Assessment Schedule – Kohia 2015 (Modified)
Mathematics and Statistics (Statistics): Apply probability distributions in
solving problems (91586)
Evidence Statement
One
(a)(i)
(ii)
(b)
(c)(i)
(ii)
Expected Coverage
~ Binomial n = 125, p = 0.4
P(X<40) = 0.02633 (4sf)
The probability that less than 40 of the clients are male on that
day is 0.02633 (4sf)
Accept Normal Dist approximation calculation.
Binomial distribution because:
• Only two outcomes for each client: either male or female.
• Fixed probability of 40% for a client being male.
• Fixed number of trials (125 customers in a day).
• Assume each client’s gender is independent from another.
(Independence could be implied by description).
P(0 clients in 15 minutes) = 0.018; e-λ=0.018
Average for 15 minutes = -ln(0.018) = 4.017 (4sf)
Average for 1 hour = 16.07 (4sf)
~ Poisson λ = 16.07 ; GC P(X<10) = 0.04183952
The probability that during a single hour, less than 10
customers arrive = 0.04184 (4sf)
Poisson distribution because:
•S: clients cannot arrive in exactly the same place and time…
•I: each client’s gender is independent …
•R: client’s wanting a haircut turn up at random…
•No. of clients per 15mins (discrete values within a
continuous interval)
Achievement (u)
Probability
Calculated
Correctly
OR
At least TWO
conditions for
binomial
distribution
identified, but
are not related to
the context.
At least TWO
conditions for
poisson
distribution
identified, and
related to the
context.
Merit (r)
Probability
Calculated
Correctly
AND
Justification of
applying this
distribution
linked to the
context (TWO
conditions.)
E(N) = 1 × 0.23 + 2 × 0.63 + 3 × 0.08 + 4 × 0.06
= 1.97
Josh works on average 1.97 (2) shifts a week
Pay for 1 or 2 shifts = N × 8 × 15.30
Pay for 3 or 4 shifts = N × 8 × 15.30 + 30
E(N) correctly
calculated.
OR
Consistent
calculation of
E(P) from an
incorrect table.
E(P) = $245.33
is given in
context .
n
p
P(P = p)
1
122.40
0.23
2
244.80
0.63
3
397.20
0.08
4
519.60
0.06
Calculation of
λ=4.017
AND
Justification of
applying this
distribution
linked to the
context (TWO
conditions.)
Excellence (t)
Calculation of
λ=16.07 and
P = 0.0418
AND
Justification of
applying this
distribution
linked to
context (TWO
conditions.)
E(P) = 122.40 × 0.23 + 244.80 × 0.63 + 397.20 × 0.08 +
519.60 × 0.06
= $245.33
Josh’s expected pay is $245.33 per week.
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
or omission)
E8
1 of t
Two
(a)
Expected Coverage
~ Normal
Mean = 28, sd = 7
P(X<30) = 0.6125 (4sf)
P(both under 30) = 0.61252 = 0.3751 (4sf)
Assuming that the amount of time for each haircut is
independent.
(b)(i)
~ Triangular
Min = 50, mode = 58, max = 60
Achievement (u)
Correct
probability of an
individual
haircut.
OR
Incorrect
probability of
one haircut,
consistently
squared and with
correct
independence
statement, in
context
Correct
proportion
found
Merit (r)
Correct
probability of
both haircuts
being less than
30 minutes,
with assumption
of independence
stated in
context.
Calculation of
P(X > 2.5) based
on an incorrect
or estimated
mean (shown).
OR
Correct z value.
Correct
calculation of
mean. Must
show working
used (z value).
OR
Correct
probability of
0.07 using
correct standard
deviation (no
working) plus
correct
discussion of a
limitation in
context.
30% of haircuts take less than 54.9 mins,
(ii)
(c)
Justification:
The triangular distribution could be used since we have
minimum, maximum and modal amounts.
~ Normal
Mean = ?, sd = 0.25, P(X<2) = 0.3
𝑋−𝜇
𝑍=
𝜎
Z = -0.5244
2−𝜇
−0.5244 =
0.25
µ = 2.1311
P(X>2.5) = 0.0700 (3sf)
There is a 7% chance that the haircut will last more than 2.5
hours.
Limitations:
The time taken to do hair will be limited by the closing time
of the salon, i.e. if the salon is only open for 8 hours, there is
a maximum time of 8 hours for cut and colour. A normal
distribution does not have an upper limit. Similarly the
amount of time taken for a cut and colour has a lower bound,
0 and the normal distribution has no lower bound. The
distribution of time taken may also not be symmetrical, since
they would be aiming to get the job done as quickly as
possible. The distribution is more likely to be positively
skewed and unimodal.
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
M5
1 of r
Excellence (t)
Correct
proportion
found
AND
Justification for
selected model
given in
context.
M6
2 of r
Correct
calculation of
P(X > 2.5) with
working shown
to get standard
deviation.
AND
Discussion
clearly identifies
one limitation
with using the
normal
distribution, and
links this to the
context to
explain why it is
a limitation.
E7
1 of t (with
minor error
or omission)
E8
1 of t
Three
(a)(i)
Expected Coverage
~ Uniform
Min = 8, Max = 20
1
1
Height =
=
20−8
P(X<13) = 5 ×
P(X>18) = 2 ×
(ii)
(b)
(c)
12
1
5
12
1
12
1
12
=
=
6
5
1
7
P(X<13 or X>18) = + = = 0.583 (3𝑠𝑓)
12
6
12
Justification:
A uniform distribution can be used because minimum and
maximum amounts are given, and there is no mode given.
~ Poisson
Mean = 8 × 15 = 120
P(X>120) = 1 – 0.6962 = 0.3038 (4sf)
Possible invalid assumptions:
• Customers arrive randomly – more likely that customers
arrive at set times e.g. lunch time for workers, after
school for students, and not at random throughout the
day.
• The mean number customers arriving per hour (or the
rate) is constant – the number of customers arriving is
more likely to vary with the time of day, eg lunch times
would expect more customers.
• Independence – people getting a haircut could influence
another person ordering one.
Accept other reasonable specific discussion around why the
Poisson model may not be suitable, linked to the conditions
required to apply the distribution.
Note: Events cannot occur simultaneously is not an incorrect
assumption, as two customers cannot arrive at EXACTLY the
same time, if you keep splitting the time up into smaller and
smaller intervals they will eventually end up in two different
intervals.
The triangular distribution would be appropriate because:
• The shape of the distribution is skewed to the right
• The variable is continuous (the time taken for a haircut)
• There appears to be a very clear minimum (cannot take
less than zero minutes) and there are never any haircuts
that take more than 30 minutes.
Using Triangular Distribution
Min = 0, max = 30, mode = 7
Approximate values are:
P(0<X<5) = 5/42 = 0.119
P(5<X<10) = 97/322 = 0.301
P(10<X<15) = 35/138 = 0.254
P(15<X<20) = 25/138 = 0.181
P(20<X<25) = 5/46 = 0.109
P(25<X<30) = 5/138 = 0.0362
Put into context.
For example:
Using the triangular distribution the probability that the
haircut takes between 10 and 15 minutes is 0.254, whereas the
probability on the histogram is 0.24. These are very similar so
this backs up that the triangular distribution is appropriate
because you would expect some difference due to random
variation.
AS91585 Kohia 2015
Page 3
Achievement (u)
Correct
probability found
OR
Diagram of
model with
incorrect or no
calculation but
correct
justification in
context.
Merit (r)
Correct
probability
found
AND
Justification for
selected model
given in
context.
Probability
correctly
calculated.
Probability
correctly
calculated.
AND
At least one
invalid
assumption
identified and
discussed in
context
At least one
correct
probability
calculated using
the triangular
distribution
OR
Two relevant
features of the
distribution are
described.
TWO relevant
features of the
distribution are
described and
compared to the
features of the
triangular
distribution.
OR
A relevant
probability is
calculated using
the triangular
distribution and
compared to the
given
distribution.
Excellence (t)
Explanation
includes
comparison of at
least 2 features
of the given
distribution and
the triangular
distribution.
AND
At least one
relevant
calculation is
performed to
allow
comparison of
probabilities
between the
given
distribution and
the triangular
distribution for
E7 and random
variation
discussed for E8
AND
conclusion
given
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
or omission)
E8
1 of t
Cut Scores
Score Range
AS91585 Kohia 2015
Not Achieved
Achieved
Achievement
with Merit
Achievement
With Excellence
0–8
9 – 13
14 – 19
20 – 24
Page 4