MAV REVISION LECTURE MATHEMATICAL METHODS UNITS 3 AND 4 Presenter: MICHAEL SWANBOROUGH Flinders Christian Community College.

MAV REVISION LECTURE

MATHEMATICAL METHODS

UNITS 3 AND 4

Presenter: MICHAEL SWANBOROUGH

Flinders Christian Community College

EXAMINATION 1

Short-answer questions (40 marks)

• Questions are to be answered without the use of technology and without the use of notes

Time Limit:• 15 minutes reading time• 60 minutes writing time

Part I: Multiple-choice questions

• 22 questions (22 marks)

Part II: Extended response questions:

• 58 marks

Time limit: • 15 minutes reading time• 120 minutes writing time

EXAMINATION 2

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 for questions worth more than one mark if appropriate working is not shown.

Examination Advice

Notes Pages

• Well-prepared and organised into topic areas.

• Prepare general notes for each topic.

• Prepare specific notes for each section of Examination 2.

• Include process steps as well as specific examples of questions.

Examination Advice

Notes Pages

• 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-84+)

Examination Advice

Strategy - Examination 1

• 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

• Use the reading time to plan an approach for the paper.

• Make sure that you answer each question in the Multiple Choice section. There is no penalty for an incorrect answer.

• It may be sensible to obtain the “working marks” in the extended answer section before tackling the multiple choice questions.

Examination Advice

Strategy - Examination 2

• 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.

• Multiple Choice questions generally require only one or two steps – however, you should still expect to do some calculations.

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

Question 122 ))()(()( 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

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)

Bonus Prize!!

Question 4

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

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

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

Question 4

a)

8)3(2

4)3(2

)56(2

10122)(

2

2

2

2

x

x

xx

xxxf

)8,3( b)

Functions and Their Graphs

Vertical line test - to determine whether a relation is a function

rule)( where,: xfBAf

A represents the DOMAIN

B represents the CODOMAIN (not the range!)

bxaxba :,

bxaxba :,

bxaxba :,

bxaxba :,

Interval Notation

Square brackets [ ] – included

Round brackets ( ) – excluded

A function is undefined when:

a) The denominator is equal to zerob) The square root of a negative number is

present.c) The expression in a logarithm results in a

negative number.

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:

Using Transformations

NATURE – Reflection, Dilation, Translation

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

)(xfy Question 6

Determine the graph of ( )y f x

-4 -2 2 4

-4

-2

2

4

1

y

x

( )y f x

Reflection about the y-axis

-4 -2 2 4

-4

-2

2

4

-1

y

x

y = f(x)y = f( – x)

ANSWER: A

Reflected in the x-axis, Translated 2 units to the right, Translated 1 unit down

ANSWER: B

Reflection: y x

Translation: 2y x

Question 7

Translation: 2 1y x

y xGraph of

Graphs of Power Functions

x-2 -1 1 2

y

-4

-3

-2

-1

1

2

3

4y = x

4

y = x6

x-2 -1 1 2

y

-4

-3

-2

-1

1

2

3

4y = x

3y = x

5

ny x

Square Root Functions

y = x – 2 + 1

x-2 -1 1 2 3 4 5

y

-2

-1

1

2

3

4

(2 , 1)

y a x b c

The graph is:

• translated 2 units in the positive x direction

• translated 1 unit in the positive y direction

Question 9

The rule of the graph shown could be

x

y

O

3

23

2 1

3 3

1

3

1) )

) )

)

a y b y xx

c y x d y x

e y x

ANSWER: D

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 10

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 12

a b

y = f(x)

x

y

The graph shown could be that of the function f whose rule is

ANSWER: A

Absolute Value Functions

Question 14

2 1 1 2 1 0

0

2 1 1 2 1 0

3 2

2

3

k k k

k

k k k

k

k

ANSWER: D

2 1 1k k

Question 15

Part of the graph of 2 3y x x is shown below.

x-1 1 2 3 4

y

2 3y x x a) Sketch the graph of

x-1 1 2 3 4

y

2 3 4x x b) Find the set of values of x for which

2 3 4x x From the graph, solve

2 3 4 0

4 1 0

: 4 : 1

x x

x x

x x x x

Composite Functions

For the composite function f g x to be defined

Range Domain g f

Dom Domf g x g

When the composite function f g x is defined

Step 1: Complete a Function, Domain, Range

(FDR) table.

Step 2: Check that the range of g is contained in

the domain of f .

Step 3: Substitute the function g(x) into the

function f (x).

Step 4: Remember that: Dom Domf g x g

Investigating Composite Functions

Question 16

F D R

4, 0,

3,

f

g R

: 4, , 4f R f x x

2: , 3g R R g x x

24

4 3

1

g f x g x

x

x

dom dom

4,

g f x f

x

x-1 1 2 3 4 5 6

y

-1

1

2

3

4

5

6

(4, 3)

Inverse Functions

Key features:

The original function must be one-to-one

Reflection about the line y = x

Domain and range are interchanged

Intersections between the graph of the function and its inverse occur on the line y = x

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 18

ANSWER: C

Question 19

x

y

x

y

Graph of the inverse function

1: \ 1 , ( ) 2

1f R R f x

x

Question 21

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

4

5

y = 2

x = 1

y

x

1f a) exists because the function f is one-to-one

1

\ 1 \ 2

\ 2 \ 1

F D R

f R R

f R R

b)

i)

Inverse:

12

1

12

1

11

21

12

xy

xy

yx

yx

b)

ii)

Algebra of Functions

Sum and Difference of Functions

• Points of intersection of the two functions.• Points where either graph crosses the x-axis• The dominant function in different parts of the

domain.

Key features:

x

y

g(x)

f(x)

f(x) + g(x)

b

cd

a

e

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.

Question 26

5 2

0.4

log 0.4

log 0.4

ax

ax

e

e

e

e

ax

xa

ANSWER: C

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 28

Change of Base Rule for Logarithms

loglog

logb

ab

xx

a

Question 29

5

log 6log 6

log 5

1.113

e

e

ANSWER: D

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 30

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 in

positive

b

b

y

2sin 22

y x

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 by a factor of 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 about 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.

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

Sum 7

x x

x

x

Question 36 Analysis Question

325 4cos , for 0 24

12

tT t

Maximum: 25 4 29 C

Minimum: 25 4 21 C

a)

Maximum is when

3

12

3 12

15

3pm

t

t

t

b)

4 8 12 16 20 24

4

8

12

16

20

24

28

T

t

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)

T

S A

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)

Revision Quiz

1 2

3

4 5

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

Bonus Prize!!

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

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 5

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

)(log xfy e)(

)(

xf

xf

dx

dy

Logarithmic Functions

)75(log xy e75

5

xdx

dy

Further Rules of Differentiation

)( xfey )()( xfexfdx

dy

Exponential Functions

)35( 2 xxey )35( 2

)52( xxexdx

dy

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

ANSWER: B

Question 38

log cos 2

2sin 2

cos 2

2 tan 2

ey x

xdy

dx x

x

Graphs of Derived Functions

ANSWER: C

Question 40

x

y

O 1 x

y

O 1

y = f '(x)

Question 42

2 2 3

2 2 3

5

25 1

When ,2 4

5 1,

2 4

dy dyx

dx dxx

x

x y

a)

b)5 1 1

, ,2 4 3

1 1 5

4 3 2

1 5

4 3 613

3 12

dy

dx

y x

xy

xy

Question 43

(0) 0 (0) 0

(4) 0 (3) 0

( ) 0 for - ,0 0,3

( ) 0 for 3,

f f

f f

f x

f x

x-5 -4 -3 -2 -1 1 2 3 4 5

y

3

4 0, 0 0

and 4 are factors

4

2 4

4 8 2

1

4

f x ax x b

f f

x x

b

f

a

a

b)

3

43

3 2

44

4

3

4, 16

0 16 4

16 64

xf x x

xx

f x x x

x f x

y x

y x

c)

Approximations

ANSWER: D

Question 45

3(2) 2

2, 0.2

1.8 2 0.2 2

f x h f x hf x

f

x h

f f f

Related Rates

2

3

By similar triangles:

2

4

2

1

3 2

12

r

hh

r

hV h

h

Question 46

2 m

4 m

dh dh dV

dt dV dt 21

3V r h

3

2

2

3

12

44

dh dh dV

dt dV dt

dV

dt

hV

dV h

dhdh

dV h

2

2

43

12

3

12

90.42m/ min

dh

dt h

hh

dh

dt

Question 47 Analysis Question

3 2: , xf R R f x x e

a)

2 2 3 2

2 2 3

2 3 2 2 3 2

3 2

3 2

As 2 3

2, 3

x x

x

x x

f x x e x e

e x x

e x x e ax bx

a b

b) Find the EXACT COORDINATES of the two stationary points and their nature.

2 3 2

2

2

0

2 3 0

0

2 3 0

30,

2

x

x

f x

e x x

e

x x

x

33 27

,2 8

e

Local maximum at:

Stationary point of inflexion at:

0,0

c) i)

2

2

2

2 2

2 2 2

2

At 1,

1 2 3

Tangent:

1

x y e

f e

e

y e e x

y e e x e

y e x

ii)

At 0, 0

0 0

Tangent:

0

x y

f

y

3 2 2 3 2

3 2 2 3 2 2 3 2

4 2 3 2 3 22 3 2

2 3 2 3 2

2 3

2 3 2 3

2 32 3

2 3 2 1

p p

p p p

p p pp

p p

y p e e p p x p

y p e e p p x pe p p

y p e p e p ee p p x

e p p x p e p

iii) Select a point

3 2, pp p e and find the equation

of the tangent at this point.

3 2

2

2 1 0

0, 1 0

p

p

p e p

p e

This tangent will pass through the origin when x = 0

Therefore the only two tangents that pass through the origin are when 0 and 1x x

3 2 2 3 2

2 2 2 3 2 3 2

2 3 2 3 2

4 3

12 2 2 4 3

8 12 2 2 2 6

x x

x x x

x x

dx px qx e kx e

dx

x px q e e x px qx kx e

e x p x p q x q kx e

d) i) Use CALCULUS to find the exact values of the constants p, q and k.

Equating coefficients:

8

12 2 0, 6

6 0, 6

k

p p

q q

d) ii)

12 3 2

0

1 12 3 2

0 0

1 12 3 2

0 0

12 12 3 2 2

00

2 2 0

2

18

8

14 6 6 3

2 8

1 14 6 6 3 3

2 8

23 3

8

x

x

x

x

A e x x e dx

e x dx x e dx

e x dx x e dx

xe x x x e

e e e

e

Antidifferentiation and Integral Calculus

cna

baxdxbax

nn

)1(

)()(

1

1,1

1 1

ncxn

dxx nn

ANSWER: B

Question 48

1

2

1

2

1

2

3 2 1

12

2

3 2 1

3

2 1

xy c

x c

cx

3

2

3

2

3

2 1

3 2 1

dy

dx x

x

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

( )log ( )

( ) e

f xdx f x c

f x

Example

4 ( )where ( ) 4 3

4 3 ( )

log 4 3e

f xdx dx f x x

x f x

x c

( ) ( )

( ) ( )

b b

aaf x dx F x

F b F a

Definite Integrals

Properties of Definite Integrals

4

1

( ) 2f x dx

ANSWER: E

Question 50

4

1

4 4

1 1

4

1

2 3

2 3

4 3

4 12 3

13

f x dx

f x dx dx

x

followsit then ),()( xgxfdx

d

cxfdxxg )()(

Integration by recognition

ANSWER: A

Question 52On the interval (a, b) the gradient of g(x) is negative.

x

y

b

O

y = f(x)

a

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 bounded by the curve and the x-axis on the interval [a, c] is given by:

( ) ( )

( ) ( )

b c

a b

b b

a c

f x dx f x dx

f x dx f x dx

ANSWER: D

Question 53

a b cO

y = f(x)

y

x

Question 54

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 b

a a

b

a

A f x dx g x dx

f x g x dx

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

O

y = 1 – e– x

y = – x + 1

y

x

Question 55

Find the solution to the equation 1 1 xx e

0.567x

0.567

0

0.567

0

0.5672

0

1 1

2

0.27

x

x

x

A x e dx

x e dx

xe

b) Use CALCULUS to find the area of the shaded region.

Numerical techniques for finding area

ANSWER: D

Question 56

1 2 3

2 9 28

39

A f f f

Question 57 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)

Write down the equation in x, the solutions of which give the x-coordinates of the stationary points of the curve

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 or binomial distributions. These values are usually obtained by counting.

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 often 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

The number of hours each day, X, spent cycling has the following probability distribution.

X 0 1 2 3 4

PrXx 330 6

30 7

30 10

30 4

30

The proportion of days for at least two hours of cycling is:

Question 59

ANSWER: D

7 10 4 21

30 30 30 30

ANSWER: B

x

0 0

1

2

3

4

Pr( )X x Pr( )x X x

Question 60

3

30

6

307

3010

304

30

6

3014

30

16

30

30

30

66

30

Pr

66

30

x X x

Markov Chains

A Markov chain is a chain of events for which the probabilities of outcomes or states depend on what has happened previously.

Tree diagrams are a useful tool for solving problems.

0.9

0.1

0.6

0.4

S

N

S

S

N

NS

0.6

0.4

Th Fr Sa

Question 61

If it has snowed the day before the probability of snow is 0.6. If it has not snowed on the previous day then the probability of snow is 0.1.If it has snowed on Thursday, what is the probability that it will not snow on the following Saturday?

Pr(no snow on Saturday)

0.6 0.4 0.4 0.9

0.24 0.36

0.60

0.9

0.1

0.6

0.4

S

N

S

S

N

NS

0.6

0.4

Th Fr Sa

Question 62

A bag contains three white ball and seven yellow balls. Three balls are drawn without replacement. The probability that they are all yellow is:

Pr( )

7 6 5

10 9 87

24

YYY

W

Y

W

W

Y

Y

710

69

310

Y

W

58 ANSWER: D

The Binomial Distribution

~ Bi( , )X n p

nxppCxX xnxx

n ......,2,1,0,)1()()Pr(

qpnX

pqqpnX

pnX

)(SD

1where,)(Var

)(E

2

A random sample of 20 tickets is taken. The probability that this sample contains exactly twelve Adult tickets is equal to:

20 12 8

12

~ Bi 20, 0.6

Pr( 12) (0.6) (0.4)

X n p

X C

ANSWER: B

Question 63

Mean: 10

SD: 3

9 10

0.9

0.1

np

npq

q

q

p

ANSWER: A

Question 65

Continuous Random Variables

Properties of probability density functions

0f x a) for all real numbers x

b) 1f x dx

c) Prb

a

a X b f x dx

2 if 0 2

0 if 0 or 2

ax x xf x

x x

Question 66

a)

2

0

22

0

232

0

2 1

2 1

13

84 1

3

ax x dx

a x x dx

xa x

a

3

4a

b) 1

2

0

122

0

12 3 2

0

1 3Pr 2

2 4

3 3

2 4

3

4 4

3 1

16 325

32

xX x dx

x xdx

x x

Continuous Random Variables

Mean: E X x f x dx

Mode: the value for which is a maximum Pr X x

Variance: 2 2 2Var X x f x dx

Median: the value m such that 1

2

m

a

f x dx

Question 67

1sin 0.25

2

1 1cos

2 4

1 1 1cos cos

2 2 4

1cos

22.09

a

a

x dx

x

a

a

a

ANSWER: E

The Normal Distribution

The mean, mode and median are the same.

The total area under the curve is one unit.

Pr( )b

a

a X b f x dx

Pr( ) 1 Pr( )Z z Z z

z

Using symmetry properties

Pr PrZ z Z z

-z z

Pr Pr Pra Z b Z b Z a

a b

Question 71

X is normally distributed with a mean of 72 and a standard deviation of 8. Use the result that to find: Pr 1 0.84Z

8072

a)

Pr 80

Pr 1

1 Pr 1

0.16

X

Z

Z

64 72

b)

Pr 64 72

Pr 1 0

0.34

X

Z

c)

Pr 64 \ 72

Pr 64 72

Pr 72

Pr 64

Pr 72

Pr 1

Pr 0

0.32

X X

X X

X

X

X

Z

Z

• Draw a diagram, clearly labelling the mean.

• Shade the region required.

• Use either the appropriate symmetry properties or a calculator to find the required probability

• Remember that:

x

z

Solving normal distribution problems

Volunteers for a weight loss program have weights which are normally distributed with a mean of 100 kg and a standard deviation of 8 kg.

One person is selected at random. The probability that this person’s weight is over 110 kg is approximately

Question 72

110 100Pr( 110) Pr

8

Pr( 1.25) 1 Pr( 1.25)

1 0.8944

0.1056

X Z

Z Z

ANSWER: E

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 function to find the required probability and the corresponding z value.

• Use the relationship to

calculate the required x value.

x

z

Question 73

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

a)

Question 77 Analysis Question

2~ 140,1.2

141.5 140Pr 141.5 Pr

1.2

Pr 1.25

0.106

X N

X Z

Z

140 141.5

b)

140 140 + d 140 – d

Pr 140 0.075

Pr 140 0.925

140 1401.4395

1.2

1.7

X d

X d

d

d

c) 2 1012

2

~ Bi 12, 0.15

Pr 2 0.15 0.85

0.292

X n p

X C

0.15 0.17 1

0.68

k

k

d) i)

ii) E 0.68 5 0 0.15 0.17 8

0.68 3.4 0.17 1.36

0.85 4.76

Y x x

x x

x

E 0

0.85 4.76 0

$5.60

Y

x

x

iii)

iv)

Pr Good ReadyPr Good/Ready

Pr Ready

0.68

0.68 0.170.8

Conditional probability

Pr( )Pr( / )

Pr( )

A BA B

B

THE FINAL RESULT