Download Apply probability distributions in solving problems

Assessment Schedule
Statistics and Modelling: Apply probability distributions in solving problems (91586)
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 = 782, σ = 158
(a) (i)
P(X < 600) = 0.1247
P(X > 850) = 0.3335
P(less than $600 or more than $850) = 0.4582
Normal distribution, µ = 782, σ = 158
(a) (ii)
P(600 < X < 820) = 0.4703
P(X < 820) = 0.5950
P(over $600 / under $820)
= P(600 < X < 820) ÷ P(X < 820)
= 0.4703 ÷ 0.5950 = 0.7904
Prepared by TEAM Solutions University of Auckland
Achievement
Merit
Proportion correctly
calculated.
Probability of tourist
spending between $600
and $820 calculated.
Conditional probability
correctly calculated.
Excellence
Normal distribution, µ = 502, σ = 83
P(X < 450) = 0.2655
P(450 < X < 600) = 0.6157
P(X > 600) = 0.1189
(b)
Probabilities for the
different levy groups
calculated.
For one tourist:
E(levy) = 0.2655 × 5 + 0.6157 × 8 + 0.1189 × 12 = 7.6799
E(levy2)= 0.2655 × 52 + 0.6157 × 82 + 0.1189 × 122 = 63.1639
SD(levy) = √(63.1639 – 7.67992) = 4.1830
At least two correct
calculations of expected
value and
variance/standard
deviation for individual
levy or the total levies
calculated
The variation of the
levies is taken into
account when
determining whether
the council should
build the reef, by
considering the size of
the standard deviation
of the total levies.
Merit
Excellence
For all 12000 tourists:
E(total levies) = 12000 × 7.6799 = $92 158.80
SD (total levies) = √(12000) × 4.1830 = $458.22
As the expected total is $92 158.80 with a low standard deviation of $458.22
it is unlikely that the required $100 000 can be reached; so the plan should
not go ahead.
Two
Expected Coverage
E(X) = 0 × 0.45 + 1 × 0.3 + 2 × 0.15 + 3 × 0.1 = 0.9
Correct calculation of
expected number of
cancellations in a oneweek interval.
Assume normal. P(X>95) = 0.05
Solves problem
(a)
(b) (i)
(b) (ii)
Achievement
We are assuming that the attendance is normally distributed.
This is unlikely to be the case, as we might expect certain
games to be fully sold out (100% attendance) e.g. semi-finals
and final, home games for a popular team, games for smaller
stadia. 100% is more than 4 standard deviations above the
mean
Solves problem and
discusses assumption of
normality.
Assuming normality, P(one game has under 90% occupancy) = 0.079
Assuming attendance of each game is independent, probability is 0.0794
(c)
= 3.9 x 10-5 (2 sf). Game attendance is unlikely to be
independent with equal probability as games will be held at
different venues with different players. As the season
progresses towards the final, attendance may also increase.
Three
Expected Coverage
Makes progress towards
solving problem e.g.
probability for single
game.
Achievement
Correctly finding f(x) for
c<x<b
2/17
Devises strategy to
solve problem.
Merit
Probability calculated
with identification of
probability distribution
and parameter
.
(a)
5
f(x) = 2(22-x)/170 , c<x<b
f(15) = 0.0824 (4dp)
12
15
22
Devises appropriate
strategy to solve
problem, discussing
assumptions.
Excellence
Probability is unshaded Area
= ½ x 0.0824x (22-15)
= 0.2882 (4dp)
(i)
Probability from Graph = 11/89 + 13/89 + 15/89 + 4/89 + 1/89 + 4/89 + 6/89 =
54/89 = 0.607 (3dp)
Probability form Part (a) = 0.41
Correctly calculates the
probability of more than
15 guests
Mode is higher (18 compared to twelve) , minimum is lower, Max is the same.
The distribution is significantly different, however a triangular distribution
would still be the best model to use.
Notes one significant
difference
Discusses at least 2
differences and
comments on
applicability of the
chosen distribution
Discusses at least 2
differences in context
and effectively
evaluates model
estimate against true
probability.
Model used in part a is a theoretical model, based on a rough estimate of
numbers. There could be an argument that it is based on experimental data
and is therefore an experimental model. This is acceptable if explained
clearly. The actual numbers of guests shown in part b shows the true
probability which was being modeled in part a. For future predictions this
would provide reliable data for an experimental model. Provided that other
conditions remain the same for example the number of guests booking into
the hotel
One of theoretical,
experimental or true
probability occurences
correctly indentified
Two or more of
theoretical,
experimental or true
probability occurences
identified with some
justification
Thorough and
reasoned discussion of
occurences of
theoretical,
experimental or true
probability within the
context of this problem
New distribution drawn with new max min and mode.
Min of 0 or 8 used,
mode of 18 and max of 22.
Or alternative appropriate parameters.
New distribution drawn
with some sensible
parameters
Calculates probability
according to distribution
chosen
AND Effectively
justifies choice of
model
(c)
(d)
(d)
P(at least 15 guests) calculated correctly according to the model.
For each question:
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.
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