Download 91586 Sample Assessment Schedule

ASSESSMENT SCHEDULE – BURGER PRINCE Kohia 2014
Mathematics and Statistics 91586 (3.14): Apply probability distributions in solving problems
Assessment Criteria
Achievement
Achievement with Merit
Achievement with Excellence
Apply probability distributions in solving problems
involves:
Apply probability distributions, using relational
thinking, in solving problems involves:
Apply probability distributions, using extended
abstract thinking, in solving problems involves:
 selecting and using methods
 selecting and carrying out a logical sequence of
 devising a strategy to investigate or solve a
 demonstrating knowledge of concepts and terms
 communicating using appropriate
representations.
steps
problem
 connecting different concepts or representations
 identifying relevant concepts in context
 demonstrating understanding of concepts
 developing a chain of logical reasoning
and also relating findings to a context or
communicating thinking using appropriate
statements.
and also, where appropriate, using contextual
knowledge to reflect on the answer.
Evidence Statement
ONE
(a)
Expected Coverage
Bacteria
levels
0 (or negligible)
10
20
30
40
Over
45
Frequency
4
72
13
0
1
0
Probability calculated
with identification of
probability distribution
and parameters.
Probability calculated
with identification of
probability distribution
and parameter
justification of
applying this
distribution linked to
the context.
Poisson distribution, λ = 11.3
Appropriate distribution to model is a Poisson distribution.
Applying this distribution because:
AS91586 Kohia 2014
Merit
AND
Using the given midpoints of intervals and reading negligible as 0.
E(X) = 11.33
Accept other values based on the categories and sensible
approximations.

Achievement
discrete (bacteria colony count) within a continuous interval (area)
1
Excellence

events cannot occur simultaneously (only one count occurs at a
time)

events are random with no set pattern (bacteria count changes
randomly between each sample)

for an interval (area) the mean number of occurrences (bacterial
count) is proportional to the size of the interval.
I am making the assumption that the samples are independent of each
other, that bacteria count of one swab will not influence the bacteria
count of the subsequent swab.
Poisson λ = 11.3
(b)
Probability correctly
calculated.
P( will fail the requirements) = P ( X > 15 ) =1 – P(x ≤ 15)
= 1 – 0.8905 = 0.1095 ≈ 0.11
Poisson λ = 11.3
Probability of being below 8 colonies/cm2 = (X ≤ 7) = 0.1249
Number of trials = 7 days of daily swabbing
(c)
Correctly calculates
probability for colony
count below 8.
Correctly calculates
probability for colony
count below 8
CAO
AND
Identifies and forms
new binomial
distribution
New Binomial distribution with p = 0.1249 and p = number of trials =
7.
Y ~ B (n = 7, p = 0.45296)
P (Y ≥ 6) = 1 – P (Y ≤ 5) = 1 – 0.99997 = 0.00003
The restaurant will almost certainly fail to meet the new requirements
at least two days a week.
N
N1
N2
A3
A4
M5
M6
E7
E8
No
relevant
evidence.
Making
progress.
1 of u
2 of u
3 of u
1 of r
2 of r
1 t with
minor
errors
1t
AS91586 Kohia 2014
2
Correct answer
supported by clear
reasoning.
TWO
Expected Coverage
Achievement
Normal distribution, µ = 15.67, σ = 7.25
Excellence
Both high and low
probabilities correctly
calculated
P(Low spender) = P (X < 5.25) = 0.075324 = 0.075
(a) (i)
Merit
P(High spender) = P (X > 20.45) = 0.25484 = 0.25
CAO
Percentage(either or high or low) = 32.5%
Accept any rounding
Normal distribution, µ = 15.67, σ = 7.25
Inverse normal
calculation correct.
Using inverse normal calculations P (X > x) = 0.10
(a) (ii)
X = 24.961 ≈ $25.00
The customers would be expected to spend $25 and over to get into
the “Super High” category. Accept any correct rounding in the final
answer.
(b)
Normal distribution is an appropriate model because:
 Shape is approximately symmetrical
 Most values in the centre, with fewer towards the edges.
 The variable being measured (amount spent per customer) is
continuous
The histogram of the customer spend provides evidence the target
has been met at the peak of the graph is at approximately
$22.50 which is over the target.
Given the pre-campaign mean was at $15.67, the new approximate
mean is now 1 standard deviation (σ = 7.25) higher.
The shape of the graph is a bit asymmetrical, which suggests that the
mean may be slightly less than $22.50 so target may not have
been met, but mean will be close to target, so success of new
campaign has been demonstrated.
As the customers were randomly selected and the sample size is also
adequate (n = 300), the findings from this sample can be seen
as reliable.
The target appears to have been met given the preliminary data.
AS91586 Kohia 2014
3
One normal
distribution features
correctly identified
Two normal
distribution features
correctly identified
Two normal
distribution features
correctly identified
AND
AND
AND
new higher mean
identified as source of
success
new higher mean
identified as source of
success with
statistical justification.
new higher mean
identified as source
of success with
statistical
justification
Accept any sensible
explanations
AND
Reference is made
to 2 of: sampling
method/sample
size/sampling
variability.
The assumption is that the histogram follows a Normal distribution.
Students may assume the mean is $22.50 and SD is still $7.25 or
estimate mean and SD from the graph
[Actual mean = $20.50, SD $6.00]
Accept any sensible estimate.
Standardising the lower end of category, $10.04:
Z=
𝑥− µ
σ
=
10.04−15.67
7.25
Assumption of Normal
distribution is made
with normal
distribution
parameters identified.
AND
= -0.77655
Sensible attempt
made to calculate new
range for Medium
spenders
Standardising the upper end of category, $20.45:
Z=
(b) (i)
𝑥− µ
σ
=
20.45−15.67
7.25
= 0.6593
With the new distribution and solving for x, the lower end will be:
𝑥− 22.50
7.25
Assumption of Normal
distribution is made
with normal
distribution
parameters identified
= -0.77655 => x = $18.61
The upper end will be:
𝑥− 22.50
7.25
= 0.6593 => x = $25.80
The new category for a Medium spender after the marketing campaign
will lie between $18.61 and 25.80.
Allow for rounding.
N
N1
N2
A3
A4
M5
M6
E7
E8
No
relevant
evidence.
Making
progress.
1 of u
2 of u
3 of u
1 of r
2 of r
1 of t
2 of t
AS91586 Kohia 2014
4
Assumption of
Normal distribution
is made with normal
distribution
parameters
identified
AND
Correct calculation
of new range for
Medium spenders
Achievement
Probability
7
Expected Coverage
1
THREE
Merit
Correct graph
sketched
Correct graph
sketched
OR
AND
correct distribution
identified.
correct distribution
identified.
(a) (i)
7
days
The probability distribution is modelled by a Uniform Distribution.
If time measured in hours vertical height is 1/168
3
(a) (ii)
Correct probability
identified.
P( Inspector arrives between 2pm and 5pm) = 168
CAO
ID card in original
drawer
0
1
2
3
P( X = x)
2
6
3
6
0
1
6
E(X) = 1 and SD (X) = 1
(b) (i)
Total number of ways to place ID hooks= 6
Correct probability
distribution table
drawn
OR
AND
correct E(X) or SD(X).
correct E(X) and
SD(X).
CAO
Let the drawers be 1, 2 and 3
No original drawer = 2,3,1 and 3,1,2
One original drawer = 1,3,2 or 2,1,3 or 3,2,1
If two in original then will be three in original drawer.
Three original drawer – 1,2,3
AS91586 Kohia 2014
Correct probability
distribution table
drawn
5
Excellence
Assumption is the independence of the rankings between the
inspectors. There should be no ‘rivalry’ nor collusion and they should
not be aware of the rankings of the each other.
Expected value
correctly calculated.
CAO
Let the first inspector have distribution X and the second have
distribution Y.
Average of the two inspectors expected rankings =
=
(b) (ii)
4 + 3.5
2
E(X+ Y)
2
=
E(X) + E(Y)
2
Expected value
correctly calculated
AND
Expected value and
Standard deviation
correctly calculated
EITHER
AND
Standard deviation
correctly calculated
assumption of
independence stated
in context
OR
= 3.75
assumption of
independence stated
in context
As SD(X) = 1.2 and SD(Y) = 2.3:
VAR(X) = 1.22 = 1.44 AND VAR(Y) = 2.32 = 5.29
Average of the two inspectors standard deviations rankings:
1
2
1
4
VAR( (X + Y) ) = [ VAR(X) + VAR(Y)] =
1
1
4
1
4
[1.44 + 5.29] = [6.73]
1
SD(½ (X + Y)) = √4 [6.73] = 2 x √6.73 = 1.297 = 1.3
The average ranking of the two inspectors is 5.29 with a SD of 1.3.
N
N1
N2
A3
A4
M5
M6
E7
E8
No
relevant
evidence.
Making
progress.
1 of u
2 of u
3 of u
1 of r
2 of r
1 t with
minor
errors
1 of t
Judgement Statement
Score
range
AS91586 Kohia 2014
Not Achieved
Achievement
Achievement
with Merit
Achievement
with Excellence
0–7
8 – 12
13 – 18
19 – 24
6