m.a.v. revision lectures mathematical methods units 3 and 4 presenter: michael swanborough flinders...
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M.A.V. REVISION LECTURES
MATHEMATICAL METHODS
UNITS 3 AND 4
Presenter: MICHAEL SWANBOROUGH
Flinders Christian Community College
EXAMINATION 1 - Facts, Skills and Applications Task
Part A - Multiple-choice questions
Part B - Short-answer questions
EXAMINATION 2 - Analysis Task
Examinations
Examination Advice
General Advice
• Answer questions to the required degree of accuracy.
• If a question asks for an exact answer then a decimal approximation is not acceptable.
• When an exact answer is required, appropriate working must be shown.
Examination Advice
General Advice
• When an instruction to use calculus is stated for a question, an appropriate derivative or antiderivative must be shown.
• Label graphs carefully – coordinates for intercepts and stationary points; equations for asymptotes.
• Pay attention to detail when sketching graphs.
Examination Advice
General Advice
• Marks will not be awarded to questions worth more than one mark if appropriate working is not shown.
Examination Advice
Notes Pages
• Well-prepared and organised into topic areas.
• Prepare two pages of general notes.
• Prepare two separate pages for each of the two examinations.
• Include process steps rather than just specific examples of questions.
Examination Advice
Notes Pages
• Some worked examples can certainly be of benefit.
• Include key steps for using your graphic calculator for specific purposes.
• Be sure that you know the syntax to use with your calculator (CtlgHelp is a useful APP for the TI-83+)
Examination Advice
Strategy - Examination 1
• Use the reading time to plan an approach for the paper.
• Make sure that you answer each question. There is no penalty for incorrect answers.
• It may be sensible to obtain the “working marks” in the short answer section before tackling the multiple choice questions.
Examination Advice
Strategy - Examination 1
• Some questions require you to work through every multiple-choice option – when this happens don’t panic!!
• Eliminate responses that you think are incorrect and focus on the remaining ones.
• Questions generally require only one or two steps – however, you should still expect to do some calculations.
Examination Advice
Strategy - Examination 2
• Use the reading time to carefully plan an approach for the paper.
• Momentum can be built early in the exam by completing the questions for which you feel the most confident.
• Read each question carefully and look for key words and constraints.
Examination Advice
Strategy - Examination 2
• If you find you are spending too much time on a question, leave it and move on to the next.
• When a question says to “show” that a certain result is true, you can use this information to progress through to the next stage of the question.
Revision Quiz
1 2 3
4 5 6
7 8 9
Question 1
sin xeThe derivative of is equal to
cos xe
cos(cos ) xx e
a)sin xeb) c)
d) e)
sin(cos ) xx e
(cos ) xx e
A
f(x)
x1 2 3 4 5 6-1-2-3
12345
-1-2-3-4-5
The range of the function with graph as shown is
Question 2
B
6,2
4,24,5
4,5
3,24,5
6,54,2
a)
b)
c)
d)
e)
Angie notes that 2 out of 10 peaches on her peach tree are spoilt by birds pecking at them. If she randomly picks 30 peaches the probability that exactly 10 of them are spoilt is equal to
Question 3
a)
d)
b)
e)
c)
2.0 2010 )8.0()2.0(
2010 )8.0()2.0( 201010
30 )8.0()2.0(C
102010
30 )8.0()2.0(C
D
Question 4
1
2
)( dxxf
1
0
0
2
)()( dxxfdxxf
1
0
0
2
)()( dxxfdxxf
2
1
)( dxxf
0
2
1
0
)()( dxxfdxxf
a)
d)
e)
c)
b) y = f(x)
-2
-1
2
1
y
x
The total area of the shaded region shown is given by
D
Question 5
What does V.C.A.A. stand for?
a) Vice-Chancellors Assessment Authority
b) Victorian Curriculum and Assessment Authority
c) Victorian Combined Academic Authority
d) Victorian Certificate of Academic Aptitude
e) None of the above
B
X1
~ N (11
, )
X2
~ N (22
, )
2
2
Which one of the following sets of statements is true?
a) 2121 and b) 2121 and c) 2121 and d) 2121 and e) 2121 and
A
Question 6
Bonus Prize!!
Question 822 ))()(()( cxbxaxxP
where a, b and c are three different positive real numbers. The equation has exactly
a) 1 real solution
b) 2 distinct real solutionsc) 3 distinct real solutions
d) 4 distinct real solutions
e) 5 distinct real solutions B
Question 9
3
5
For the equation 03sin2 x π2,0
the sum of the
solutions on the interval is
a) b)
c) d)
e)
23
7
3E
EXAMINATION 1 - FACTS, SKILLS AND APPLICATIONS TASK
• Part A
– 27 multiple-choice questions (27 marks)
• Part B
– short-answer questions (23 marks)
• Time limit:
– 15 minutes reading time
– 90 minutes writing time
• Extended response questions
– 4 questions (55 marks)
• Time limit: – 15 minutes reading time
– 90 minutes writing time
EXAMINATION 2 - ANALYSIS TASK
4 3 2
3 2
2
3 3
3 3
1 3
1 3 3
x x x x
x x x x
x x x
x x x x
Question 1
ANSWER: B
The linear factors of the polynomial
are4 3 23 3x x x x
xxxx
xxxx
xxxx
37545018024
)125150608(3
)5()5)(2(3)5()2(3)2(3
234
23
3223
3)52(3 xxa) Expand fully
Question 4
22 23 axxx
7
2140
22)2(220
0)2(23
a
a
a
P
b) is exactly divisible by
.2x Find the value of a.
y
x
10
2
105 23 xxxy
)53)(2(
105
105
63
3
2
531052
105
2
2
2
23
2
23
23
xxx
x
x
xx
xx
xx
xxxxxx
xxxy
Question 5
a)
3 2
2
2 2 2
2
5 10
( 2)( )
2 10 2 5
5 3
( 2)( 3 5)
y x x x
y x x bx c
c x bx x
c b
y x x x
2
293,
2
293,2
2
293
2
)5)(1(433
053,2
0)53)(2(
2
2
2
x
x
x
xxx
xxxb)
3
27
4320160
4320)()2(
3
3
333
6
a
a
a
aC
ANSWER: B
Question 6 Coefficient of 63 )2(in axx
6048
)32)(9(21
)2()3( 522
7
C
ANSWER: D
Question 7 Coefficient of 72 )23(in xx
Functions and Their Graphs
Vertical line test - to determine whether a relation is a function
rule)( where,: xfBAf
A represents the DOMAIN
bxaxba :,
bxaxba :,
bxaxba :,
bxaxba :,
Interval Notation
Square brackets [ ] – included
Round brackets ( ) – excluded
2, 3
3, 3
3, 1 2, 3
2, 1 2, 3
2, 0 2, 3
a)
b)
c)
d)
e)
The range of the function with graph as shown is
ANSWER: D
Question 9
f(x)
x1 2 3 4-1-2-3
1
2
3
4
-1
-2
-3
A function is undefined when:
a) The denominator is equal to zerob) The square root of a negative number is
present.
Maximal (or implied) Domain
The largest possible domain for which the function is defined
32)( xxfConsider the function
032 x
,
2
3or
2
3:xx
So the maximal domain is:
)4()( 2 xxxf
)1)(4)(2()( 2 xxxxxf
)16)(3()( 4 xxxf
)4)(6()( 2 xxxxf
)12)(6()( 22 xxxxxf
This question requires EVERY option to be checked carefully.
a)
b)
c)
d)
e)
Question 10
placesfour in axis theCuts
)3)(4)(2)(3(
)12)(6()( 22
xxxx
xxxxxf
ANSWER: E
2( ) ( )( )f x x a x b 2( ) ( )( )f x x a x b 2( ) ( )( )f x x a x b 2( ) ( )( )f x x a x b
2( ) ( ) ( )f x x a x b
a)
b)
c)
d)
e)
Question 11
a b
y = f(x)
x
y
The graph shown could be that of the function f whose rule is
ANSWER: A
Using Transformations
NATURE - Translation, Dilation, Reflection
MAGNITUDE (or size)
DIRECTION
When identifying the type of transformation that has been applied to a function it is essential to state each of the following:
1.Translations
a) Parallel to the x-axis – horizontal translation.
b) Parallel to the y-axis – vertical translation.
To avoid mistakes, let the bracket containing x equal zero and then solve for x.
If the solution for x is positive – move the graph x units to the RIGHT.
If the solution for x is negative – move the graph x units to the LEFT.
2. Dilations
a) Parallel to the y-axis – the dilation factor is the number outside the brackets. This can also be described as a dilation from the x-axis.
b) Parallel to the x-axis – the dilation factor is the reciprocal of the coefficient of x. This can also be described as a dilation from the y-axis.
Note: A dilation of a parallel to the y-axis is the
same as a dilation of 1
aparallel to the x-axis.
3. Reflections
)(xfy a) Reflection about the x-axis
)( xfy b) Reflection about the y-axis
)( xfy c) Reflection about both axes
xy d) Reflection about the line
y
x
Reflection about the x-axis
y
x
Reflection about the y-axis
Reflection about both axes
y
x
y
x
2
)(xfy Question 13
Determine the graph of )(1 xfy
y
x-2
)(xfy
Reflection about the x-axis
y
x-1
)(1 xfy
Translation of 1 unit parallel to the y-axis
ANSWER: A
EXTRA QUESTION
The graph of the function f is obtained from the graph of the function with equation y xby a reflection in the y-axis followed by a dilation of 2 units from the x-axis. The rule for f is:
a)
b)
c)
d)
e)
2f x x
2f x x
0.5f x x
0.5f x x
2f x x ANSWER: E
Reflection: f x x
Dilation: 2f x x
1 2 – 1 – 2
y
x
2
2
32
32
2
2
1
1
2
2
– 1
– 1
– 2
– 2
y = f(x)
y = g(x)
Question 15
Dilation by a factor of 0.5 from the y-axis
Dilation by a factor of 2 from the x-axis
Transform f(x) to g(x)
Graphs of Rational Functions
The equations of the horizontal and vertical asymptotes of the graph with equation
23
4y
x
Vertical: 4 0
4
x
x
Horizontal: 3y
ANSWER: E
Question 16
Inverse Functions
Key features:
Domain and range are interchanged
Reflection about the line y = x
The original function must be one-to-one
1 domran ff ff domran 1
To find the equation of an inverse function
Step 1: Complete a Function, Domain, Range (FDR) table.
Step 2: Interchange x and y in the given equation.
Step 3: Transpose this equation to make y the subject.
Step 4: Express the answer clearly stating the rule and the domain.
)1(log2
1
2)1(log
1
1
:Inverse
2
2
xy
yx
ex
ex
e
e
y
y
1)( where,: )2( xexfRRf
Rf
Rf
RDF
,1
,11
ANSWER: A
Question 17
ANSWER: C
Question 18
x
y
x
y
Graph of the inverse function
f(x)
x 2 4 6 8-2-4-6-8
2
4
6
8
-2
-4
-6
-8
)2(log4)(,),2(: xxfRf e
Question 20
places) decimal (3649.3
2
)2(log42)2,(
)2(log4)(
2
1
k
ek
kk
xxf
e
e
f(x)
x
x = 2
= 2y
(3, 0)
(0, 3)
Label asymptotes
Label coordinates
Approach asymptotes
y
x
2
1
axy
ANSWER: E
Question 21
The equation relating x and y is most likely:
y
x
0.5
0.5
1
1
1.5
1.5
2
2
2.5
2.5
3
3
– 0.5
– 0.5
0.5
0.5
1
1
1.5
1.5
2
2
– 0.5
– 0.5
– 1
– 1
g(x)
h(x)
f(x)
)()()( xhxgxf
)()()( xhxgxf
)(2)( xgxf
)1()( xhxg
)(2)( xgxh
a)
b)
c)
d)
e)
ANSWER: B
Question 22
Solving indicial equations
Step 1: Use appropriate index laws to reduce both sides of the equation to one term.
Step 2: Manipulate the equation so that either the bases or the powers are the same.
Step 3: Equate the bases or powers. If this is not possible then take logarithms of both sides to either base 10 or base e.
places) decimal2(25.3
3
1997log
2
13
1997
19973
2
2
x
x
e
e
e
x
x
Question 23
Step 1: Use the logarithmic laws to reduce the given equation to two terms – one on each side of the equality sign.
Step 2: Convert the logarithmic equation to indicial form.
Step 3: Manipulate the given equation so that either the bases or the powers are the same.
Solving logarithmic equations
Step 4: Equate the bases or powers. If this is not possible then take logarithms of both sides to either base 10 or base e.
Step 5: Check to make sure that the solution obtained does not cause the initial function to be undefined.
2
2
24
2 4 2
2
2log log 16 4
log log 16 4
log 416
16
16 , 4
but 0 4
a a
a a
a
x
x
x
xa
x a x a
x x a
ANSWER: A
Question 26
yx
y
x
y
x
yx
yxx
yxx
10
10
1log
1loglog
log1log2log3
log1loglog3
10
1010
101010
102
1010
ANSWER: D
Question 27
Circular (Trigonometric) Functions
dcxbaxf ))(sin()(
dcxbaxf ))(cos()(
Amplitude: a
Period: b
2
Horizontal translation: c units in the negative x-direction
Vertical translation: d units in the positive y-direction
ANSWER: C
Question 29
1
1
2
2
3
3
4
4
1
1
2
2
3
3
4
4
– 1
– 1
y
x
Amplitude: 2
2Period: 4
2Translation: 2 units up
b
b
2sin 22
y x
: , ( ) 2cos(3 ) 1f R R f x x
Amplitude: 2
Period: 3
2
Range:
2 1 1
2 1 3
3, 1
ANSWER: B
Question 30
0 where,)2sin()( pqxpxf
y
x
p
-p
0q
qp qp
pqp
qp2
1
a)
b)
c)
d)
e)
ANSWER: C
Question 31
Question 32
1 2 – 1 – 2
y
x
2
2
32
32
2
2
1
1
2
2
– 1
– 1
– 2
– 2
y = f (x)
y = g(x)
Dilation of factor 2 from the x-axis
ANSWER: C
1 2 – 1 – 2
y
x
2
2
32
32
2
2
1
1
2
2
– 1
– 1
– 2
– 2
y = f (x)
y = g(x)
Reflection in the x-axis
1 2 – 1 – 2
y
x
2
2
32
32
2
2
1
1
2
2
– 1
– 1
– 2
– 2
y = f (x)
y = g(x)
Solving Trigonometric Equations
• Put the expression in the form sin(ax) = B
• Check the domain – modify as necessary.
• Use the CAST diagram to mark the relevant
quadrants.
• Solve the angle as a first quadrant angle.
• Use symmetry properties to find all solutions
in the required domain.
• Simplify to get x by itself.
Question 33
a) 4
44cos 16
12
14 16
214
t
C
20 4cos 1612
cos 112
1212
4 pm
t
t
t
t
t
b)
ANSWER: E
Question 34
sin 2 1 0,4x x
sin 2 1 0 2 8
2 , 2 , 4 , 62 2 2 2
5 9 13, , ,
4 4 4 4
x x
x
x
3
3tan
6
)2tan(
)2cos()2sin(
a
a
x
ax
xax
ANSWER: E
Question 35
Question 36 Analysis Question
325 4cos , for 0 24
12
tT t
Maximum: 25 4 29 C
Minimum: 25 4 21 C
a)
cos is maximum when
3
12
3 12
15
3pm
t
t
t
b)
325 4cos 23
123 1
cos12 23 5
,12 3 3
3 4, 20
7, 23
7am, 11pm
t
t
t
t
t
t
c)
Maximum at 15
Interval: 15 2, 15 2
13 28.46 C
17 28.46 C
Minimum temp: 28.46 C
t
t T
t T
d)
34sin
12 123
sin3 12
tdT
dtt
e) i)
3sin 0.2
3 123 0.6
sin12
30.192, 2.949
123.73, 14.27
Interval is: 3.73, 14.27
t
t
t
t
e) ii)
DIFFERENTIAL CALCULUS
dx
du
du
dy
dx
dyChain Rule:
dx
dvu
dx
duvuv
dx
d)(Product Rule:
2vdx
dvu
dx
duv
v
u
dx
d
Quotient Rule:
Further Rules of Differentiation
( )y f x( )
2 ( )
dy f x
dx f x
Square Root Functions
Further Rules of Differentiation
sin ( )y f x ( ) cos ( )dy
f x f xdx
cos ( )y f x ( )sin ( )dy
f x f xdx
Trigonometric Functions
tan ( )y f x 2( )sec ( )dy
f x f xdx
Further Rules of Differentiation
xy elogxdx
dy 1
)(log xfy e)(
)(
xf
xf
dx
dy
Logarithmic Functions
)75(log xy e
75
5
xdx
dy
)(sinlog xy e
cos
sincot
dy x
dx xx
Examples:
Further Rules of Differentiation
xey xedx
dy
)(xfey )()( xfexfdx
dy
Exponential Functions
)35( 2 xxey )35( 2
)52( xxexdx
dy
xey cos xexdx
dy cossin
Examples:
3
3 3
2 3
3 2
( 4)
( 4) ( 4)
(3 ) ( 4)( )
( 3 4)
x
x x
x x
x
y e x
dy d de x x e
dx dx dx
e x x e
e x x
ANSWER: D
Question 37
4
2
22
22
2
)3cos()2()3sin(3
)3cos()3cos(
)3cos(
t
tttt
dt
dy
t
tdt
dtt
dt
dt
dt
dy
t
ty
ANSWER: A
Question 39
Graphs of Derived Functions
ANSWER: A
Question 40
1
1
2
2
3
3
– 1
– 1
– 2
– 2
1
1
– 1
– 1
f(x)
x
1
1
2
2
3
3
– 1
– 1
– 2
– 2
1
1
– 1
– 1
f(x)
f(x)
x
ANSWER: C
Question 42
1
2
When 4, 16
3
4, 6
16 6 4
6 8
x y
dyx
dxdy
xdx
y x
y x
3 2 2 2y x x x
ANSWER: B
Question 43
( 1.215, 0.548)Positive gradient for
Approximations
ANSWER: B
Question 44
(16) 16
0.04
16.04 16 0.04 16
f x h f x hf x
f
h
f f f
Question 46 Analysis Question
2: , 2 3x xf R R f x e ke
3a a) b)
0 0
at 0, ( ) 0
0 2 3
0 1 2 3
2 4
2
x f x
e ke
k
k
k
2
2
2
4 3
2 4
0 2 4
2 ( 2) 0
2 as 0
log 2
x x
x x
x x
x x
x x
e
y e e
dye e
dx
e e
e e
e e
x
2 log 2 log 2( ) 4 3
4 8 3
1
log 2, 1
e e
e
f x e e
c) Use CALCULUS to find the EXACT values of the COORDINATES of the turning point.
2
When , 0
0 4 3
( 3)( 1) 0
3, 1
log 3, 0
but 0
log 3
c c
c c
c c
e
e
x c y
e e
e e
e e
c
c
c
d)i)
log 3
2
0
log 32
0
2 log 3 log 3 0 0
log 9
4 3
14 3
2
1 14 3log 3 4
2 2
1 14 3 3log 3 4
2 24 3log 3
e
e
e e
e
x x
x x
e
e
e
A e e dx
e e x
e e e e
e
ii)
y = a
y = f(x)
c – c
y = g(x)
y
x
2( ) 4 3x xg x e e
Antidifferentiation and Integral Calculus
cna
baxdxbax
nn
)1(
)()(
1
1,1
1 1
ncxn
dxx nn
cx
cx
dxx
35
)35(
)5(7
)35()35(
7
76
Examples
cx
cx
cx
dxxdx
x
5
3
5
3
5
3
5
2
5
2
)72(2
5
)72(6
53
25
3)72(
3
)72(3
)72(
3
3
2
1
2
1
2
1
2
2 (4 1)
(4 1)2
14
2
(4 1)
1
(4 1)
x dx
xc
x c
x
3
2
2
(4 1)
dy
dxx
ANSWER: E
Question 47
Trigonometric Functions
Rules of Antidifferentiation
)cos()sin( kxkkxdx
d
)sin()cos( kxkkxdx
d
ckxk
dxkx
)cos(1
)sin(
ckxk
dxkx )sin(1
)cos(
Rules of Antidifferentiation
Exponential Functions
kxkx
xx
keedx
d
eedx
d
cek
dxe kxkx 1
Rules of Antidifferentiation
Logarithmic Functions
)(
)()(log
xf
xfxf
dx
de
cxfdxxf
xfe )(log
)(
)(
Examples
cx
xxfdxxf
xfdx
x
e
)34(log
34)( where)(
)(
34
4
cx
xxfdxxf
xf
dxx
xdx
x
x
e
)5(log2
1
5)( where)(
)(
2
15
2
2
1
5
2
2
22
)()(
)()(
aFbF
xFdxxfb
a
ba
Definite Integrals
22
0
23 2
0
( 3 4)
34
3 2
8 128 (0)
3 2
2
3
x x dx
x xx
Example
Properties of Definite Integrals
4
1
(2 ( ) 1)f x dx
ANSWER: D
Question 49
4 4
1 1
44
11
4
1
2 ( ) 1
2 ( )
2 ( ) 3
f x dx dx
f x dx x
f x dx
4 4 4
1 1 1
44
11
2 3 2 3
2 3
4 12 3
13
f x dx f x dx dx
f x dx x
EXTRA QUESTION
4
1
( ) 2f x dx If then is equal to:
4
1
2 ( ) 3f x dx
followsit then ),()( xgxfdx
d
cxfdxxg )()(
Integration by recognition
cxxxdxx
ddxxxdxx
dxxdxxdx
dxxdxx
xxxdx
d
ee
ee
ee
ee
ee
loglog
1loglog
loglog1
log)log1(
log1log
ANSWER: B
Question 50
ANSWER: B
Question 52
a b O
y = f(x)
y
x
On the interval (a, b) the gradient of g(x) is positive.
Calculating Area
• Sketch a graph of the function, labelling all x-intercepts.
• Shade in the region required.• Divide the area into parts above the x-axis and
parts below the x-axis.• Find the integral of each of the separate sections,
using the x-intercepts as the terminals of integration.
• Subtract the negative areas from the positive areas to obtain the total area.
The total area of the shaded region is given by:
0 1
2 0
( ) ( )f x dx f x dx
ANSWER: C
Question 53
y = f(x)
y
x -2 1
2
ANSWER: D
Question 54
The total area bounded by the curve and the x-axis is given by:
a b cO
y = f(x)
y
x
b b
a cf x dx f x dx
Question 55
log 2
2log 2 1 1
2
log 2
e
e
e
y x x x
dyx x
dx x
x
a)
b) Hence, find the exact area of the shaded region
1
2 e
2
y
x
2212
1 22
log 2 log 2
1 1log log 1
2 2 2 2
1
2
ee
e
e e
x dx x x x
e ee
Area between curves
b
a
b
a
b
a
dxxgxf
dxxgdxxfA
)()(
)()(a b
f(x)
g(x)
x
y
• Sketch the curves, locating the points of intersection.
• Shade in the required region.
• If the terminals of integration are not given – use the points of intersection.
• Check to make sure that the upper curve remains as the upper curve throughout the required region. If this is not the case then the area must be divided into separate sections.
• Evaluate the area.
Method
y
xb c
y = f(x)
y = g(x)
c
dxxgxf0
))()((
The area of the shaded region is given by:
BOS 1997 CAT 2 Q. 18
Question 56
Find the exact area of the shaded region
4
4
2
2
34
34
54
54
32
32
74
74
2
2
1
1
– 1
– 1
y = cosx
y = sinx
y
x
5
4
4
5
4
4
sin cos
cos sin
5 5cos sin cos sin
4 4 4 4
2 2
A x x dx
x x
0 1 2
1 2
(0) (1) (2)
1
A f f f
e e e
e e
Numerical techniques for finding area
ANSWER: A
Question 57
1
1
2
2
3
3
f(0) f(1)
f(2)
x
y
Question 58 Analysis Question
4 3 212 5 3
2y x x x x
23
23
3 34 5
2 2
3 34 5 0
2 2
dy xx x
dx
xx x
a)
b) i)
23 3 3
4 52 2
when 1, 1
1
11,
21
1 12
1.5
normal
dy xx x
dxdy
xdx
m
x y
y x
y x
b) ii)
4 3 2
4 3 2
2
1.5 0.5 2 5 3
0.5 2.5 0.5 1.5 0
1 1.5 1 0
x x x x x
x x x x
x x x
A repeated root at x = -1 indicates that the normal is a tangent to the curve at this point.
5 5When 1, 1,
2 2x y B
A
B
x
yc) i)
c) i)
14 3 2
1
14 3 2
1
0.5 2 5 3 1.5
0.5 2.5 0.5 1.5
A x x x x x dx
x x x x dx
c) ii)
Discrete Random VariablesA discrete random variable takes only distinct or discrete values and nothing in between.
Discrete variables are treated using either discrete, binomial or hypergeometric distributions.
A continuous random variable can take any value within a given domain. These values are usually obtained through measurement of a quantity.
Continuous variables are treated using normal distributions.
Expected value and expectation theorems
)Pr(
)Pr(.....)Pr()Pr(
)(E
2211
xXx
xXxxXxxXx
X
nn
bXabaX
XaaX
)(E)(E
)(E)(E
Variance and Standard Deviation
22
2
)(E)(E
)(Var
XX
X
)(Var)(SD XX
)(Var)(Var 2 XaaX
Melissa constructs a spinner that will fall onto one of the numbers 1 to 5 with the following probabilities.
Number 1 2 3 4 5
Probability 0.3 0.2 0.1 0.1 0.3
The mean and standard deviation of the number that the spinner falls onto are, correct to two decimal places,
Question 60
ANSWER: E
x
1 0.3 0.3 0.3
2 0.2 0.4 0.8
3 0.1 0.3 0.9
4 0.1 0.4 1.6
5 0.3 1.5 7.5
2.9 11.1
Pr( )X x Pr( )x X x 2 Pr( )x X x
22
2
Var( ) E( ) E( )
11.1 (2.9)
2.69
SD( ) 1.64
X X X
X
The Binomial Distribution
),(Bi~ pnX
nxppCxX xnxx
n ......,2,1,0,)1()()Pr(
qpnX
pqqpnX
pnX
)(SD
1where,)(Var
)(E
2
In a two-week period of ten school days, the probability that the traffic lights have been green on exactly nine occasions is:
10 9 1
9
Bi 10,0.4
Pr( 9) (0.4) (0.6)
X
X C
ANSWER: A
Question 61
3 7103Pr 3 0.2 0.8
Bi 10, 0.2
Mean
2
Variance
1.6
X C
X n p
np
npq
ANSWER: A
Question 63
The Hypergeometric Distribution
),,(Hg~ NDnX
,)Pr(n
Nxn
DNx
D
C
CCxX
)(Var)(SD
11)(Var
)(E
2
XX
N
nN
N
D
N
DnX
N
DnX
6 41 310
4
Pr( 1)C C
XC
Question 64
A team of four is selected from six women and four men. What is the probability that the team consists of exactly one woman and three men.
ANSWER: A
12 3 12 33 1 4 015 15
4 4
Pr( 3) Pr( 3) Pr( 4)
0.846
X X X
C C C C
C C
Question 65
A jar contains fifteen jellybeans of which twelve are green. Four jelly beans are taken from the jar at random and eaten, calculate Pr( 3)X
Calculator program
The Normal Distribution
The mean, mode and median are the same.
The total area under the curve is one unit.
b
adxxfbXa )()Pr(
1
2
Same Different
Same Different
Which one of the following sets of statements is true?
ANSWER: A
Question 67
1 2 1 2, X1
~ N (11
, )
X2
~ N (22
, )
2
2
• Draw a diagram, clearly labelling the mean.
• Shade the region required.
• Determine the z value which corresponds to the value of x by using
• Use the cumulative normal distribution table to find the required probability.
x
z
Method
Using the cumulative normal distribution table
8413.0)1Pr( Z
1
1587.0
8413.01
)1Pr(1)1Pr(
ZZ
1
1587.0
8413.01
)1Pr(1
)1Pr()1Pr(
Z
ZZ-1
8185.0
0228.08413.0
)9772.01(8413.0
)2Pr(1)1Pr(
)2Pr()1Pr(
)2Pr()1Pr()12Pr(
ZZ
ZZ
ZZZ
-2 1
The mass of fruit jubes, in a packet labelled as containing 200 grams, has been found to be normally distributed with a mean of 205 grams and a standard deviation of 4 grams.
The percentage of packets that contain less than 200 grams is, correct to one decimal place,
Question 68
1056.0
8944.01
)2.1Pr(1
)2.1Pr()2.1Pr(
4
205200Pr)200Pr(
Z
ZZ
ZX
ANSWER: C
The eggs laid by a particular breed of chicken have a mass which is normally distributed with a mean of 61 g and a standard deviation of 2.5 g. The probability, correct to four decimal places, that a single egg has a mass between 60 g and 65 g is
Question 71
6006.0
)6554.01(9452.0
)4.0Pr(1)6.1Pr(
)4.0Pr()6.1Pr(
6.14.0Pr)6560Pr(
ZZ
ZZ
ZX
ANSWER: C
Applications of the normal distribution
• Draw a diagram, clearly shading the region that corresponds to the given probability.
• Use the symmetry properties of the curve to write down the appropriate z value.
• Use the inverse normal distribution table (or graphic calculator) to find the required probability and the corresponding z value.
• Use the relationship to
calculate the required x value.
x
z
Question 72
Black Mountain coffee is sold in packets labeled as being of 250 grams weight. The packing process produces packets whose weight is normally distributed with a standard deviation of 3 grams.
In order to guarantee that only 1% of packets are under the labeled weight, the actual mean weight (in grams) would be required to be closest to
a) 243 b) 247 c) 250 d) 254 e) 257
Pr 250 0.01
2502.33
3257
X
250
ANSWER: E
Question 74
82
6745.03
80
75.0)Pr(
25.0)Pr(
d
d
dX
dX
108
)12(2842
a)
b)
802.0
)85.0Pr(
12
842.94Pr)2.94Pr(
Z
ZX
Question 75 Analysis Question
c)
%23
2266.0
)75.0Pr(1
)75.0Pr(
12
8475Pr)75Pr(
Z
Z
ZX
d)
98
1.1484
175.112
84
88.012
84Pr
88.0)Pr(
12.0)Pr(
a
a
a
aX
aX
aX
e)
052.012.0
0062.0
)JumboPr(
)114Pr(
)Jumbo/114Pr(
X
X
Pr( )Pr( / )
Pr( )
A BA B
B
Conditional probability
f)
360$
02.182000Income
02.18
12.03065.01923.09)(
0.1230Jumbo
0.6519Standard
0.239Small
ProbPrice
XE
g)
6 0 6 6 1 5
0 1
Pr( 2) 1 Pr( 0) Pr( 1)
1 (0.12) (0.88) (0.12) (0.88)
1 0.4644 0.3780
0.156
X X x
C C
THE FINAL RESULT
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