KOD : HBMT4403
DATE / TARIKH :
DURATION / TEMPOH :
TIME / MASA :
INSTRUCTIONS TO CANDIDATES / ARAHAN KEPADA CALON
1. This question paper is set in English (in bold print) and Bahasa Melayu. ANSWER using ONE
LANGUAGE ONLY either ENGLISH or BAHASA MELAYU.
Kertas soalan ini disediakan dalam Bahasa Inggeris (dalam cetakan bold) dan Bahasa Melayu. JAWAB dalam SATU
BAHASA SAHAJA sama ada BAHASA INGGERIS atau BAHASA MELAYU.
2. This question paper consists of THREE Parts - PART A, PART B and PART C. Read
CAREFULLY the instructions for each PART.
Kertas soalan ini terdiri daripada TIGA Bahagian - BAHAGIAN A, BAHAGIAN B dan BAHAGIAN C. Baca
DENGAN TELITI arahan bagi setiap BAHAGIAN.
3. Write your answers in the Answer Booklet provided.
Tulis jawapan anda dalam Buku Jawapan yang dibekalkan.
This question paper consists of TWELVE PAGES of questions printed on both sides of the paper,
excluding this page.
Kertas soalan ini mengandungi DUA BELAS MUKA SURAT soalan yang dicetak pada kedua-dua belah muka surat, tidak
termasuk muka surat ini.
HBMT4403_SEM JAN10/F_AH
2…
PART A / BAHAGIAN A
INSTRUCTIONS / ARAHAN
Part A contains FIVE questions. Answer ALL questions.
Bahagian A mengandungi LIMA soalan. Jawab SEMUA soalan.
QUESTION/ SOALAN: Marks/ Markah
1. Using the first principles, show and explain that if ,uvy = then
dx
duv
dx
dvu
dx
dy+=
Menggunakan Prinsip Pertama, tunjuk dan terangkan jika diberi ,uvy =
maka dx
duv
dx
dvu
dx
dy+=
( 4 )
2. Given that xy cos= , show that 014
4
2
23 =++
y
dx
ydy
Diberi xy cos= , tunjukkan bahawa 014
4
2
23 =++
y
dx
ydy
( 4 )
3. With the help of a diagram, explain to your students the definition of
a parabola and state the general equation.
Dengan bantuan rajah, terangkan kepada pelajar anda definisi parabola
dan nyatakan persamaan am bagi parabola.
( 4 )
HBMT4403_SEM JAN10/F_AH
3…
4. Explain with the help of a suitable example the integration by parts.
Terangkan dengan menggunakan contoh yang sesuai kaedah pengamiran
bahagian demi bahagian.
( 4 )
5. Illustrate with an example each, the difference between permutation
and combination.
Tunjukkan dengan satu contoh untuk setiap kes, perbezaan di antara
permutasi dan kombinasi.
( 4 )
(Total/ Jumlah: 20)
HBMT4403_SEM JAN10/F_AH
4…
PART B / BAHAGIAN B
INSTRUCTIONS / ARAHAN
Part B contains FIVE questions. Answer THREE questions ONLY.
Bahagian B mengandungi LIMA soalan. Jawab TIGA soalan SAHAJA.
QUESTION/ SOALAN: Marks/ Markah
1. (a) You are teaching your class the theory of integration.
(i) Explain indefinite integration as the reverse process of
differentiation and give a simple example.
(ii) If ).( ,52
)(2
xfx
xxf ′
−= find Hence, evaluate
( )( )
dxx
xx∫−
−
−2
1 252
5
(b) Sketch on the same coordinate system, the curves
176 2 ++= xxy and 134 2 +−−= xxy . Calculate the area of the
region bounded by the two curves.
( 5 )
( 5 )
( 10 )
(Total/ Jumlah: 20)
(a) Anda sedang mengajar kelas anda teori pengamiran.
(i) Jelaskan pengamiran tak tentu sebagai proses berbalik bagi
pembezaan dan berikan satu contoh yang ringkas.
(ii) Jika ).( cari,52
)(2
xfx
xxf ′
−= Kemudian, dapatkan nilai
( )( )
dxx
xx∫−
−
−2
1 252
5.
(b) Lakarkan lengkung 176 2 ++= xxy dan 134 2 +−−= xxy pada
satu sistem koordinat yang sama. Kira luas kawasan yang dibatasi
oleh dua lengkung tersebut.
HBMT4403_SEM JAN10/F_AH
5…
2. (a) Explain to your students the difference between exclusive and
non-exclusive sets. Use Venn diagrams to illustrate the
difference.
(b) Show to your students how to solve this problem step-by-step
using probability tree:
In a small town where Ameer stays, the rain falls randomly three days
in a week (*a week = 7 days). Of the total number of days that rain,
80% of the days Ameer takes bus to school. If it does not rain, Ameer
cycles to school. For every 5 days Ameer cycles to school, 4 days he
saves the pocket money.
For one particular day, find the probability that:
(i) Ameer does not take bus to school.
(ii) Ameer saves his pocket money because cycling to school.
( 6 )
( 14 )
(Total/ Jumlah: 20)
(a) Terangkan kepada pelajar anda perbezaan di antara set-set yang
eksklusif dan tidak eksklusif. Tunjukkan perbezaan tersebut dengan
menggunakan gambar rajah Venn.
(b) Tunjukkan kepada pelajar anda bagaimana menyelesaikan masalah
di bawah langkah demi langkah menggunakan gambar rajah pokok
kebarangkalian:
Di pekan kecil tempat Ameer tinggal, hujan turun secara rawak tiga hari
dalam seminggu (*seminggu = 7 hari). Daripada keseluruhan jumlah hari
yang hujan, 80% daripadanya Ameer pergi ke sekolah dengan menaiki
bas. Jika tidak hujan, Ameer akan ke sekolah menaiki basikal. Bagi setiap
5 hari Ameer menaiki basikal ke sekolah, dia dapat menyimpan duit poket
selama 4 hari.
HBMT4403_SEM JAN10/F_AH
6…
Cari kebarangkalian bagi satu hari tertentu bahawa:
(i) Ameer tidak menaiki bas ke sekolah
(ii) Ameer menyimpan duit poket kerana menaiki basikal ke
sekolah.
3. You are planning to do a classroom project with your students. The
project consists of seven activities. The duration of each activity is
shown in the table below. You would like to apply the network model
for this project.
Activity Preceding activities Duration (days)
A
B
C
D
E
F
G
-
-
-
B
A,D
C
E,F
2
5
1
10
3
6
8
(a) Show and explain how to draw a network diagram for the
project and to determine the minimum completion time of the
project.
(b) Calculate the total float for each activity and state the critical
path of the project. Then, explain the importance of “float” for
project management.
( 10 )
( 10 )
(Total/ Jumlah: 20)
HBMT4403_SEM JAN10/F_AH
7…
Anda bercadang untuk menjalankan satu projek bersama-sama pelajar di
dalam sebuah kelas. Projek tersebut mengandungi tujuh aktiviti. Tempoh
masa untuk setiap aktiviti ditunjukkan di dalam jadual di bawah. Anda
ingin mengaplikasikan model jaringan untuk projek berkenaan.
Aktiviti Aktiviti terdahulu Tempoh (hari)
A
B
C
D
E
F
G
-
-
-
B
A,D
C
E,F
2
5
1
10
3
6
8
(a) Tunjukkan dan terangkan gambar rajah jaringan bagi projek di atas
dan bagaimana untuk menentukan masa penyelesaian terpendek
bagi projek tersebut.
(b) Kirakan jumlah apungan bagi setiap aktiviti dan nyatakan laluan
kritikal bagi projek tersebut. Kemudian terangkan kepentingan
apungan dalam pengurusan projek.
4. How would you guide your students to solve the following problems?
(a) Box A contains 3 cards numbered 1,2 and 3 ; Box B contains 4
cards numbered 1,2,3, and 4. The cards are mixed thoroughly in
each box, and a card is then selected at random from each box.
Let X denotes the sum of numbers obtained in each trial.
i) Tabulate the distribution of X. Show that it is a
probability distribution, that is X is a discrete random
variable.
ii) Find )5( >XP
( 7 )
( 3 )
HBMT4403_SEM JAN10/F_AH
8…
(b) The discrete random variable X has a probability function given
by ( )
=
=
otherwise
xifx
xf
0
5,4,3,2,115
Find mean, variance and standard deviation of X.
( 10 )
(Total/ Jumlah: 20)
Bagaimana anda membimbing pelajar anda menyelesaikan masalah-
masalah berikut?
(a) Kotak A mengandungi 3 kad bernombor 1, 2 dan 3 ; kotak B
mengandungi kad bernombor 1, 2, 3 dan 4. Kad-kad tersebut
digoncang bebas dalam setiap kotak masing-masing, kemudian satu
kad dikeluarkan secara rawak daripada setiap kotak.
Jadikan X mewakili jumlah nombor yang diperolehi dalam setiap
cubaan.
i) Bina jadual taburan X. Tunjukkan bahawa ia adalah taburan
kebarangkalian, di mana X ialah pembolehubah rawak
diskrit.
ii) Cari )5( >XP
(b) Suatu pembolehubah rawak diskrit X mempunyai fungsi
kebarangkalian berikut:
( )
=
=
selainnya
xjikax
xf
0
5,4,3,2,115
Cari min, varian dan sisihan piawai bagi X.
HBMT4403_SEM JAN10/F_AH
9…
5. You are showing in the class the steps taken to solve the following
problems. Explain in detail.
(a) In a mathematics examination, 55% of the students do not pass.
If 20 students takes the examination, find the probability that:
i) Exactly 15 students pass
ii) at least two students pass
iii) more than 16 students pass
(b) Given that ).16,100(~ NX Use the standardised normal table to
find:
(i) )112108( << XP
(ii) ( )4100 <−XP
(10 )
(10 )
(Total/ Jumlah: 20)
Anda sedang menunjukkan di dalam kelas anda langkah-langkah
menyelesaikan masalah-masalah berikut. Terangkan dengan terperinci.
(a) Dalam satu ujian Matematik, 55% daripada pelajar yang menduduki
ujian tersebut tidak lulus. Jika satu kelas yang terdiri daripada 20
orang pelajar mengambil ujian tersebut, cari kebarangkalian
bahawa:
i) 15 orang pelajar lulus
ii) Sekurang-kurangnya dua orang pelajar lulus
iii) Lebih daripada 16 orang pelajar lulus.
(b) Diberi ).16,100(~ NX Dengan menggunakan jadual taburan
normal piawai, cari:
(i) )112108( << XP
(ii) ( )4100 <−XP
HBMT4403_SEM JAN10/F_AH
10…
PART C / BAHAGIAN C
INSTRUCTIONS / ARAHAN
Part C contains TWO questions. Answer ONE question ONLY.
Bahagian C mengandungi DUA soalan. Jawab SATU soalan SAHAJA.
QUESTION/ SOALAN: Marks/ Markah
1. Linear programming deals with problems of maximization and
minimization in real life situation. As a teacher, you should be able to
explain and transform the problems into mathematical model as well
as creating a simple problem (question) using a model.
Referring to the mathematical model given below, answer the
following questions.
Maximise 21 53 xxZ +=
Subject to:
0,
1823
122
4
21
21
2
1
≥
≤+
≤
≤
xx
xx
x
x
(a) Create a simple real life problem / question to explain how the
above model is developed.
(b) Transform the mathematical model to standard form and
explain how to solve the linear programming problem using
simplex method.
( 4 )
(16)
(Total/ Jumlah: 20)
HBMT4403_SEM JAN10/F_AH
11…
Pengaturcaraan linear berkait dengan penyelesaian masalah memaksima
dan meminima dalam situasi kehidupan sebenar. Sebagai guru, anda
sepatutnya mampu menerangkan dan menukar masalah tersebut kepada
bentuk model matematik dan juga membina soalan ringkas daripada model
matematik sedia ada.
Merujuk kepada model matematik yang diberikan di bawah, jawab soalan-
soalan berikut:
Maximise 21 53 xxZ +=
Subject to:
0,
1823
122
4
21
21
2
1
≥
≤+
≤
≤
xx
xx
x
x
(a) Bina satu soalan / masalah ringkas dalam situasi kehidupan sebenar
untuk menjelaskan bagaimana model di atas diperolehi.
(b) Tukarkan model matematik yang diberikan kepada bentuk piawai
dan terangkan bagaimana untuk menyelesaikan masalah
pengaturcaraan linear tersebut dengan menggunakan kaedah
simpleks.
HBMT4403_SEM JAN10/F_AH
12…
2. Linear regression attempts to model the relationship between two
variables by fitting a linear equation to observed data. One variable is
considered to be an independent (explanatory) variable, and the other
is considered to be a dependent variable. The most common method
for fitting a regression line is the method of least-squares.
Based on the information above and the data given below, explain to
your students and show in detail how to answer the following
questions:
X 1 2 3 4 4 6
Y 62 78 70 90 93 103
a) Find the least square regression line equation of y on x for the
above data.
b) Explain the meaning of correlation coefficient ( r ) and coefficient
of determination ( r2 ). Then, find and interpret the values for r
and r2 of the linear regression equation above.
( 12 )
( 8 )
(Total/ Jumlah: 20)
Regresi linear adalah kaedah untuk memodelkan hubungan di antara dua
pembolehubah dengan memadankan suatu garis linear kepada data yang
diperolehi. Satu pembolehubah disebut sebagai pembolehubah tak
bersandar dan satu lagi disebut pembolehubah bersandar. Kaedah yang
paling kerap digunakan untuk memadankan garis regresi linear ialah
kaedah kuasa dua terkecil.
Berdasarkan maklumat di atas dan set data yang diberikan di bawah,
jelaskan kepada pelajar anda dan tunjukkan dengan terperinci cara-cara
bagaimana untuk menjawab soalan-soalan berikut:
X 1 2 3 4 4 6
Y 62 78 70 90 93 103
SCH BSMG
ACCP 1015 AU12003A - HR
(a) Dapatkan persamaan garis regresi kuasa dua terkecil y melawan x
bagi data di atas.
(b) Terangkan apakah yang dimaksudkan dengan pekali korelasi, r dan
pekali penentu, r2. Kemudian dapatkan dan jelaskan nilai r dan r
2
bagi persamaan garis regresi linear di atas.
END OF QUESTION PAPER / KERTAS SOALAN TAMAT
SCH BSMG
ACCP 1015 AU12003A - HR
KURSUS: HBMT 4403 TEACHING MATHEMATICS IN FORM SIX
MARKING SCHEME FINAL EXAMINATION
SEMESTER: JANUARY 2010
Note: Students are not allowed to get this marking scheme.
PART A
INSTRUCTIONS
Part A contains FIVE questions. Answer ALL questions.
QUESTION 1 (Chapter 1)
From the first principles, we know that ( )
x
xfxxfxf
x δδ
δ
)(lim)(
0
−+=′
→
Given that y=uv,
x
xvxuxxvxxuuv
dx
d
x δδδ
δ
)()()()(lim)(
0
−++=∴
→
to change this fraction into an equivalent one that contains difference quotients for the
derivatives of u and v, we subtract and add )()( xvxxu δ+ in the numerator:
( ) ( )
( ) ( )x
xuxxuxv
x
xvxxvhxu
x
xuxxuxv
x
xvxxvxxu
x
xvxuxvxxuxvxxuxxvxxuuv
dx
d
xxx
x
x
δδ
δδ
δδ
δδ
δ
δδδδδ
δδδ
δ
δ
)(lim)(
)()(limlim
)()()(
)(lim
)()()()()()()()(lim)(
000
0
0
−+•+
−+•+=
−++
−++=
−+++−++=∴
→→→
→
→
as xδ approaching zero, )( xxu δ+ approaches )(xu because u , being differentiable at x , is
continuous at .x The two fractions approach the value of dxdv / at x and dxdu / at .x
∴dx
duv
dx
dvuuv
dx
d+=)(
(proving steps : 3 marks + 1 mark explanation)
(4 marks)
SCH BSMG
ACCP 1015 AU12003A - HR
QUESTION 2 (Chapter 1)
( )
( )
014
1
cos1
cos
cos1
4
1cos44
cos2
sincos2
2
1
cos
sincos2
1sincoscos
2
1
cos
sin
2
1
sincos2
1
cos
4
2
23
4
2
2
3
2
2
3
2
23
2
3
22
2
1
2
1
2
1
2
1
2
1
=++
∴
−−=
−−=
+
−
=
∴
+
−=
−−
−=∴
−=
−=∴
=
−
−
ydx
ydy
y
x
x
xx
dx
ydy
x
xx
x
xxxxx
dx
dy
x
x
xxdx
dy
xy
(4 marks)
SCH BSMG
ACCP 1015 AU12003A - HR
QUESTION 3 (Chapter 2)
(diagram : 1 mark)
A parabola is the locus points which move in such a way that its distance from a fiexd point
(focus) is always equal to its perpendicular distance from a fixed straight line (directrix) not
containing the focus.
the fixed point Q(a,0) is called the focus
the fixed line x = -a is called the directrix
0 is called the vertex. (explanation : 2 marks)
General equation of a parabola axy 42 = (equation :1 mark)
(4 marks)
y
x
P(x,y)
Q(a,0)
R
x=-a
0
SCH BSMG
ACCP 1015 AU12003A - HR
QUESTION 4 (Chapter 3 )
Using the formula ∫∫ ′−=′ dxxfxgxgxfdxxgxf )().()().()().(
one part is to be differentiate and the other part is to be integrated.
Example :
Integrate ( )dxx∫ +1ln 2 . Let
( )xxgdxx
xxf
dxxgandxxf
=+
=′
=′+=
)(21
1)(
)(1ln)(
2
2
( ) ( )
( )( ) cxxxx
dxx
xx
dxx
xxxxdxx
++−+=+
−−+=
+
−+=+∴
−
∫
∫∫
12
2
2
2
22
tan221ln
1
1121ln
1
21ln1ln
Correct formula + explanation = 1 mark
Suitable example = 3 marks
(4 marks)
QUESTION 5 (Chapter 4)
A permutation is an arrangement of a group of objects in a particular order.
Example of permutation problem: How many number can be formed using digit 1, 3,
and 8?
A combination is a selection of a group of objects where the order of objects selected is
immaterial.
Example of combination problem: How many different committees of three can be
formed by selecting the members from 6 persons?
Explanations = 2 marks
Suitable examples = 2 marks
(4 marks)
[Total: 20 marks]
SCH BSMG
ACCP 1015 AU12003A - HR
BAHAGIAN B
INSTRUCTION
Part B contains FIVE questions. Answer THREE questions ONLY.
QUESTION 1 (Chapter 3 )
(a)
(i) If we differentiate )(xf , we will write )()(
xfdx
xdf ′= , which is called the derivative
of )(xf . Conversely, in integration, )(xf is the anti derivative of )(xf ′ and the process
of finding ∫ dxxf )( for a given function )(xf is called integration.
If )(xfdx
dy=
Then ∫ ∫ ∫=⇒= dxxfythereforedxxfdy )(,)( CxF += )(
Example :
If ( ) 23 34 xxdx
d=+ . Then dxx∫ 23 = Cx ++ )4( 3
(Explanation = 3 marks)
(Example : 2 mark)
(ii)
( )( )( )
( )( )
( )22
2
2
2
2
52
52
52
102
52
)2(522)(
52)(
−
−=
−
−=
−
−−=′
−=
x
xx
x
xx
x
xxxxf
x
xxf
-1
SCH BSMG
ACCP 1015 AU12003A - HR
( )( )
( )
( )
6
11
3
1
2
1)4(
2
1
522
1
52
)5(
522
1
52
52
2
1
52
)5(
2
1
22
1 2
2
22
−=
+
−−
+−=
+
−=
−
−∴
+
−=
−
−=
−
−∴
∫∫
∫
∫∫
CC
Cx
xdx
x
xx
Cx
x
x
xxdx
x
xx
(5 marks)
(b)
12
7 when
24
25 -of valueminimum thehas curve This
24
25
12
76
176
2
2
−=⇒−
+=
++=
xx
xxy
8
3 when
16
25 of valuemaximum thehas curve This
16
25
8
34
134
2
2
−=⇒+
+−=
+−−=
xx
xxy
Solve the simultaneous equation to find the intersection points of the two curves:
( )( )2134
1176
2
2
+−−=
++=
xxy
xxy
( )10
0110
01010
134176
2
22
−==∴
=+∴
=+∴
+−−=++=∴
xorx
xx
xx
xxxxy
SCH BSMG
ACCP 1015 AU12003A - HR
(Procedure + graph = 5 marks)
Area of the region bounded by the two curves:
( ) ( )
2
0
1
3
0
1
2
20
1
2
3
55
3
10
53
10
1010
176134
units
xx
dxxx
dxxxxx
=
+−=
−−=
−−=
++−+−−=
−
−
−
∫
∫
(5 marks)
(Total: 20 marks)
1
-1
1762 ++= xxy
134 2 +−−= xxy
SCH BSMG
ACCP 1015 AU12003A - HR
QUESTION 2 (Chapter 7 )
(a) Set A and Set B are exclusive sets if their intersection is empty, thus φ=∩ BA .
If Set A and Set B are not exclusive, then Set A and Set B have some elements
in common. The following Venn Diagrams show the difference of exclusive sets
and not exclusive sets.
(3 marks)
Exclusive Sets Not exclusive sets
(1.5 marks each diagram = 3 marks)
(b) Let
moneypocket his savenot doesAmeer event the:
moneypocket his savesAmeer event the:
school tocyclesAmeer event the:
school tobus sAmeer takeevent the:
rainnot doesit event that the:
rainsit event that the:
S
S
C
B
R
R
SCH BSMG
ACCP 1015 AU12003A - HR
The probability tree diagram :
`
(6 marks)
a)
( ) ( )
657.035
23
7
6.4
7
4
7
6.0
0.17
42.0
7
3
school) tobus not take does(
atau
BRPBRP
P
=
=
+=
×+×=
∩+∩=
(4 marks)
7
3
R
R
7
4
0.1C
8.0
2.0
B
C
S
S
S
S
5
4
5
4
5
1
5
1
SCH BSMG
ACCP 1015 AU12003A - HR
0
A
D
B
C F
G
E
3
b)
( ) ( )
5257.0350
184
35
4.18
35
16
35
4.2
5
40.1
7
4
5
42.0
7
3
money)pocket his savesAmeer (
atau
SCRPSCRP
P
=
=
+=
××+××=
∩∩+∩∩=
(4 marks)
(Total: 20 marks)
QUESTION 3 (Chapter 5 )
(a)
Minimum completion time = 26 days
(Correct diagram = 4 marks)
(Calculation = 6 marks)
1
0 0
2
5 5
4
15 15
3
12 1
5
18 18
6
26 26
3
8 5
10 2
SCH BSMG
ACCP 1015 AU12003A - HR
(b)
Activity Total float
A
B
C
D
E
F
G
13
0
9
0
0
19
0
Critical path : B-D-E-G
(4 marks)
Floats are useful for management because they indicate how much an activity can be
delayed without delaying the whole project. Resources can be redeployed from an
activity with float to be used elsewhere, possibly to ensure that a critical activity finishes
on time.
(6 marks)
(Total: 20 marks)
QUESTION 4 (Chapter 8 )
(a) (i)
(4 marks)
So the distribution of X is:
X=x 2 3 4 5 6 7
P(X=x)
12
1
12
2
12
3
12
3
12
2
12
1
X=x Outcomes corresponding to
X=x
P(X=x)
2
3
4
5
6
7
(1,1)
(1,2), (2,1)
(1,3), (3,1), (2,2)
(2,3), (3,2), (1,4)
(3,3), (2,4)
(3,4)
1/12
2/12
3/12
3/12
2/12
1/12
SCH BSMG
ACCP 1015 AU12003A - HR
This is a probability distribution, that X is a discrete random variable, because P(X=x) ≥ 0
for every value of x and P(S)=1
(4 marks)
(ii) P(X>5) = P(X=6) + P(X=7)
= 4
1
12
1
12
2=+ (2 marks)
(c)
∑ ==== )()( xXxPXEXMean µ
3
11
15
55
15
44
15
33
15
22
15
11 =
+
+
+
+
=
(4 marks)
9
14
3
1115
1515
525
15
416
15
39
15
24
15
11)(
)()(
2
2
222
=
−=∴
=
+
+
+
+
=
−==
Varians
XE
XEXVar µσ
(4 marks)
3
14
9
14)(,, ofdeviation Standard ==XSDX σ (2 mark)
(Total: 20 marks)
SCH BSMG
ACCP 1015 AU12003A - HR
QUESTION 5 (Chapter 9 )
(5
markah)
(a) Let q to be the proportion of students fail, th
(i)
−<<
−=<<
16
100112
16
100108)112108( ZPXP
)32( <<= ZP
02145.0
00135.00228.0
)3()2(
=
−=
(5 marks)
(ii) ( ) ( )410044100 <−<−=<− XPXP
( )10496 <<= XP
−<<
−=
16
100104
16
10096ZP
( ) ( )
( ) ( )
( )
( )[ ][ ]
( )
( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )
0003.0
45.055.045.0)20(55.045.0)190(55.045.0)1140(
45.055.045.055.045.055.045.0
)20()19()18()17(16)
9999.0
0001.01
)55.0)(45.0)(20()55.0(1
)1()0(1
11)2()
0049.0
55.045.0!5!5
!20
55.045.0)15()
)45.0,20(~ then pass, whostudents ofnumber theis If
45.055.01
55.0
20119218317
20
20
20119
19
20218
18
20317
17
20
1920
515
515
15
20
=
+++=
+++=
=+=+=+==>
=
−=
+−=
=+=−=
≤−=≥
=
=
==
=−=
=
CCCC
XPXPXPXPXPiii
XPXP
XPXPii
CXPi
BXX
pSo
q
marks 3
marks 2
SCH BSMG
ACCP 1015 AU12003A - HR
6826.0
)1587.0(21
)1(21
)11(
=
−=
−=
<<−=
Q
ZP
(5 marks)
(Total: 20 marks)
SCH BSMG
ACCP 1015 AU12003A - HR
PART C
INSTRUCTIONS
Part C contains TWO questions. Answer ONE question ONLY.
QUESTION 1 (Chapter 4 )
(a) Accept any SUITABLE and CORRECT answer.
(4 marks)
(b) Change to standard form:
053 21 =−− xxZ
411 =+ Sx
122 22 =+ Sx
1823 321 =++ Sxx
Simplex tableau:
iteration
Basic
variable
Coefficient of Right
Side Z
1x 2x 1S 2S 3S
0
Z
1S
2S
3S
1
0
0
0
-3
1
0
3
-5
0
2
2
0
1
0
0
0
0
1
0
0
0
0
1
0
4
12
18
iteration
Basic variable
Coefficient of Right Side
Z 1x 2x 1S 2S 3S
1
Z
1S
2x
3S
1
0
0
0
-3
1
0
3
0
0
1
0
0
1
0
0
5/2
0
1/2
-1
0
0
0
1
30
4
6
6
SCH BSMG
ACCP 1015 AU12003A - HR
The table is optimal since there are no more negative values entries in the z row.
Optimum solution : 1x =2 ; 2x =6 and profit (Z) to be maximized is = RM36,000.
(workout & answer : 16 marks)
iteration
Basic
variable
Coefficient of Right
Side Z
1x 2x 1S 2S 3S
2
Z
1S
2x
1x
1
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
3/2
1/3
1/2
-1/3
1
1/3
0
1/3
36
2
6
2
(Total: 20 marks)
QUESTION 2 (Chapter 10)
(a)
y = ax+b as a regression line:
x y x² y² xy
1 62 1 3844 62
2 78 4 6084 156
3 70 9 4900 210
4 90 16 8100 360
4 93 16 8649 372
6 103 36 10609 618
∑x = 20 ∑y = 496 ∑x²=82 ∑y²= 42186 ∑xy = 1778
( )
33.15
6
2082
2
2
2
=
−=
−=∑ ∑n
xxSxx
SCH BSMG
ACCP 1015 AU12003A - HR
( )
33.1183
6
49642186
2
2
2
=
−=
−=∑ ∑n
yySyy
( )( )
66.124
6
)496)(20(1778
=
−=
−=∑ ∑∑n
yxxySxy
59.55
)33.3(13.867.82
13.8
33.15
66.124
=
−=
−=
=
==
xayb
Sxx
Sxya
→ regression line equation : y= 55.59+8.13x
(workout and answer : 12 marks)
(b)
The correlation coefficient, r, is the strength of the relationship between the two variables in
the regression equation and is always a value between -1 and 1, inclusive. The regression
coefficient is the slope of the line of the regression equation.
(2 marks)
The coefficient of determination, r2, measures how good the estimated regression equation is.
The higher the r2, the more confidence one can have in the equation. Statistically, the
coefficient of determination represents the proportion of the total variation in the y variable
that is explained by the regression equation. It has the range of values between 0 and 1
(2 marks)
correlation coefficient , r = 9256.033.118333.15
66.124=
×=
SxxSyy
Sxy
Indicates a strong positif linear relationship between X and Y
SCH BSMG
ACCP 1015 AU12003A - HR
(2 marks)
(c) coefficient of determination , r²
8567.0
33.1183
33.15/66.124/ 22
=
=
=
Syy
SxxSxy
Indicates that 85.67% of the total variation is explained by the regression line.
(2 marks)
(Total: 20 marks)
SCH BSMG
ACCP 1015 AU12003A - HR
COURSE: HBMT4403 TEACHING MATHEMATICS IN FORM SIX
1. DIFFERENTIATION FORMULA
2. INTEGRATION FORMULA
Basic functions
=
Basic Functions
Addition
Linearity
Product Rule
Quotient Rule
Chain Rule
Inverse
Trig Functions
SCH BSMG
ACCP 1015 AU12003A - HR
Trigonometric functions
3. DISCRETE PROBABILITY DISTRIBUTION
∑ ==== )()()( xXxPXEXMean µ
222 )()( µσ −== XEXVar
2)(deviation Standard σ=X
SCH BSMG
ACCP 1015 AU12003A - HR
4. REGSRESSION ANALYSIS
( )∑ ∑−=
n
xxSxx
2
2 ; ( )
∑ ∑−=n
yySyy
2
2 ; ( )( )
∑ ∑∑−=n
yxxySxy
xaybandSyy
Sxxawherebaxy −==+= ;
SxxSyy
Sxyr =
=
Syy
SxxSxyr
/2
2