Assessment Schedule – Kohia 2015
Mathematics and Statistics (Statistics): Apply probability concepts in solving
problems (91585)
Evidence Statement
One
(a)(i)
Expected Coverage
Cut
and
Colour
Normal
Cut
Total
(ii)
(b)
Male
Female
Total
0.03
0.21
0.24
0.28
0.48
0.76
0.31
0.69
1
Achievement (u)
The number of
people is
correctly
determined in
part (i).
One correct
probability is
calculated in part
(ii).
A female is approximately 3.1 times more likely to get a cut
and colour than a male.
P(OHD | Male) = 0.5992 (4sf)
P(OHD | Female) = 0.6843 (4sf)
0.5992
0.6843
Excellence (t)
OR
P (cut and colour) = 0.24
Number of people who get cut and colour =
0.24 × 500 = 120
P(Cut and Colour | Male) = 0.097
P(Female | Cut and Colour) = 0.304
Relative Risk =
Merit (r)
The correct
probabilities are
compared in
part (ii).
= 0.8756 (4sf)
The risk for a male developing occupational hand dermatitis
is 0.8756 [times] greater than the risk for a female.
(c)
The correct
conditional
probabilities are
identified for the
comparison, e.g.
through the use
of probability
statements.
At least one
correct
probability
relevant to the
problem is
calculated.
a + b = 0.2
a = 0.32 × x
b = 0.68 × 0.1x
0.32 × x + 0.68 × 0.1x = 0.2
x = 0.5155 (4sf)
0.16496
P(OHD | Dropped Out) =
= 0.8248
A statement is
made that
compares the
risk using
numerical
values.
A reasonable
attempt to
model the
situation using
appropriate
methods and /
or diagrams is
demonstrated,
including at
least one correct
probability
relevant to the
problem being
calculated.
The correct
conditional
probability is
calculated,
supported by a
clear
communication
of strategy used
to obtain this
probability,
including use of
correct
probability
statements.
0.2
N0
N1
No response / Reasonable
No relevant
attempt at
evidence
one part of
the question.
AS91585 Kohia 2015
N2
Almost
complete
correct
answer
A3
1 of u
A4
2 of u
Page 1
M5
1 of r
M6
2 of r
E7
1 of t (with
minor error)
E8
1 of t
Two
(a)(i)
Expected Coverage
270
The proportion of customers who got a $10 discount was
500
= 0.54.
This is different to the theoretical probability of 0.5
(ii)
However, we would expect a difference between the observed
proportion and the theoretical probability, [due to chance
variation.]
True probability is the unknown actual probability that the
spinner will land on the $10 discount. The actual probability
of getting a $10 discount could be affected by where the
spinner starts, how fast it is spun, if the spinner is evenly
weighted etc. So while we expect the true probability to be
close to the theoretical probability of 0.5 and the experimental
probability of 0.54, we cannot know exactly what it is.
Students could use a standard error calculation to discuss the
difference between the two numbers.
(b)(i)
Customer
Happy
Customer Not
Happy
Total
(ii)
Served by
Terrance
Not Served by
Terrance
Total
1%
89%
90%
7%
3%
10%
8%
92%
100%
1% of customers are served by Terrance and Happy so they
are not mutually exclusive.
i.e. P(A⋂B) ≠ 0
P(Happy | Served by Terrance) = 0.125
P(Happy | Not Served by Terrance) = 0.9674 (4sf)
0.9674
0.125
= 7.739 (4sf)
Customers not served by Terrance are 7.74 (or 8) times more
likely to be happy than customers served by Terrance, so yes
his boss should be worried.
N0
N1
No response / Reasonable
No relevant
attempt at
evidence
one part of
the question.
AS91585 Kohia 2015
N2
Almost
complete
correct
answer
A3
1 of u
A4
2 of u
Page 2
Achievement (u)
A vague
explanation
about how the
observed
proportion is
unlikely to be the
same as the
theoretical
probability for
part (a).
Merit (r)
A clear
explanation
about how the
observed
proportion is
unlikely to be
the same as the
theoretical
probability for
part (a).
Excellence (t)
A clear
explanation
about how the
observed
proportion is
unlikely to be
the same as the
theoretical
probability for
part (a).
OR
OR
AND
Some discussion
about true
probability in
part (b).
Discussion
given about
what true
probability is
and what could
affect it in part
(b).
Correct
statement made
to justify not
being mutually
exclusive
Discussion
given about
what true
probability is
and what could
affect it in part
(b).
Table
constructed
The correct
conditional
probabilities are
identified for the
comparison, e.g.
through the use
of probability
statements.
M5
1 of r
A statement is
made about his
boss that is
backed up by
numerical
values.
M6
2 of r
E7
1 of t (with
minor error)
E8
1 of t
Three
(a)
Expected Coverage
5+ years
266
394
660
Owns Salon
Didn’t Own Salon
Total
< 5 years
30
810
840
Achievement (u)
Proportion
correctly
calculated for
part (a).
Total
296
1204
1500
OR
(b)
P(working less than 5 years and don’t own salon)
= 0.54 (4sf)
266
P(own salon | working 5+ years) =
(0.3424)
660
266
(c)
265
P(both own salon | both working 5+ years) =
×
660
659
= 0.1628 (4sf)
660
P(5+ years) =
1600
296
P(owns salon) =
1500
P(5+ years and owns salon) =
266
1500
660
P(5+ years) × P(owns salon) =
= 0.177
1600
×
296
1500
= 0.0814
Conditional
probability for
one hairdresser
correctly
calculated for
part (b).
Correct
probabilities
calculated as part
of a reasonable
attempt to use an
independence
rule.
Merit (r)
Conditional
probability for
both
hairdressers
correctly
calculated for
part (b).
Excellence (t)
Independence
rule used with
correct
probabilities to
determine
events are not
independent.
Therefore the events are not independent
as P(A) × P(B) ≠ P(A ∩ B)
(d)
A reasonable
attempt is made
to organise
information (e.g.
calculates at least
four values
correctly) and
arrives at a
consistent
incorrect
probability.
A venn diagram
(or similar) is
completed
Probability
correctly
calculated.
P(female who has been working less than 5 years and owns
15
their own salon) =
= 0.01
1500
Accept other valid chains of reasoning
N0
N1
No response / Reasonable
No relevant
attempt at
evidence
one part of
the question.
N2
Almost
complete
correct
answer
A3
1 of u
A4
2 of u
M5
1 of r
M6
2 of r
E7
1 of t (with
minor error)
E8
1 of t
Cut Scores
Score Range
AS91585 Kohia 2015
Not Achieved
Achieved
Achievement
with Merit
Achievement
With Excellence
0–6
7 – 12
13 – 19
20 – 24
Page 3