ASSESSMENT SCHEDULE – BURGER PRINCE - CGHS - 2014
Mathematics and Statistics 91586 (3.14): Apply probability distributions in solving problems
Assessment Criteria
Achievement
Achievement with Merit
Apply probability distributions in solving problems involves:
Achievement with Excellence
Apply probability distributions, using relational thinking, in
solving problems involves:
Apply probability distributions, using extended abstract
thinking, in solving problems involves:
 demonstrating knowledge of concepts and terms
 selecting and carrying out a logical sequence of steps
 devising a strategy to investigate or solve a problem
 communicating using appropriate representations.
 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.
 selecting and using methods
Evidence Statement
ONE
(a)
Expected Coverage
Bacteria
levels
0
10 20 30 40 Over
45
Frequency
4
72 13 0
1
0
Poisson => X = # bacteria per cm2
then using the given midpoints of
intervals and reading negligible as 0.
E(X) = λ = 11.33 (11.44 if 2.5 used for
1st grp). Accept other values based on
the categories selected.
Binomial Chosen:
X = # of swabs/90 with ≥16 bacteria
Parameters are n = 90, p = 14/90 = 0.1556
Applying the distribution because:
 Discrete # outcomes fail or pass test
 Two outcomes fail or don’t fail the test
 Fixed # of trials = 90 tests
 Assume independence ie the bacteria
Applying this distribution because:
count of one swab will not influence
 discrete (bacteria colony count)
2
the bacteria count of the another swab
within a continuous interval cm

Allone
outcomes
have the
 events cannot occur simultaneously (only
count occurs
at asame
time)prob’ of
success
and
failure
 events are random with no set
pattern (bacteria count changes
randomly between each sample)
AS91586 Kohia 2014
Achievement
Merit
Probability
distribution selected
and correct
parameters
evaluated
Probability
distribution selected
and correct
parameters
evaluated
Or
AND
Probability
3 or more correct
distribution selected reasons with
and 3 or more
context.
correct reasons with
context.
1
Excellence


for an interval (area) the mean number of occurrences (bacterial count) is
proportional to the size of the interval.
Assume independence ie the bacteria count of one swab will not influence the
bacteria count of the another swab
Poisson λ = 11.3 or consistent
P( will fail the requirements) =
(b)
P ( X > 15 ) =1 – P(x ≤ 15) = 1 –
0.8905 = 0.1095 ≈ 0.11 (0.1133 for
λ = 11.33, 0.1180 if λ = 11.44)
Binomial Probability = 14/90 = 0.1556,
the same as the parameter as this is the
probability of failing.
Poisson Situation
Binomial Situation
Poisson λ = 11.3 X = # of colonies of
bacteria in 1cm2
Probability of being below 8 colonies/cm 2
= (X ≤ 7) = 0.1249
New Binomial Dist’ n=7days & p =
0.2889 (assume 72 evenly spread)
p could be as low as 10/90 = 0.1111
Number of trials = 7 days daily swabbing
(c)
New Binomial distribution, p = 0.1249
and n = # of trials = 7.
Y = # of times in 7 swabs that less than 8
colonies were found
Y ~ B (n = 7, p = 0.1249)
P (Y ≥ 6) = 1 – P (Y ≤ 5) = 1 – 0.999976
= 0.000024
X = # of swabs/90 with ≤ 7
bacteria
Parameters are n = 90, p = 26/90
= 0.2889
Probability correctly
calculated.
Correctly calculates
probability for
colony count below
8.
Correctly calculates
probability for
colony count below
8
CAO
AND
Identifies and forms
new binomial
distribution
Y = # of times in 7 swabs that
less than 8 colonies were found.
Y~ B (n = 7, p = 0.2889)
P (Y ≥ 6) = 1 – P (Y ≤ 5)
= 1 – 0.9969 = 0.0031
but could be as low as 0.000012 if
p=0.1111
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
(a) (i)
Expected Coverage
Achievement
Normal distribution, µ = 15.67, σ = 7.25
P(Low spender) = P (X < 5.25) = 0.075324 = 0.0753
accept 0.0599 (lower = 0)
P(High spender) = P (X > 20.45) = 0.25484 = 0.2548
Percentage(either high or low) = 0.3301 = 33% (or 31.5%)
Accept any rounding
Excellence
Both high and low
probabilities
correctly calculated
CAO
Normal distribution, µ = 15.67, σ = 7.25
Inverse normal
calculation correct.
Using inverse normal calculations P (X > x) = 0.10
(a) (ii)
Merit
X = 24.961 ≈ $25.00 (rounded to $25 is OK)
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.
(bi)
Normal distribution is an appropriate model because:
 Shape is approximately bell shaped & symmetrical about
 or most values in the centre, with fewer towards edges.
 No fixed upper or lower limit to endpoint boundaries.
 The variable being measured (amount spent per
customer) is continuous
The histogram of the amount spent per customer provides
evidence the target has been met at the peak of the graph is at
approximately $22.50 which is over the target or similar (con).
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.
AS91586 Kohia 2014
3
One normal
distribution features
correctly identified
Two normal
distribution features
correctly identified
AND
AND
new higher/lower
mean identified as
source of
success/failure
new higher/lower
mean identified as
source of
success/failure with
statistical
justification.
AND
Accept any sensible
explanations
AND
OR
Two normal
distribution features
correctly identified.
Two normal
distribution
features correctly
identified
new higher/lower
mean identified as
source of
success/failure
with statistical
justification
Reference is made
to 1 of sampling
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.
(b) (ii)
The assumption is that the histogram follows a Normal dist’n.
Students may assume the mean is $22.50 and SD is still $7.25
or estimate mean and SD from the graph. The actual mean =
$20.59, SD $5.90] Accept any sensible estimate.
With original mean = $15.67 & sd = $7.25 =>
P(X≤10.04) = 0.2187 and P(X≥20.45) = 0.2548
Using INVN and mean = $20.59, SD $5.90 and the above
probabilities we get a medium customer spending in the range
$16.00 and $24.48.
Using INVN and mean = $22.50, SD $7.25 and the above
probabilities we get a medium customer spending in the range
$16.87 and $27.28.
method/sample
size/sampling
variability.
Assumption of
Normal distribution
is made with normal
distribution
parameters
identified.
Assumption of
Normal distribution
is made with normal
distribution
parameters
identified
Assumption of
Normal distribution
is made with
normal distribution
parameters
identified
AND
AND
Sensible attempt
made to calculate
new range for
Medium spenders
Correct calculation
of new range for
Medium spenders
The new category for a Medium spender after the marketing
campaign will lie between $16.87 and 27.28.
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
1 1
Expected Coverage
Achievement
168 7
THREE
1 1
168 7
(a) (i)
Merit
Correct graph
sketched
Correct graph
sketched
OR
AND
correct distribution
identified.
correct distribution
identified.
Excellence
7
168
The probability distribution is modelled by a Uniform Distribution.
Days
hours
If time measured in hours vertical height is 1/168
3
(a) (ii)
P( Inspector arrives between 2pm and 5pm) = 168 = 0.01786
Correct probability
identified. CAO
Area of a triangle = ½ base × height
Correct diagram of
the situation and an
attempt made to
solve.
Correct modal
serving time found.
P(X = 1) = 0.0766
calculated or
attempt to calculate
combined
parameters.
Parameters correct
and probability
calculated.
2×(60−30)
Base = 60-30 = 30, height = (150−30)(𝑐−30)
(b)
2×(60−30)
Area = ½×30×(150−30)(𝑐−30)
900
0.166667 = 120×(𝑐−30)
Solving gives c = 75 seconds, the modal serving time.
Assume both people, A and B, are served independently with
N(74s,9.8s)
(c)
T = A+B, Mean of T = 148s, sd of T = √(9.82 + 9.82 ) = 13.86s
P(T≤120) = 0.0217
AS91586 Kohia 2014
only around 2% of the time.
5
Assumption,
parameters and
probability
calculated.
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