Download 91586 Sample Assessment Schedule

NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 1 of 7
SAMPLE ASSESSMENT SCHEDULE
Mathematics and Statistics 91586 (3.14): Apply probability distributions in solving problems
This assessment schedule has been updated (14 June 2013), along with the assessment task, to reflect the standard. Linear transformations of
continuous random variables have been removed from the curriculum and so are no longer in the standard.
Student work exemplars were completed before this update and have not been amended.
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 steps
 demonstrating knowledge of concepts and terms
 connecting different concepts or representations
 devising a strategy to investigate or solve a
problem
 communicating using appropriate representations.
 demonstrating understanding of concepts
 identifying relevant concepts in context
and also relating findings to a context or communicating
thinking using appropriate statements.
 developing a chain of logical reasoning
and also, where appropriate, using contextual
knowledge to reflect on the answer.
Evidence Statement
One
Expected Coverage
Normal distribution, µ = 215, σ = 13.2
(a) (i)
P(X < 200) = 0.1279
P(X > 220) = 0.3524
P(less than 200 g or 220 g and over) = 0.4803
Achievement
Probability correctly
calculated.
Merit
Excellence
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 2 of 7
Normal distribution, µ = 215, σ = 13.2
P(X < 220) = 0.64757
P(205 < X < 220) = 0.42322
(a) (ii)
P(over 205g/ under 220g) =
P(205 < X < 220) / P(X < 220) = 0.42322 / 0.64757
= 0.6536
Probability of guava
weight between 205 g
and 220 g calculated.
Conditional probability
correctly calculated.
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 3 of 7
Using the normal distribution, with µ = 12.4 mins and σ = 3.0 mins
P(X < 10) = 0.192
P(10 < X < 15) = 0.767
P(X > 15) = 0.041
Time to
change flat
tyre (T)
Fee (F)
Probability
T < 10
10 < T < 15
T > 15
$35
0.212
$40
0.595
$60
0.193
Probability distribution
table (or equivalent
probability statements)
is developed for the
distribution of fees for
individual flat tyres.
E(F) = 35 x 0.212 + 40 x 0.595 + 60 x 0.193 = $42.80
E(F2) = 352 x 0.212 + 402 x 0.595 + 602 x 0.193 = 1906.5
SD(F) =
(b)
= $8.641
The expected total fees for 48 tyres = 48 x $42.80 = $2 054.40
The standard deviation of the total fees for 48 tyres =
x 8.641 = $59.87
(assuming independence of the times to change the flat tyres of each of the
48 tyres).
The expected total fees for 48 tyres per week is $2054.40, which is $445.60
below the centre’s expectations. When considering this difference of $445.60
in terms of the standard deviation of $59.87, the centre’s target is over 7
standard deviations above the expected total fees. Therefore, it is not
reasonable that the centre could generate fees of at least $2 500 from
Andrew changing 48 flat tyres per week.
Not Achieved
NØ
No response; no relevant evidence.
N1
Candidate gives a partial solution to ONE part of the question.
N2
Candidate gives partial solutions to TWO parts of the question.
A3
Candidate gives ONE opportunity from the Achievement criteria.
A4
Candidate gives TWO opportunities from the Achievement criteria.
Achievement
Correct calculation of
expected value and
standard deviation for F
OR
correct calculation of the
expected value for the
total fees for 48 flat tyres
and the use of the
expected value and
relevant justification to
determine whether the
centre’s target is
reasonable
The variation of the
total fees from Andrew
changing 48 tyres per
week is taken into
account when
determining whether
the centre’s target is
reasonable, by
calculating and
considering the
expected value and
the standard deviation
for the total fees from
Andrew changing 48
tyres per week
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 4 of 7
M5
Candidate gives ONE opportunity from the Merit criteria.
M6
Candidate gives TWO opportunities from the Merit criteria.
E7
Candidate meets the Excellence criteria except for minor errors in calculation.
E8
Candidate meets the Excellence criteria.
Merit
Excellence
Two
Expected Coverage
For 15-minute interval
(a)
E(X) = 0 x 0.3 + 1 x 0.35 + 2 x 0.2 + 3 x 0.15 = 1.2
Poisson distribution, λ = 2.4
Applying this distribution because:
(b) (i)

discrete (queue jumpers) within a continuous interval (time)

it cannot occur simultaneously (only one queue jumper at a time)

queue jumping is a random event with no pattern to it

for a small interval (eg half an hour) the mean number of occurrences
(queue jumpers) is proportional to the size of the interval.
Achievement
Merit
Excellence
Correct calculation of
expected number of
jumpers per 15-minute
interval.
Probability calculated
with identification of
probability distribution
and parameter.
Probability calculated
with identification of
probability distribution
and parameter
AND
justification of applying
this distribution linked to
the context.
Assumption is that each queue jumper is independent of other queue
jumpers.
P(X ≤ 1) = 0.3084
Poisson distribution
P(X = 0) = 0.9
e-λ = 0.9
(b) (ii)
λ = 0.105 (per 10 minutes)
λ = 0.316 (per 30 minutes)
P(X ≤ 1) = 0.959
The probability of there being more than one queue jumper per half an hour
is less than 5%, so this could be evidence of the store being ‘queue-jump
free’.
Not Achieved
NØ
No response; no relevant evidence.
Calculation of λ for the
four weeks.
Calculation probability
of no more than one
queue jumper per half
an hour using λ for the
four weeks, with
discussion of store
manager’s claim.
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 5 of 7
N1
Candidate gives a partial solution to ONE part of the question.
N2
Candidate gives partial solutions to TWO parts of the question.
A3
Candidate gives ONE opportunity from the Achievement criteria.
A4
Candidate gives TWO opportunities from the Achievement criteria.
M5
Candidate gives ONE opportunity from the Merit criteria.
M6
Candidate gives TWO opportunities from the Merit criteria.
E7
Candidate meets the Excellence criteria except for minor errors in calculation.
E8
Candidate meets the Excellence criteria.
Achievement
Merit
Excellence
Three
Expected Coverage
Binomial distribution
n = 4, p = 0.67
Applying this distribution because:
(a)

fixed number of trials (four seasons/years)

fixed probability (67% success rate for pollination)

two outcomes (pollinated, not pollinated)

independence of events (a plant being pollinated or not does not affect
chances of the same plant being pollinated or not in another
season/year).
Achievement
Probability calculated
with identification of
probability distribution
and parameters.
Merit
Probability calculated
with identification of
probability distribution
and parameter
AND
justification of applying
this distribution linked to
the context.
Assumption is that the bees visit all the plants.
P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.4015 = 0.5985
Binomial distribution
n = 12, p = 0.91
P(X = 11) = 0.3827
(b)
P(11 plants produce fruit for both seasons/years)
= 0.38272 = 0.1465
Assuming whether a plant is pollinated or not in one season does not affect
the chances of the same plant being pollinated or not in another season/year,
and assuming whether a plant is pollinated or not in one season/year does
Probability calculated
for one season.
Probability calculated
for both seasons with at
least one assumption of
independence stated.
Excellence
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 6 of 7
not affect the chances of another plant being pollinated or not in the same
season/year.
The distribution of amounts of individual sales can be modelled by the
triangular distribution (assuming that sales are always between $0 and
$1000). The shape of the distribution of the data provided supports this model
as it has a clear mode at $700 (or modal class of $650 - $750). The random
variable “amount per sale” can be treated as continuous.
This gives the following model:
Triangular distribution
identified as appropriate
model and probability of
an individual sale of at
least $800 correctly
calculated using this
model.
Triangular distribution
identified and justified
as an appropriate model
using features of the
supplied data and the
probability of an
individual sale of at
least $800 correctly
calculated using this
model.
Triangular distribution
identified and justified
as an appropriate
model using features
of the supplied data
and the probability of
an individual sale of at
least $800 correctly
calculated using this
model
AND
(c)
The value of f(x) at 800 = 2(1000 – 800)/[(1000 – 0)(1000 – 700)] = 0.0013
P(X ≥ 800) = 0.5 x 200 x 0.0013 = 0.13
(using the formula for the area of a triangle)
Assuming that the amount of each individual sale is independent from each
other, the binomial distribution can be used, with n = 8 (fixed number trials, 8
sales), and p = 0.13 (fixed probability of success, probability of sale being at
least $800):
P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – 0.7206= 0.2794
NØ
No response; no relevant evidence.
N1
Candidate gives a partial solution to ONE part of the question.
Not Achieved
Identification and
justification of the
binomial distribution as
an appropriate model
and the probability of
at least two sales
being at least $800
correctly calculated
using this model.
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 7 of 7
N2
Candidate gives partial solutions to TWO parts of the question.
A3
Candidate gives ONE opportunity from the Achievement criteria.
A4
Candidate gives TWO opportunities from the Achievement criteria.
M5
Candidate gives ONE opportunity from the Merit criteria.
M6
Candidate gives TWO opportunities from the Merit criteria.
E7
Candidate meets the Excellence criteria except for minor errors in calculation.
E8
Candidate meets the Excellence criteria.
Achievement
Merit
Excellence