Download 2013 EOY Exam File

NAME:
TEACHER:
Mathematics and Statistics
2013
Level 2
Standards:
AS 91261
Apply algebraic methods in solving problems
Page 2
AS 91262
Apply calculus methods in solving problems
Page 13
AS 91267
Apply probability methods in solving problems
Page 24
All Assessment schedules
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Page 1
2013 EOY Level 2 23
Writer: David Fortune
NAME:
TEACHER:
Level 2 Mathematics and Statistics
2013
91261 Apply algebraic methods in solving problems
Credits: Four
Achievement
Apply algebraic methods in
solving problems.
Achievement with Merit
Apply algebraic methods, using
relational thinking, in solving
problems.
Achievement with Excellence
Apply algebraic methods, using
extended abstract thinking, in solving
problems.
You should answer ALL parts of ALL the questions in this booklet.
You should show ALL your working for ALL questions.
The questions in this booklet are NOT in order of difficulty.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
Check that you have a copy of the formulae sheet.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE
END OF THE EXAMINATION.
TOTAL
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Page 2
2013 EOY Level 2 23
Writer: David Fortune
You are advised to spend 60 minutes answering the questions in this booklet.
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QUESTION ONE:
(a)
Solve:
(i)
log 9 x  3
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(ii)
x 8  65536
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(iii)
6x  3  2x 2
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(b)
1
2
and x  
2
5
2
Write the quadratic equation in standard form ax  bx  c  0
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A quadratic equation has solutions x 
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(c)
Cody has three wooden spherical balls.
The radius of the middle sized ball is twice the radius of the small ball.
The radius of the largest ball is 5 cm larger than the radius of the smallest ball.
The surface area of the largest ball is equal to the sum of the surface areas of the two
smaller balls.
Find the radius of the middle sized ball.
Note: The surface area of a sphere is given by the formula SA  4 r 2
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2013 EOY Level 2 23
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(d)
Find the set of values for k for which the equation kx 2  x  k  1  0 has real roots, one
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of which is positive and one of which is negative.
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2013 EOY Level 2 23
Writer: David Fortune
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QUESTION TWO
(a)
Simplify
15 x 2 y 2 (3 y ) 2

7 wx
5w
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(b)
(i) Simplify by writing as a single fraction
4
8

x  2 2x  1
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(ii) Solve
4
8

 10
x  2 2x  1
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(c)
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Hemi buys a new car for $25000.
The value of her second hand car, V , after t years, can be modelled by the equation
V  C  (0.75) t
where C is the Cost price of the car in dollars.
(i) How much has the car lost in value after Hemi has owned the car for 5 years?
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(ii) After how long would Hemi’s car be worth 40 % of the purchase price?
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(d)
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For what value(s) of q does the straight line y  q  2 x intersect with the hyperbola
xy  5
in two separate places.
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QUESTION THREE
(a)
(i) Factorise
3x 2  5 x  2
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(ii) Solve
3x 2  5 x  2  0
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(b)
Simplify
64 x
14
 4 x 2


1
4
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(c)
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Solve log 4 ( x  4)  log 4 ( x  4)  2
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(d)
Two numbers differ by 5.
Prove that the difference of their squares is a multiple of 5.
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(e)
Solve the equation 4 2 x  9  4 x  20 .
Hint: Let u  4 x .
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Extra paper for continuing your answers, if required.
Clearly number the question(s).
Question
Number
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Page 12
2013 EOY Level 2 23
Writer: David Fortune
NAME:
TEACHER:
Level 2 Mathematics and Statistics
2013
91262 Apply calculus methods in solving problems
Credits: Four
Achievement
Apply calculus methods in
solving problems.
Achievement with Merit
Apply calculus methods, using
relational thinking, in solving
problems.
Achievement with Excellence
Apply calculus methods, using
extended abstract thinking, in
solving problems.
You should answer ALL parts of ALL the questions in this booklet.
You should show ALL your working for ALL questions.
The questions in this booklet are NOT in order of difficulty.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
Check that you have a copy of the formulae sheet.
YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE
END OF THE EXAMINATION.
TOTAL
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Page 13
2013 EOY Level 2 23
Writer: David Fortune
You are advised to spend 60 minutes answering the questions in this booklet.
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QUESTION ONE:
(a)
A function f is given by f ( x)  3x 2  4 x  7
Find the gradient at the point where x  2 .
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(b)
The gradient at a point on a curve is given
dy
 3x 2  4
dx
Find the equation of the curve given that the curve passes through the point (1, 2) .
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(c)
The equation of a parabola is y  3x 2  2 x  4 .
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Find the equation of the tangent to the parabola at the point (1, 1) .
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(d)
If g ( x)  2 , g (2)  2 , g ( 2)  2 and g (0)  2 then find g (x ) .
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2013 EOY Level 2 23
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(e)
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A rectangle is formed with two sides on the x and
y axes as shown.
One corner of the rectangle is at the origin and the
opposite corner is on the curve y  9  2 x 2 .
Find the maximum area of the rectangle.
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2013 EOY Level 2 23
Writer: David Fortune
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QUESTION TWO:
(a)
The graph shows the function
y  f (x) .
On the axes below, sketch the graph of the gradient function y  f  (x) .
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2013 EOY Level 2 23
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(b)
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A curve y  f (x) passes through the origin and has a gradient function
dy
 2 x  3x 2
dx
Find the coordinates of the point on the curve where x = 3.
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(c)
The velocity, v, in m s-1, of a train is v(t )  0.2t  3
0  t  120
where t is the time in seconds after leaving the station.
Three seconds after starting it is 9 m from the station.
Find the distance from the station after 20 seconds.
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2013 EOY Level 2 23
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(d)
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The profit function, P (t ) , of the Acme Manufacturing Company is given by
P(t )  0.01t 2  0.75kt  1000
where t is time in months after starting the company and k is a constant.
Given the profit is decreasing for t > 45 months find the value of k.
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(e)
When a person coughs the windpipe (trachea) contracts, allowing air to be expelled
at maximum velocity.
The velocity v of airflow during a cough can be modelled by the function v  kr 2 (R  r )
where r is the trachea’s radius, in cm, during a cough and
R is the normal radius of the trachea, in cm and
k is a positive constant that depends on the length of the trachea.
Find the radius for which the airflow is greatest during a cough.
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2013 EOY Level 2 23
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QUESTION THREE:
(a)
Find the coordinates of the function h( x)  2.5x 2  3.5x  7 where the gradient is 1.5.
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(b)
The surface area, SA, of a cube changes as the length, x, of the sides change.
SA  6x 2
Find the rate of change of the surface area, with respect to the length of the side,
when the side is 3 units.
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2013 EOY Level 2 23
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(c)
The graph shows the derivative function h(x) .
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Sketch the graph of h(x) .
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2013 EOY Level 2 23
Writer: David Fortune
(d)
A rocket starts from rest.
Its acceleration, in m s-2, is given by a (t ) 
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1 2
(t  4) .
4
How far does it travel in the tenth second?
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(e)
Find the maximum vertical distance between these curves y  4 x 3  15x 2  20 x  7 and
y  3x 2  16 x  10 .
[NOTE: Only consider  5  x  1.4 , the region between the intersections]
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2013 EOY Level 2 23
Writer: David Fortune
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Extra paper for continuing your answers, if required.
Clearly number the question(s).
Question
Number
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Page 23
2013 EOY Level 2 23
Writer: David Fortune
NAME:
TEACHER:
Level 2 Mathematics and Statistics
2013
91267 Apply probability methods in solving problems
Credits: Four
Achievement
Apply probability methods
in solving problems.
Achievement with Merit
Apply probability methods, using
relational thinking, in solving
problems.
Achievement with Excellence
Apply probability methods, using
extended abstract thinking, in solving
problems.
You should answer ALL parts of ALL the questions in this booklet.
You should show ALL your working for ALL questions.
The questions in this booklet are NOT in order of difficulty.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
Check that you have a copy of the formulae sheet.
YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE
END OF THE EXAMINATION.
TOTAL
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Page 24
2013 EOY Level 2 23
Writer: David Fortune
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You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
A manufacturer of digital watches finds that the watches vary in how accurately they keep the
time. The time gained per day is modelled by a normal distribution.
On average they gain one second per day, with a standard deviation of 2.5 seconds per day.
The questions below relate to this manufacturer’s digital watches
(a)
(i) What is the probability that a watch, selected at random, gains between 1 and 4 seconds
per day?
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(ii) What percentage of watches gain at least 8 seconds per day?
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(iii) What is the probability a watch, selected at random, loses time?
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(iv) In a batch of 1200 watches what is the expected number of watches that are out by
at least 1.5 seconds per day?
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2013 EOY Level 2 23
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(v) Because of the issue of warranties the manufacturer wants to limit the number of
watches gaining more than 3 seconds per day to 10 % of the total number produced.
The manufacturer can reset the standard deviation of the time lost per day, whilst
keeping the same mean.
What should the new standard deviation be after resetting?
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2013 EOY Level 2 23
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(b)
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A garden centre purchases potting mix from two different suppliers, Growquick and
Readyfast. The percentage moisture content in the potting mix from each supplier is
normally distributed with mean and standard deviation as shown.
Mean
Growquick
Readyfast
25.1%
23.5%
Standard
deviation
4.2%
4.9%
The garden centre wishes to purchase a large amount of potting mix with moisture content
under 30%. Which supplier should they choose? Give statistical reasons.
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(c)
The time that people take to complete a puzzle is normally distributed with mean of
18 minutes and a standard deviation of 5 minutes. A large group of people were given
the puzzle. If 40 people completed the puzzle in under 10 minutes then estimate how
many people took longer than 10 minutes.
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2013 EOY Level 2 23
Writer: David Fortune
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QUESTION TWO
(a)
The table is below is from the Ministry of Transport “CRASH FACT SHEET 2012”.
BAC
0
30
50
80
Relative risk of fatal crash by
blood alcohol level
30+
20-29
15-19
years
years
years
1
3
5.3
2.9
8.7
15
5.8
17.5
30.3
16.5
50.2
86.8
BAC is the driver’s Blood Alcohol Concentration, in mg of alcohol per 100mL of blood
The calculation of risk is in relation to that of a sober driver aged 30+ years.
(i) Explain what the 5.3 in the first row means.
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(ii) Estimate the relative risk, for all 3 age groups, for a driver with a BAC of 80
compared to a driver with a BAC of 50.
Comment on your findings.
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(b)
The Newtown district kept statistics on driving licence applicants.
They recorded whether the applicants had professional (paid) driving lessons from an
instructor, as well as whether they passed or failed.
The results are shown in the table below:
Passed test Failed test
Had paid
lessons
Didn’t have
paid lessons
TOTALS
TOTAL
53
8
61
47
27
74
100
35
135
(i) What proportion of applicants in Newtown passed the driving test?
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(ii) What percentage of those people in Newtown, who paid for driving lessons, passed
their driving test?
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(iii) Show that the risk of failing the driving test in Newtown is about 1 in 4.
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(iv) ACE driving school claimed in an advertisement that people that had paid for driving
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lessons were one third more likely to pass than if they had not had paid for lessons.
State whether or not you agree, stating full reasons and showing relevant calculations.
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(v) A researcher looked at the gender of all applicants in the above group of applicants.
There were 56 male applicants of whom 38 passed.
A newspaper headline stated that the risk for males failing their driving test in
Newtown was approximately the same as the risk of females failing.
Comment on the claim.
You should provide suitable reasons, backed by calculations from the results table.
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2013 EOY Level 2 23
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QUESTION THREE
(a)
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On average 1 in 20 people have the particular antibody mGF in their blood.
The machine that tests people for this antibody does not always give the correct result.
Of the people with mGF 94% test positive on the machine (ie test as having mGF), 4% test
negative and 2 % are inconclusive.
Of the people without the antibody mGF, 92 % test negative, 5% test positive and the rest are
inconclusive.
(i) What is the probability that a person, selected at random, had the mGF antibody and
the test was positive.
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(ii) What is the percentage of people with a negative test?
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(iii) If a person selected at random had a positive test what is the probability that they in
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fact had the mGF antibody?
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(iv) In one town where everybody was tested there were 710 inconclusive results.
What is the expected number of people in this town?
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(b)
Venture Tours takes tours to popular destinations. The leader of a tour to Kingstown tells
the group members that the weather in Kingstown is “very reliable”. He explains “very
reliable” by stating that if the weather is good one day there is an seventy five percent chance
it will also be good the next day, however if the weather is bad one day there is only 40%
chance it will be bad the next day. The tour arrives on Tuesday and the weather is good.
(i) What is the probability that the weather will also be good on Friday (3 days later)?
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(ii) The tour has one outdoor experience day planned.
What is the probability that at least 1 of the next 3 days is a good day?
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Writer: David Fortune
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Extra paper for continuing your answers, if required.
Clearly number the question(s).
Question
Number
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ASSESSMENT SCHEDULE
91261 Apply algebraic methods in solving problems
Achievement
Achievement with Merit
Achievement with Excellence
Apply algebraic methods in solving
problems involves:
Apply algebraic method, using relational
thinking, in solving problems involves
one or more of:
Apply algebraic methods, using extended
abstract thinking, in solving problems
involves one or more of:
• selecting and using methods
• demonstrating knowledge of algebraic
concepts and terms
• communicating using appropriate
representations.
• selecting and carrying out a logical
sequence of steps
• connecting different concepts or
representations
• demonstrating understanding of
concepts
• forming and using a model;
and also relating findings to a context, or
communicating thinking using
appropriate mathematical statements.
• devising a strategy to investigate or solve
a problem
• identifying relevant concepts in context
• developing a chain of logical reasoning, or
proof
• forming a generalisation;
and also using correct mathematical
statements, or communicating mathematical
insight.
Sufficiency for each question:
N0: No response, no relevant evidence.
N1: Attempt at ONE question with some algebraic knowledge.
N2: 1 u
A3: 2 u
A4: 3 u
M5: 1 r
M6: 2 r
E7: 1 t
E8: 2 t
Judgement Statement
Score range
Produced by SINCOS
Not Achieved
Achievement
Achievement with
Merit
Achievement with
Excellence
0-7
8 - 13
14 - 18
19 -24
Schedule Page 35
2013 EOY Level 2
Writer: David Fortune
23
Question
ONE
Evidence
Achievement (u)
Apply algebraic
methods in solving
problems.
Merit (r)
Apply algebraic
methods, using
relational thinking,
in solving
problems.
Excellence (t)
Apply algebraic methods
using extended abstract
thinking, in solving
problems.
1a(i)
x  9 3  729
Correct solution.
1a(ii)
x4
Correct solution.
1(iii)
2x 2  6x  3  0
x  2.366 x  0.634
Written as a
quadratic
equation = 0 and
ONE solution
found.
Both correct
solutions.
(2 x  1)(5 x  2)  0
Correct equation
written in any
form.
Correct equation
found in correct
form.
Correct quadratic
equation found.
Quadratic
equation solved.
Problem solved
correctly.
Found solutions
to quadratic
equation.
Found
1  4k (k  1)  1
Problem solved.
1b
Equation must be
10 x 2  x  2  0
1c
[or multiple]
Let r be radius of small ball.
4 r  4 (2r )  4 (r  5)
5 r 2  (r  5) 2  r 2  10r  25
4r 2  10r  25  0
r  4.045 r  1.545
2
2
2
Radius of middle ball = 8.1 cm
1d
Using quadratic formula
 1  1  4k (k  1)
x
2k
[Accept with  signs.]
For one solution to be positive
1  4k (k  1)  1 (numerator positive)
k (k  1)  0
0  k 1
Produced by SINCOS
Schedule Page 36
2013 EOY Level 2
Writer: David Fortune
23
Question
TWO
Evidence
Achievement (u)
Apply algebraic
methods in solving
problems.
2a
Merit (r)
Apply algebraic
methods, using
relational thinking, in
solving problems.
Excellence (t)
Apply algebraic methods
using extended abstract
thinking, in solving
problems.
15 x 2 y 2 (3 y ) 2 15 xy 2 5w


 2
7 wx
5w
7w
9y
25 x

21
Correct
expression,
fully simplified.
2b(i)
4
8
20


x  2 2x  1
( x  2)( 2 x  1)
Correct
expression,
fully simplified.
2b(ii)

20
 10
( x  2)( 2 x  1)
 2  ( x  2)( 2 x  1)
2 x 2  3x  0
x(2 x  3)  0
x  0 x  1.5
Correct quadratic
equation set up.
Both correct
solutions.
2c(i)
25000  25000(0.75) 5
Value after 5
years found.
Correct solution.
Correct simplified
equation (line 2).
Problem solved but
not in context.
OR
Set up equation t =
Problem solved in
context.
Correct equation
set up to solve.
q 2  40 obtained
Both correct solutions.
[Accept with  ]
= $19 067.38
[Accept $19 067]
2c(ii)
10000  25000  (0.75) t
0.4  (0.75) t
log 0.4
t
 3.185
log 0.75
ie has value 40% after 3.185 years
2d
x( q  2 x)  5
2 x 2  qx  5  0
q  q 2  40
4
For 2 solutions q 2  40
q  6.32 q  6.32
but not correct
solution.
x
Produced by SINCOS
Schedule Page 37
2013 EOY Level 2
Writer: David Fortune
23
Question
THREE
Evidence
Achievement (u)
Apply algebraic
methods in solving
problems.
3a(i)
3a(ii)
3b
(3x  1)( x  2)
Correct
factorisation.
(3x  1)( x  2)  0
1
x
x  2
3
Both correct
solutions.
64 x
14
 4x




 16 x 16
1

16 x 16

1
4

1
2  4
 64 x14 
  2 
 4x 

1
4

1
4
Merit (r)
Apply algebraic
methods, using
relational thinking, in
solving problems.
Expression
simplified to line
2
OR
equivalent.
Correct expression.
fully simplified.
Correct
expression for
x2.
Both correct
solutions.
Excellence (t)
Apply algebraic methods
using extended abstract
thinking, in solving
problems.
1
2x 4
3c
log 4 ( x 2  16)  2
x 2  16  4 2
x 2  32
x   32  5.66
3d
x 2  y 2  ( x  y)( x  y)
 5( x  y )
as ( x  y ) =5
Almost correct
proof with mainly
justified steps
shown.
Correct proof with
justified steps shown.
One correct
solution.
Both correct solutions.
(numbers differ by 5)
Thus x 2  y 2 is a multiple of 5
OR equivalent
3e
u 2  9u  20  0
(u  5)(u  4)  0
u  5 or u  4
4x  5 4x  4
log 5
x
= 1.161
log 4
Produced by SINCOS
Quadratic
equation solved.
x=1
Schedule Page 38
2013 EOY Level 2
Writer: David Fortune
23
ASSESSMENT SCHEDULE
91262 Apply calculus methods in solving problems
Achievement
Achievement with Merit
Achievement with Excellence
Apply calculus methods in solving
problems involves:
Apply calculus methods, using relational
thinking, in solving problems involves
one or more of:
Apply calculus methods, using extended
abstract thinking, involves one or more of:
• selecting and using methods
• demonstrating knowledge of calculus
concepts and terms
• communicating using appropriate
representations.
• selecting and using a logical sequence
of steps
• connecting different concepts or
representations
• demonstrating understanding of
concepts
• forming and using a model;
and also relating findings to a context, or
communicating thinking using
appropriate mathematical statements.
• devising in solving problems a strategy to
investigate a problem
• demonstrating understanding of abstract
concepts
• developing a chain of logical reasoning, or
proof
• forming a generalisation;
and also using correct mathematical
statements, or communicating mathematical
insight.
Sufficiency for each question:
N0: No response, no relevant evidence.
N1: Attempt at ONE question demonstrating limited knowledge of calculus techniques.
N2: 1 u
A3: 2 u
A4: 3 u
M5: 1 r
M6: 2 r
E7: 1 t
E8: 2 t
Judgement Statement
Score range
Produced by SINCOS
Not Achieved
Achievement
Achievement with
Merit
Achievement with
Excellence
0-7
8 - 14
15 - 20
21 -24
Schedule Page 39
2013 EOY Level 2
Writer: David Fortune
23
Question
ONE
Evidence
Achievement (u)
Apply calculus
methods in solving
problems.
1a
1b
f ( x)  6 x  4
Merit (r)
Excellence (t)
Apply calculus
methods, using
relational thinking, in
solving problems.
Apply calculus methods,
using extended abstract
thinking, in solving
problems.
Gradient correct.
At x = 2 gradient = 8
y  x 3  4x  c
At (1, 2) 2  1  4  c
Correct function.
y  x 3  4x  1
1c
1d
1e
dy
 6x  2
dx
At (1, 1) the gradient =  8
Tangent is y  1  8( x  1)
y  8 x  7
g ( x)  2
Thus g ( x)  2 x  c
But g ( 2)  2 ie 2  4  c and
c  2 ie g ( x)  2 x  2
Thus g ( x)  x 2  2 x  c
But g ( 2)  2 ie 2  4  4  c
c  2 ie g ( x)  x 2  2 x  2
1
Thus g ( x)  x 2  x 2  2 x  c
3
But g (0)  2 ie c  2
1
g ( x)  x 2  x 2  2 x  2
3
Gradient correct.
Area = xy  x(9  2 x 2 )
A found.
 9x  2x3
A  9  6 x 2
Correct tangent
found.
Correct function for
g (x ) found.
Correct function for
g (x ) found.
Values of x OR y
found at maximum.
Maximum area
correctly found.
3
2
y6
= 0 when x 2 
Max area =
Produced by SINCOS
3
 6  7.35
2
Schedule Page 40
2013 EOY Level 2
Writer: David Fortune
23
Question
TWO
Achievement
(u)
Evidence
Apply calculus
methods in
solving
problems.
2a
2b
Merit (r)
Apply calculus
methods, using
relational thinking, in
solving problems.
Excellence (t)
Apply calculus methods,
using extended abstract
thinking, in solving
problems.
Sketch is a
straight line
with negative
gradient and x
intercept
between 1 and
2.
Correct
coordinates.
dy
 2 x  3x 2
dx
y  x 2  x3  c
Through origin so c = 0
Equation y  x 2  x 3
So coordinates are (3, -18)
2c
v(t )  0.2t  3
s(t )  0.1t  3t  c
At t = 3 9  0.9  9  c c =  0.9
s(t )  0.1t 2  3t  0.9
After 20 sec distance = 99.1 m
2
2d
P  0.02t  0.75k
Decreasing when
 0.02t  0.75k  0 t
75
k
ie for t  
2
Correct
antiderivative.
Correct distance.
P(x) correctly
differentiated
and made less
than zero..
Correct inequality
in terms of t and k.
Correct value for k.
Also decreasing for t > 45
45  
2e
75
k
2
k  1.2
v  kr 2 (R  r )  kRr 2  kr 3
v   2kRr  3kr 2
2
= 0 when r = 0 or r 
3
v correctly
differentiated.
Correct maximum.
Correct maximum, with
justification for
maximum.
r = 0 is a minimum (zero velocity) and
r
2
R gives maximum velocity.
3
Produced by SINCOS
Schedule Page 41
2013 EOY Level 2
Writer: David Fortune
23
Question
THREE
Evidence
Achievement (u)
Apply calculus
methods in solving
problems.
3a
h( x)  5 x  3.5
1.5  5x  3.5
dSA
 12 x
dx
Graph x < 0 any
straight line with
gradient 3
OR
Graph for x > 0 an
“upside down”
parabola with
turning point at
x = 3.
3c
v(t ) 
Apply calculus methods,
using extended abstract
thinking, in solving
problems.
Rate of change
correct.
At x = 3 rate of change = 36.
3d
Apply calculus
methods, using
relational thinking, in
solving problems.
Excellence (t)
Coordinates
correct.
ie x = 1 and y = 6
Coordinates are (1,6)
3b
Merit (r)
1 3
t t c
12
Correct velocity,
showing constant
has been
calculated.
At t = 0 v = 0 so c = 0
1 3
t t
12
1 4 1 2
s (t ) 
t  t c
48
2
v(t ) 
Graph x < 0 any
straight line with
gradient 3
AND
Graph for x > 0 an
“upside down”
parabola with
turning point at
x = 3.
[No graph at x = 0.]
Correct distance
function, showing
constant has been
calculated.
OR
Distance in 9th
second calculated.
Correct distance,
showing constants have
been calculated.
Found max
distance but not
showed the value
found is a
maximum
OR
Found incorrect
distance.
Found max distance and
showed the value found
is a maximum using any
method.
Distance travelled in 10th second
1
3e
1

10
=  t 4  t 2  = 62.1 m
2 9
 48
Vertical distance
= 4 x 3  15 x 2  20 x  7  3x 2  16 x  10
 4 x  12 x  36 x  17
dV
 12 x 2  24 x  36
dx
dV
0
When
dx
( x  1)( x  3)  0
x  1 x  3
3
2
Derivative
correct and
equated to zero.
dV 2
 24 x  24 so max at x  3
d 2x
Max distance = 125
Produced by SINCOS
Schedule Page 42
2013 EOY Level 2
Writer: David Fortune
23
ASSESSMENT SCHEDULE
91267 Apply probability methods in solving problems
Achievement
Achievement with Merit
Achievement with Excellence
Apply probability methods in solving
problems involves:
Apply probability methods, using
relational thinking, in solving problems
involves one or more of:
Apply probability methods, using extended
abstract thinking, in solving problems
involves one or more of:
• selecting and using methods
• demonstrating knowledge of
probability concepts and terms
• communicating using appropriate
representations.
• selecting and carrying out a logical
sequence of steps
• connecting different concepts or
representations
• demonstrating understanding of
concepts;
and also relating findings to a context, or
communicating thinking using
appropriate mathematical statements.
• devising a strategy to investigate or solve
a problem
• identifying relevant concepts in context
• developing a chain of logical reasoning
• making a statistical generalisation;
and also, where appropriate, using
contextual knowledge to reflect on the
answer.
.
Sufficiency for each question:
N0: no relevant evidence
N1: Attempt at one question with some knowledge shown.
N2: 1 u
A3: 2 u
A4: 3 u
M5: 2 r
M6: 3 r
E7: 1 t
E8: 2 t
Judgement Statement
Score range
Produced by SINCOS
Not Achieved
Achievement
Achievement with
Merit
Achievement with
Excellence
0-8
9 - 14
15 - 19
20 -24
Schedule Page 43
2013 EOY Level 2
Writer: David Fortune
23
Question
One
Evidence
Achievement (u)
Apply probability
methods in solving
problems.
Merit (r)
Apply probability
methods, using
relational thinking, in
solving problems.
1a(i)
0 < z < 2.5
Probability = 0.38493
Correct probability
calculated.
1a(ii)
z > 3.2
Probability = 0.0025551
0.26 percent
Correct probability
calculated and
answer written as
percentage.
1a(iii)
z < - 0.4
Probability = 0.34457
Correct probability
calculated.
1a(iv)
z > 0.6 and z < - 1
Probability = 0.5794
Expected number 695 (or 696)
Correct probability
calculated.
Expected number
calculated.
1a(v)
10 % is z = 1.281
Inverse normal
value of z found.
CAO
Found the % value
for one grower.
Correct supplier
with explanation /
justification.
x

3 1

 1.281
Excellence (t)
Apply probability
methods, using
extended abstract
thinking, in solving
problems.
Correct new sd
with justification.
 1.281
New sd is   1.56
1b
1c
For Growquick 87.8 % of potting mix
has less than 30% water content.
For Readygrow 90.7 % of potting mix
has less than 30% water content.
Thus preferred supplier is Readygrow.
Prob ( People taking less than 10 mins)
= 0.054799
40
x

0.054799 0.945201
CAO – no credit.
Prob 0.054779
found.
Whole group
number found (730
people).
Correct justified
solution.
The number taking more than 10
minutes = x = 689 (or 690)
OR
40  0.054799  p
p population
p
40
 729.9
0.054799
Other people = 729.9 – 40 = 689
NOTE:
There will be differences between probability answers quoted if students use normal distribution
tables rather than GC. Either should be accepted.
Values used here are all GC values
Produced by SINCOS
Schedule Page 44
2013 EOY Level 2
Writer: David Fortune
23
Question
TWO
Achievement
(u)
Evidence
Apply probability
methods in
solving problems.
2a(i)
2a(ii)
A 15-19 year old driver is 5.3 times as likely to
have a fatal crash as a 30+ year old.
30+ yo relative risk =
20-29 yo relative risk = 2.87
15-19 yo relative risk = 2.86
Increasing the alcohol amount increases the
risk for all drivers by a factor of 2.8, regardless
of their age.
2b(i)
100
135
 0.741
2b(ii)
53
61
 0.869
ie 87 % of those who paid, passed their test.
2b(iii)
P (failing) =
35
135
 0.259
1 in 4 is 0.25 so statement correct.
2b(iv)
Risk (paid and passing) = 0.869
Risk (not paid and passing) = 0.635
Relative risk paid to not paid 1.369
So 37 % more likely to pass if they paid.
Ad under rates relative risk – actually 37%
more likely rather than 1/3
2b(v)
Male
Female
TOTAL
passed
failed
Tot
38
62
100
18
17
35
56
79
135
Excellence (t)
Apply probability
methods, using
relational
thinking, in
solving problems.
Apply probability
methods, using
extended abstract
thinking, in solving
problems.
Correct
understanding.
Correct relative
risks calculated.
16.5
= 2.84
5.8
Merit (r)
Correct relative
risks calculated
AND
at least one
correct
statement.
Correct
probability
calculated.
Correct
percentage
calculated.
Correct
probability
calculated.
Correct
probability with
justification
statement.
A correct
probability
calculated.
Both
probabilities
calculated
correctly.
A correct
probability
calculated.
Both
probabilities
calculated
correctly..
Probabilities
correctly compared
and a justified
statement made
about the claim
being close to
correct.
Probabilities
correctly compared
and a justified
statement made
about the claim
being wrong.
18
 0.321
56
17
 0.215
Risk of female failing =
79
Risk of male failing =
So relative risk of male failing to female
failing is 1.49 ie males are 50 % more likely to
fail than females
The claim is wrong.
Produced by SINCOS
Schedule Page 45
2013 EOY Level 2
Writer: David Fortune
23
Question
Three
Evidence
Achievement (u)
Apply probability
methods in solving
problems.
3a(i)
0.05  0.94  0.047
Correct
probability.
3a(ii)
0.05  0.04  0.95  0.92  0.876
Correct
percentage.
= 87.6 %
3a(iii)
Pr (positive test)
= 0.05  0.94  0.95  0.05  0.0945
Pr had mGF if they had a positive test
=
3a(iv)
Pr(positive test)
correctly
calculated.
Merit (r)
Apply probability
methods, using
relational thinking, in
solving problems.
Excellence (t)
Apply probability
methods, using
extended abstract
thinking, in solving
problems.
Correct probability.
0.05  0.94
 0.497
0.0945
P (inconclusive)
= 0.05  0.02  0.95  0.0.03
= 0.0295
A relevant correct
probability
calculated [eg
pr(inconclusive)].
0.0295  x  710
Expected number
correctly calculated.
x = 24067 people (or 24068)
3b(i)
Prob
= 0.753  .75  .25  .6  2  .25  .4  .6
= 0.706875
A correct relevant
probability
calculated.
Correct total
probability
calculated.
3b(ii)
1  0.25  0.4 2 = 0.96
A correct relevant
probability
calculated.
One error in total
probability
calculated.
OR
sum of 7 probabilities.
Produced by SINCOS
Schedule Page 46
Correct total
probability
calculated with
explanation.
2013 EOY Level 2
Writer: David Fortune
23