NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 1 of 8
Level 3 Mathematics and Statistics
(Statistics)
91585 (3.13): Apply probability concepts in solving
problems
Credits: Four
Check that you have completed ALL parts of the box at the top of this page.
You should answer ALL parts of ALL questions in this booklet.
If you need more room for any answer, use the space provided at the back of this booklet.
Check that this booklet has pages 2–10 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE END OF THE ALLOTTED TIME.
You are advised to spend 60 minutes answering the questions in this booklet.
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 2 of 8
QUESTION ONE
(a)
At the time of the 2006 Census, 65% of people aged 15 years and over were employed.
Of the people employed at the time of the 2006 Census, 77% were in full-time work and the rest were in
part-time work.
Of the people employed full-time:

18 % had no qualification

33% had a school qualification as their highest qualification

49% had a post-school qualification as their highest qualification.
Of the people employed part-time:
(b)

20% had no qualification

43% had a school qualification as their highest qualification

37% had a post-school qualification as their highest qualification.
(i)
Calculate the proportion of people from the 2006 Census who were employed part-time with no
qualification.
(ii)
If two people were randomly selected from the 2006 Census, calculate the probability that both
were employed full-time. Justify any assumptions that you have made in your calculation of this
probability.
It was found that of a group of 120 people from Wellington:

32 are at least 65 years old

83 earn at least $50 000

17 are under 65 years old and earn under $50 000.
(i)
Calculate the probability that a randomly selected person from this group earns at least $50 000 and
is under 65 years old. Hence determine with reasoning if the events ‘a person from this group earns
at least $50 000’ and ‘a person from this group is under 65 years old’ are mutually exclusive.
It was also found that of this group:

98 have a post-school qualification

everyone is either at least 65 years old, or earns at least $50 000, or has a post-school qualification

19 are at least 65 years old but do not have a post-school qualification

4 are at least 65 years old, earn at least $50 000, and have a post-school qualification.
(ii)
Calculate the probability that a randomly selected person from this group has a post-school
qualification and earns at least $50 000. You should explain your reasoning in calculating your
answer.
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 3 of 8
QUESTION TWO
(a) The data below shows internet access by household type for the West Coast of the South Island.
Household of a
Household of a
privately
rented house
owned house
Access to
the internet
No access to
the internet
(b)
26.1 %
40.4 %
13.7 %
19.8 %
(i)
Consider the events ‘a household of a privately owned house’ and ‘a household has access to the
internet’. Explain whether these events are independent.
(ii)
Is the household of a privately owned house more likely to have access to the internet than the
household of a rented house? Support your answer with appropriate statistical statements.
The Newborn Metabolic Screening Programme screens newborn babies for different metabolic disorders.
Screening attempts to identify babies who have a metabolic disorder. These babies are then referred for
diagnostic testing to confirm or rule out having a metabolic disorder.
For a particular metabolic disorder:

screening identifies about 18 in 20 000 babies as being more likely than others to have this disorder

subsequent diagnostic testing (including retesting) finds around 10% of babies identified through
screening actually have this disorder

approximately 6 in 66 400 babies actually have this disorder.
If a baby actually has this disorder, what is the approximate risk of not being diagnosed with this disorder
through the screening process?
QUESTION THREE
Emma was given a set of six different keys to her new house. One of the keys opened the dead lock on the front
door, and a different key opened the door lock on the front door.
Emma did not know which keys were the correct keys for each lock, and was able to open both locks after four
key attempts.
Emma’s friend Sene thought this was a low number of key attempts, and wondered what process Emma used to
find the correct keys.
Sene designed and carried out a simulation to estimate how many key attempts it would take before both locks
were open, using a ‘trial and error’ process, to see if this might have been the process Emma used.
For the ‘trial and error’ process, Sene assumed:

that a key was selected at random to try to open the dead lock
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 4 of 8

once Emma had tried a key for the dead lock, she did not try it again

once Emma found the correct key for the dead lock, she removed this from the set of keys and tried the
same process with the door lock.
The results of Sene’s simulation are shown below:
Number of key
attempts before
both locks are open
Frequency
2
29
3
70
4
103
5
136
6
164
7
176
8
134
9
74
10
76
11
38
(a)
Calculate the theoretical probability of a person using a 'trial and error’ process taking more than two key
attempts before both locks are open. Compare this probability with the results from Sene’s simulation and
discuss any differences.
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 5 of 8
SAMPLE ASSESSMENT SCHEDULE
Mathematics and Statistics 91585 (3.13): Apply probability concepts in solving problems
Assessment Criteria
Achievement
Achievement with Merit
Apply probability concepts in solving problems involves:
Achievement with Excellence
 selecting and using methods
Apply probability concepts, using relational thinking, in
solving problems involves:
Apply probability concepts, using extended abstract
thinking, in solving problems involves:
 demonstrating knowledge of concepts and terms
 selecting and carrying out a logical sequence of steps
 communicating using appropriate representations.
 connecting different concepts or representations
 devising a strategy to investigate or solve a
problem
 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
 making a statistical generalisation
and also where appropriate, using contextual
knowledge to reflect on the answer.
Evidence Statement
One
Expected Coverage
P(employed) = 0.65
P(part-time/employed) = 0.23
Achievement
Merit
Candidate calculates
probability correctly.
P(no qualification/part-time/employed) = 0.2
(a) (i)
[equivalently probabilities are correctly drawn on a probability tree]
P(employed ∩ part-time ∩ no qualification)
= 0.65 x 0.23 x 0.2 = 0.030 (3 s.f.)
P(employed ∩ full-time)
(a) (ii)
= P(employed) x P(full-time/employed)
Candidate calculates
probability correctly
Candidate calculates
probability correctly
= 0.65 x 0.77 = 0.5005
AND
AND
P(both full-time) = 0.50052 = 0.251 (3 s.f.)
states assumption of
independence for both
people being employed
part-time.
explains why
assumption of
independence is
reasonable.
The assumption is that of independence of both people being employed parttime, which is reasonable given the very large number of people involved in
the census (the proportion of 0.5005 will not change significantly with one
person taken out).
Excellence
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 6 of 8
Student constructs two-way table:
At least
65
Under
65
Candidate calculates
probability correctly
Total
At least
$50 000
12
71
83
Under
$50 000
20
17
37
Total
32
88
120
[equivalently set notation or probability statements are used]
P(at least $50 000 ∩ under 65) = 71/120 = 0.592 (3 s.f.)
(b) (i)
and
(b) (ii)
As this is not zero, the two events are not mutually exclusive.
OR student constructs Venn diagram:
[equivalently other diagrams or probability statements are used]
P(at least $50 000 ∩ post-school qualification) = 72/120
AND
states the events are not
mutually exclusive for
part (i).
Candidate determines
the count / probability of
another combined event
not given in the question
(partial completion of
Venn diagram) for part
(ii).
Candidate calculates
the probability
correctly for part (ii).
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 7 of 8
Two
Expected Coverage
P(household of a privately owned house) = 0.398
P(access to internet) = 0.665
P(household of a privately owned house ∩ access to internet) = 0.261
(a) (i)
P(household of a privately owned house) x P(access to internet) = 0.398 x
0.645 = 0.265 (3 s.f.)
Accept alternatives, eg
P(B/A) ≠ P(B).
P(internet/household of a privately owned house)
= 0.261/0.398 = 0.656 (3 s.f.)
One conditional
probability correctly
calculated.
No, the household of a privately owned house is not more likely to have
access to the internet than the household of a rented house.
P(have disorder) = 6/66400
P(identified through screening and have disorder) =
18/20000 x 0.1 = 0.00009
(b)
P(not identified through screening and have disorder) = 6/66400 – 0.00009
P(not identified through screening/have disorder) =
(6/66400 – 0.00009)/(6/66400) = 0.004
Risk of not being diagnosed with this disorder through the screening process
is approximately 0.4%.
Merit
Excellence
Independence rule used
with correct probabilities
to determine events are
not independent.
Therefore the events are not independent as P(A) x P(B) ≠ P(A ∩ B)
P(internet/household of a rented house)
= 0.404/0.602 = 0.671 (3 s.f.)
(a) (ii)
Achievement
Both conditional
probabilities calculated
and compared to reach
conclusion.
Accept alternative
reasoning, eg relative
ratios.
Proportion of babies
identified through
screening and who have
this disorder correctly
calculated.
Proportion of babies not
identified through
screening and who have
this disorder correctly
calculated.
Approximate risk of not
being diagnosed with
this disorder through
the screening process
correctly calculated.
NCEA Level 3 Mathematics and Statistics 91585 (3.13) — page 8 of 8
Three
Expected Coverage
P(two key attempts) = 1/6 x 1/5 = 1/30
Both theoretical and
experimental
probabilities calculated
with comment on
similarity.
The central 90% of simulation results are between 3 and 10 key attempts, so
4 attempts is within this range. As the simulation results have a unimodal
distribution it is appropriate to use the central 90%.
Range for central 90%
calculated.
Range for central 90%
calculated with
discussion of whether
this supports
conclusion.
P(four attempts) = P(one attempt dead lock and three attempts door lock) +
P(two attempts dead lock and two attempts door lock) + P(three attempts
dead lock and one attempt door lock)
One possibility for
taking four key attempts
identified
All possibilities for
taking four key attempts
identified
AND
AND
one probability
calculated.
at least one probability
calculated.
= 29/30 (or 96.7%)
97.1% of the results from the simulation are above two key attempts, which is
similar to theoretical probability.
(b)
P(one attempt dead lock and three attempts door lock)
= 1/6 x 4/5 x ¾ x 1/3 = 1/30
(c)
Merit
Either theoretical or
experimental probability
calculated.
P(more than two key attempts) = 1 – 1/30
(a)
Achievement
P(two attempts dead lock and two attempts door lock)
= 5/6 x 1/5 x 4/5 x ¼ = 1/30
P(three attempts dead lock and one attempt door lock)
= 5/6 x 4/5 x ¼ x 1/5 = 1/30
P(four attempts) = 1/10
Excellence
Probability of
taking four key
attempts
calculated.