M.A.V. REVISION LECTURES

M.A.V. REVISION LECTURES

MATHEMATICAL METHODS

UNITS 3 AND 4

Presenter: MICHAEL SWANBOROUGHFlinders 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 34 5 67 8 9

Question 1sin 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 isQuestion 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( 2010

1030 )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 5What 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 exactlya) 1 real solutionb) 2 distinct real solutionsc) 3 distinct real solutionsd) 4 distinct real solutionse) 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)

2 37

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

72140

22)2(220

0)2(23

aa

a

P

b) is exactly divisible by

.2x Find the value of a.

y

x

10

2

105 23 xxxy

)53)(2(

105105

63

3

2

531052

105

2

2

2

23

2

23

23

xxx

xxxx

xx

xx

xxxxxx

xxxy

Question 5

a)

3 2

2

2 2 2

2

5 10

( 2)( )

2 10 2 55 3

( 2)( 3 5)

y x x x

y x x bx c

c x bx xc b

y x x x

2293,

2293,2

2293

2)5)(1(433

053,2

0)53)(2(

2

2

2

x

x

x

xxx

xxxb)

327

4320160

4320)()2(

3

3

333

6

aa

a

aC

ANSWER: B

Question 6 Coefficient of 63 )2(in axx

6048)32)(9(21)2()3( 52

27

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

,

23or

23: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

1a parallel 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

2 34

yx

Vertical: 4 04

xx

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(log21

2)1(log1

1

:Inverse

2

2

xy

yxex

ex

e

e

y

y

1)( where,: )2( xexfRRf

Rf

RfRDF

,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 (3649.32

)2(log42)2,()2(log4)(

21

kek

kkxxf

e

e

f(x)

x

x = 2

= 2y

(3, 0)

(0, 3)

Label asymptotes

Label coordinates

Approach asymptotes

y

x

21

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(25.33

1997log21

3199719973

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

1616 , 4

but 0 4

a a

a a

a

x

x

x

xa

x a x a

x x a

ANSWER: A

Question 26

yxyx

yx

yxyxxyxx

10

10

1log

1logloglog1log2log3log1loglog3

10

1010

101010

102

1010

ANSWER: D

Question 27

Circular (Trigonometric) Functions

dcxbaxf ))(sin()(

dcxbaxf ))(cos()(

Amplitude: a

Period: b2

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: 22Period: 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 12 1 3

3, 1

ANSWER: B

Question 30

0 where,)2sin()( pqxpxfy

x

p

-p

0q

qp qp

pqp

qp21

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) 444cos 1612

14 162

14

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

axxax

ANSWER: E

Question 35

Question 36 Analysis Question

325 4cos , for 0 24

12t

T t

Maximum: 25 4 29 CMinimum: 25 4 21 C

a)

cos is maximum when

312

3 1215

3pm

t

tt

b)

325 4cos 23

123 1cos

12 23 5,

12 3 3

3 4, 207, 237am, 11pm

t

t

t

ttt

c)

Maximum at 15

Interval: 15 2, 15 2

13 28.46 C17 28.46 C

Minimum temp: 28.46 C

t

t Tt T

d)

34 sin12 12

3sin

3 12

tdTdt

t

e) i)

3sin 0.2

3 123 0.6sin

12

30.192, 2.949

123.73, 14.27

Interval is: 3.73, 14.27

t

t

t

t

e) ii)

DIFFERENTIAL CALCULUS

dxdu

dudy

dxdy

Chain Rule:

dxdvu

dxduvuv

dxd

)(Product Rule:

2vdxdvu

dxduv

vu

dxd

Quotient Rule:

Further Rules of Differentiation

( )y f x( )

2 ( )dy f xdx 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)()(

xfxf

dxdy

Logarithmic Functions

)75(log xy e

755

xdx

dy

)(sinlog xy e

cossincot

dy xdx x

x

Examples:

Further Rules of Differentiation

xey xedxdy

)(xfey )()( xfexfdxdy

Exponential Functions

)35( 2 xxey )35( 2

)52( xxexdxdy

xey cos xexdxdy cossin

Examples:

3

3 3

2 3

3 2

( 4)

( 4) ( 4)

(3 ) ( 4)( )

( 3 4)

x

x x

x x

x

y e xdy d de x x edx 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(

ttttt

dtdy

t

tdtdtt

dtdt

dtdy

tty

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

12

When 4, 16

3

4, 6

16 6 4

6 8

x y

dy xdx

dyxdx

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) 160.04

16.04 16 0.04 16

f x h f x hf x

fh

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 30 1 2 32 4

2

x f x

e kek

kk

2

2

2

4 3

2 4

0 2 4

2 ( 2) 0

2 as 0log 2

x x

x x

x x

x x

x x

e

y e edy e edx

e e

e e

e ex

2 log 2 log 2( ) 4 34 8 3

1log 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, 1log 3, 0

but 0

log 3

c c

c c

c c

e

e

x c y

e e

e e

e ec

c

c

d)i)

log 3

2

0

log 32

0

2 log 3 log 3 0 0

log 9

4 3

1 4 321 14 3log 3 42 2

1 14 3 3log 3 42 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

cnabaxdxbax

nn

)1()()(

1

1,1

1 1

ncxn

dxx nn

cx

cxdxx

35)35(

)5(7)35()35(

7

76

Examples

cx

cx

cx

dxxdxx

53

53

53

52

52

)72(25

)72(653

253

)72(3

)72(3)72(

3

32

12

12

12

2 (4 1)

(4 1)21 4

2

(4 1)1

(4 1)

x dx

x c

x c

x

32

2

(4 1)

dydx

x

ANSWER: E

Question 47

Trigonometric Functions

Rules of Antidifferentiation

)cos()sin( kxkkxdxd

)sin()cos( kxkkxdxd

ckxk

dxkx

)cos(1)sin(

ckxk

dxkx )sin(1)cos(

Rules of Antidifferentiation

Exponential Functions

kxkx

xx

keedxd

eedxd

cek

dxe kxkx 1

Rules of Antidifferentiation

Logarithmic Functions

)()()(log

xfxfxf

dxd

e

cxfdxxfxf

e )(log)()(

Examples

cx

xxfdxxfxfdx

x

e

)34(log

34)( where)()(

344

cx

xxfdxxfxf

dxx

xdxx

x

e

)5(log21

5)( where)()(

21

52

21

5

2

2

22

)()(

)()(

aFbF

xFdxxfb

a

ba

Definite Integrals

22

0

23 2

0

( 3 4)

3 43 2

8 12 8 (0)3 2

23

x x dx

x x x

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 QUESTION4

1

( ) 2f x dx If then is equal to:

4

1

2 ( ) 3f x dx

followsit then ),()( xgxfdxd

cxfdxxg )()(

Integration by recognition

cxxxdxx

ddxxxdxx

dxxdxxdx

dxxdxx

xxxdxd

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

2 log 2 1 12

log 2

e

e

e

y x x x

dy x xdx x

x

a)

b) Hence, find the exact area of the shaded region

12

e2

y

x

2212

1 22

log 2 log 2

1 1log log 12 2 2 2

12

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

54

4

54

4

sin cos

cos sin

5 5cos sin cos sin4 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 21 2 5 32

y x x x x

23

23

3 34 52 2

3 34 5 02 2

dy xx xdx

xx x

a)

b) i)

23 3 34 5

2 2

when 1, 1

111,

21 1 12

1.5

normal

dy xx xdx

dyxdx

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 2

x 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

xXxxXxxXxxXx

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 7.52.9 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

Mean2

Variance1.6

X C

X n p

np

npq

ANSWER: A

Question 63

The Hypergeometric Distribution

),,(Hg~ NDnX

,)Pr(n

Nxn

DNx

D

CCC

xX

)(Var)(SD

11)(Var

)(E

2

XX

NnN

ND

NDn

X

NDn

X

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 CC 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.

xz

Method

Using the cumulative normal distribution table

8413.0)1Pr( Z1

1587.0

8413.01)1Pr(1)1Pr(

ZZ1

1587.0

8413.01)1Pr(1

)1Pr()1Pr(

Z

ZZ-1

8185.00228.08413.0

)9772.01(8413.0)2Pr(1)1Pr(

)2Pr()1Pr()2Pr()1Pr()12Pr(

ZZZZZZZ

-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

1056.08944.01

)2.1Pr(1)2.1Pr()2.1Pr(

4205200Pr)200Pr(

ZZZ

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(

ZZZZ

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.

xz

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

2502.333

257

X

250

ANSWER: E

Question 74

82

6745.03

8075.0)Pr(

25.0)Pr(

d

ddX

dX

108)12(2842

a)

b)

802.0)85.0Pr(12

842.94Pr)2.94Pr(

Z

ZX

Question 75 Analysis Question

c)

%232266.0

)75.0Pr(1)75.0Pr(

128475Pr)75Pr(

ZZ

ZX

d)

981.1484

175.112

84

88.012

84Pr

88.0)Pr(12.0)Pr(

aa

a

aX

aXaX

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

02.1812.03065.01923.09)(

0.1230Jumbo0.6519Standard0.239SmallProbPrice

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 0.37800.156

X X x

C C

THE FINAL RESULT

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