NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 1 of 7
Assessment Schedule – 13MAS PAKURANGA COLLEGE 2015 Common Test (August)
Statistics and Modelling: Apply probability distributions in solving problems (91586)
Achievement
Apply probability distributions in solving problems involves:
 selecting and using methods
 demonstrating knowledge of concepts and terms
 communicating using appropriate representations.
Achievement with Merit
Apply probability distributions, using relational thinking, in solving problems involves:
 selecting and carrying out a logical sequence of steps
 connecting different concepts or representations
 demonstrating understanding of concepts and also relating findings to a context or communicating thinking using appropriate statements.
Achievement with Excellence
Apply probability distributions, using extended abstract thinking, in solving problems involves:
 devising a strategy to investigate or solve a problem
 identifying relevant concepts in context
 developing a chain of logical reasoning and also, where appropriate, using contextual knowledge to reflect on the answer.
ONE
(a)
(b)
Expected Coverage
Binomial Distribution
 fixed number of trials (8 movies in a day)
 fixed probability (30% of DVDs are romantic comedies)
 two outcomes (romantic comedy, not a romantic comedy)
 independence of events (someone handing in a romantic comedy one day does not affect another romantic comedy being handed in each day).
Assumption is that the videos are not rented in groups so that the genres remain independent when being handed in (or some such statement) n = 8 p = = 0.3
P(X = 3) = 0.2541
Binomial distribution
Only boxes with two or fewer pre-1990 movies are rented. n = 5, p = 0.28
P(X ≤ 2) = 0.8624
Achievement
Probability calculated with identification of probability distribution and parameters.
Probability correctly calculated.
Merit
Probability calculated with identification of probability distribution and parameter
AND justification of applying this distribution linked to the context.
Probability calculated
AND
States at least one assumption being made
Excellence

(c)
TWO
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 2 of 7
P(a randomly selected box is rented) = 0.8624
Assuming boxes are independent of each other and a box set having a 90s film does not affect another box set having a 90s film.
Model can be triangular or taken directly from the graph.
Triangular with continuity correction P( x >2.5)
0 2.5 3 8
A = 1 h = 2/8 = 0.25
Area of Red triangle = 0.260412
Let w be the number of movies rented
P( w < 2.5) = 0.260412
P( w > 2.5) = 0.739583
Probability that one customer rents 3 or more movies is 0.7396
Let x be number of movies rented.
Binomial n = 5, p = 0.7396
, x ≥ 2
P( x ≥ 2) = 1 – P( x ≤ 1) = 1 – 0.018205 = 0.9818
Probability of at least 2 people in the next 5 customers renting 3 or more movies.
OR
Model direct from distribution. Height of every column read from the graph and proportion of total calculated for 3 or more movies, then binomial as above.
Eg (240 + 200 + 165 + 120 + 80 + 35)/ (240 + 200 + 165 + 120 + 80 + 35 +
150 + 70 + 5) = 0.7887
Accept estimates between 0.75 and 0.80.
Attempts to model distribution as triangular or by reading from graph, using a correct method but with errors, resulting in an incorrect probability between 0.5 and 1 for a person renting 3 or more movies.
Solves problem using triangular model without continuity correction.
OR
Correctly calculates probability of one customer renting 3 or more movies.
OR
CAO
Correct answer supported by clear reasoning.
Expected Coverage Achievement Merit Excellence

(a) (i)
(a) (ii)
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 3 of 7
Normal distribution, µ = 72.3, σ = 8.9
P(between 40 and 75) = P(40 < X < 75) = 0.61905
61.91% of games have mixed reviews.
Accept any rounding
Normal distribution, µ = 72.3, σ = 8.9
P(X > 85) = 0.07679
P(favourable review) = P(X > 75) = 0.3808
P(over 85 / favourable review) =
P(X > 85) / P(X > 75) = 0.07679 / 0.3808
= 0.2017
Accept any rounding in the final answer.
Probability correctly calculated (proportion or percentage)
P(X > 85) or P(X > 75) correctly calculated.
Conditional probability correctly calculated.

(b)
NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 4 of 7
Using the no rmal distribution, with µ = 8.5 mins and σ = 3.2 mins
P(X < 5) = 0.1370
P(5 < X < 10) = 0.5433
P(X > 10) = 0.3196
Time spent on
Beta Critic’s
Website
T < 5 5 < T < 10 T > 10
Amount Paid
Probability
E( X ) = 0.1296 = $0.13
$0.05
0.1370
SD( X ) = 0.04159 = $0.04
$0.12
0.5433
$0.18
0.3196
The expected amount that Beta Critic will be paid per day:
5000 x 0.1296 = $648 (accept $650)
The standard deviation of the amount Beta Critic is expected to be paid per day:
√5000 × 0.04159 = $2.94 (accept $2.83 or other rounded answers).
(assuming independence of website visitors; all visitors are unique visitors)
The expected amount that Beta Critic will be paid per month:
E( T ) $648 x 30 = $19440
SD( T) √30 × 2.94 = $16
This value is about $500 less than Beta Critic’s target of $20 000 per month.
This difference of $560 is over 3 SD away from the expected earnings therefore it appears to be an unreasonable monthly target for Beta Critic.
However, Beta Critic should consider other factors such as the fact that the
5000 visitors is only an estimate, or that they should expect their website traffic to increase considering their website is new or the fact that this calculation is based on 30 days and some months have 31. Because of these other sources of variation, $20 000 might be a reasonable target to aim for.
Probability distribution table (or equivalent probability statements) is developed for the distribution of amount paid for time spent on
Beta Critic’s website.
Correct calculation of expected value per month
OR correct calculation of the expected value for the total amount paid per day and the use of the expected value and relevant justification to determine whether the monthly target is appropriate
Calculates the expected value of earnings per month, supported by some working.
AND
Comments on the suitability for the target
AND addressing other contextual factors that contribute to the suitability of the target.

NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 5 of 7
THREE
(a) (i)
Expected Coverage
Number of profanities per hour
Frequency
0
20
1
28
2
40
3
35
4
22
5
10
6 or more
5
Assuming that most of those in the 6 or more category are exactly 6.
E(X) = 2.38 profanities
Accept other values based on some of the 6 or more categories assumed to be 7 or 8 eg 2.4
Poisson distribution, λ = 2.38
Applying this distribution because:
 discrete (profanities) within a continuous interval (time)
 events cannot occur simultaneously (only one profanity occurs at a time)
 events are random with no set pattern (profanities occur randomly within each episode)
 for an interval (hour) the mean number of occurrences (profanities) is proportional to the size of the interval.
We make the assumption that profanities are independent of each other.
Poisson distribution, λ = 2.38
P(X = 6) = 0.023361
0.023361×160 hours = 3.738 hrs
Achievement
Correct calculation of expected number of profanities per episode.
AND
Identification of Poisson
Distribution with value for λ.
Probability correctly calculated or 3.738 hours stated.
Merit
Justification of applying this distribution linked to the context.
AND recognizes independence of occurence of profanity
(a) (ii)
Excellence

NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 6 of 7
(b) (i)
Poisson distribution
λ = 2.38
P(x ≤ 2) = 0.5749
14 hours in a week between 8:30 and 10 :30pm
14 × 0.5749 = 8.0486
I estimate there will be 8 hours in a week between 8:30 and 10:30 with no more than two profanities per hour.
Accept 8 hours and 3 minutes.
Calculates correct probability
(b) (ii)
Inverse Poisson distribution, x is the number of profanities per hour.
P( x = 0) = 7/50 = 0.14
λ = -ln(0.14)
λ = 1.966 expected mean = 1.966 profanities per hour
The average number of profanities per hour in the sample is less than 2.
50 is a very small sample to get information about a distribution. The difference between the sample proportions before and after the campaign are consistent with chance variability due entirely to sampling variation. Juan
Quay should not claim that his campaign has been a success based on this sample alone.
NØ No response; no relevant evidence.
Not Achieved N1 Candidate gives a partial solution to ONE part of the question.
Achievement
Merit
Excellence
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.
Estimates number of hours, answer sensibly rounded.
Calculates average profanities per hour.
Interprets average profanities per hour in terms of the stated target, including an assessment of the findings in terms of the small sample size and sampling variability.

NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 7 of 7
Cut Scores
0 – 8 9 – 13
14 – 19
20 – 24