NCEA Level 3 Mathematics and Statistics 91586 (3.14) — page 1 of 5

SAMPLE ASSESSMENT SCHEDULE

Mathematics and Statistics 91586 (3.14): Apply probability distributions in solving problems

Assessment Criteria

Achievement / Achievement with Merit / Achievement with Excellence
Apply probability distributions in solving problems involves:
  • selecting and using methods
  • demonstrating knowledge of concepts and terms
  • communicating using appropriate representations.
/ 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. / 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.

Evidence Statement

One / Expected Coverage / Achievement / Merit / Excellence
(a) (i) / Normal distribution, µ = 215, σ = 13.2
P(X < 200) = 0.1279
P(X > 220) = 0.3524
P(less than 200 g or 220 g and over) = 0.4803 / Probability correctly calculated.
(a) (ii) / Normal distribution, µ = 215, σ = 13.2
P(X < 220) = 0.64757
P(205 < X < 220) = 0.42322
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.
(b) / Distribution of weights of pineapples
Normal distribution, µ = 1.3, σ = 0.234
Distribution of total weight of 5 000 pineapples
Normal distribution, µ = 6500
σ = √5000 x 0.234 = 16.5463
Distribution of income
Normal distribution, µ = 17550
σ = 2.7 x 16.5463 = 44.68
Since the expected value for the income if $17 550, with a standard deviation of only approximately $50, it is reasonable that the grower could gainan income close enough to $17 500 (the coefficient of variation is only 0.3%).
Alternatively
P(X > 17 500) = 0.8684
So it is reasonable to expect an income of at least $17 500 as this is a very high probability.
OR
Range for central 95% of incomes
$17 462 < Income < $17 638
So it is reasonable to expect an income of $17 500 as this amount lies within the range for the central 95% of incomes. / Calculation of variance or standard deviation for total weight of pineapples.
OR
Calculation of variance or standard deviation for the income using incorrect calculation of variance or standard deviation for total weight of pineapples. / Calculation of variance or standard deviation for the income. / The variation of the pineapple weights is taken into account when determining whether the grower’s expectations are reasonable, either through considering the practical significance of the standard deviation of the income, the probability of obtaining an income at least as high as what is desired, or the range for the central 95% of incomes.
Not Achieved / NØ / No response; no relevant evidence.
N1 / Candidate gives a partial solution to ONE part of the question.
N2 / Candidate gives partial solutionsto TWO parts of the question.
Achievement / A3 / Candidate gives ONE opportunity from the Achievement criteria.
A4 / Candidate gives TWO opportunities from the Achievement criteria.
Merit / M5 / Candidate gives ONE opportunity from the Merit criteria.
M6 / Candidate gives TWO opportunities from the Merit criteria.
Excellence / E7 / Candidate meets the Excellence criteria except for minor errors in calculation.
E8 / Candidate meets the Excellence criteria.
Two / Expected Coverage / Achievement / Merit / Excellence
(a) / For 15 minute interval
E(X) = 0 x 0.3 + 1 x 0.35 + 2 x 0.2 + 3 x 0.15 = 1.2 / Correct calculation of expected number of jumpers per 15-minute interval.
(b) (i) / Poisson distribution, λ = 2.4
Applying this distribution because:
  • 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.
Assumption is that each queue jumper is independent of other queue jumpers.
P(X ≤ 1) = 0.3084 / 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.
(b) (ii) / Poisson distribution
P(X = 0) = 0.9
e-λ = 0.9
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’. / 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.
Not Achieved / NØ / No response; no relevant evidence.
N1 / Candidate gives a partial solution to ONE part of the question.
N2 / Candidate gives partial solutionsto TWO parts of the question.
Achievement / A3 / Candidate gives ONE opportunity from the Achievement criteria.
A4 / Candidate gives TWO opportunities from the Achievement criteria.
Merit / M5 / Candidate gives ONE opportunity from the Merit criteria.
M6 / Candidate gives TWO opportunities from the Merit criteria.
Excellence / E7 / Candidate meets the Excellence criteria except for minor errors in calculation.
E8 / Candidate meets the Excellence criteria.
Three / Expected Coverage / Achievement / Merit / Excellence
(a) / Binomial distribution
n = 4, p = 0.67
Applying this distribution because:
  • 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).
Assumption is that the bees visit all the plants.
P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.4015 = 0.5985 / Probability calculated with identification of probability distribution and parameters. / Probability calculated with identification of probability distribution and parameter
AND
justification of applying this distribution linked to the context.
(b) / Binomial distribution
n = 12, p = 0.91
P(X = 11) = 0.3827
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 not affect the chances of another plant being pollinated or not in the same season/year. / Probability calculated for one season. / Probability calculated for both seasons with at least one assumption of independence stated.
(c) / Bee pollination
For n = 90, shape of distribution of number of plants pollinated appears to be normal, with estimates for mean of 60.3 and standard deviation of 4.46
Hand pollination
For n = 60, shape of distribution of number of plants pollinated appears to be normal, with estimates for mean of 54.6 and standard deviation of 2.22
The total number of plants can be modelled by the Normal distribution, assuming independence of the two variables (eg bees do not have access to plants pollinated by hand), with µ = 114.9 σ = √(4.462 + 2.222) = 4.98
P(X > 99.5) = 0.999 [using continuity correction] OR
Range for central 95%, 105 < T < 124
So it is very likely that the grower will fulfil his contract using this combination of methods.
Note: Mean and standard deviation for normal distributions calculated using np and √npq respectively. Reasonable estimates of these parameters from the graphs could also be accepted. / Mean and standard deviations of each distribution calculated
OR
reasonably estimated from the graphs. / Normal distribution selected and justified as model by referring to shape of distributions
AND
mean and standard deviation of total number of plants pollinated calculated. / Use of Normal distribution to calculate probability of 100 plants (or central 95% of number of plants), with model selection justified with reference to the shape of the distributions and the assumption of independence
AND
a conclusion made that high likelihood of contract being fulfilled.
Not Achieved / NØ / No response; no relevant evidence.
N1 / Candidate gives a partial solution to ONE part of the question.
N2 / Candidate gives partial solutionsto TWO parts of the question.
Achievement / A3 / Candidate gives ONE opportunity from the Achievement criteria.
A4 / Candidate gives TWO opportunities from the Achievement criteria.
Merit / M5 / Candidate gives ONE opportunity from the Merit criteria.
M6 / Candidate gives TWO opportunities from the Merit criteria.
Excellence / E7 / Candidate meets the Excellence criteria except for minor errors in calculation.
E8 / Candidate meets the Excellence criteria.