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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