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ASSESSMENT SCHEDULE – BURGER PRINCE Kohia 2014 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: 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 devising a strategy to investigate or solve a demonstrating knowledge of concepts and terms communicating using appropriate representations. steps problem connecting different concepts or representations identifying relevant concepts in context demonstrating understanding of concepts developing a chain of logical reasoning and also relating findings to a context or communicating thinking using appropriate statements. and also, where appropriate, using contextual knowledge to reflect on the answer. Evidence Statement ONE (a) Expected Coverage Bacteria levels 0 (or negligible) 10 20 30 40 Over 45 Frequency 4 72 13 0 1 0 Probability calculated with identification of probability distribution and parameters. Probability calculated with identification of probability distribution and parameter justification of applying this distribution linked to the context. Poisson distribution, λ = 11.3 Appropriate distribution to model is a Poisson distribution. Applying this distribution because: AS91586 Kohia 2014 Merit AND Using the given midpoints of intervals and reading negligible as 0. E(X) = 11.33 Accept other values based on the categories and sensible approximations. Achievement discrete (bacteria colony count) within a continuous interval (area) 1 Excellence events cannot occur simultaneously (only one count occurs at a time) events are random with no set pattern (bacteria count changes randomly between each sample) for an interval (area) the mean number of occurrences (bacterial count) is proportional to the size of the interval. I am making the assumption that the samples are independent of each other, that bacteria count of one swab will not influence the bacteria count of the subsequent swab. Poisson λ = 11.3 (b) Probability correctly calculated. P( will fail the requirements) = P ( X > 15 ) =1 – P(x ≤ 15) = 1 – 0.8905 = 0.1095 ≈ 0.11 Poisson λ = 11.3 Probability of being below 8 colonies/cm2 = (X ≤ 7) = 0.1249 Number of trials = 7 days of daily swabbing (c) Correctly calculates probability for colony count below 8. Correctly calculates probability for colony count below 8 CAO AND Identifies and forms new binomial distribution New Binomial distribution with p = 0.1249 and p = number of trials = 7. Y ~ B (n = 7, p = 0.45296) P (Y ≥ 6) = 1 – P (Y ≤ 5) = 1 – 0.99997 = 0.00003 The restaurant will almost certainly fail to meet the new requirements at least two days a week. N N1 N2 A3 A4 M5 M6 E7 E8 No relevant evidence. Making progress. 1 of u 2 of u 3 of u 1 of r 2 of r 1 t with minor errors 1t AS91586 Kohia 2014 2 Correct answer supported by clear reasoning. TWO Expected Coverage Achievement Normal distribution, µ = 15.67, σ = 7.25 Excellence Both high and low probabilities correctly calculated P(Low spender) = P (X < 5.25) = 0.075324 = 0.075 (a) (i) Merit P(High spender) = P (X > 20.45) = 0.25484 = 0.25 CAO Percentage(either or high or low) = 32.5% Accept any rounding Normal distribution, µ = 15.67, σ = 7.25 Inverse normal calculation correct. Using inverse normal calculations P (X > x) = 0.10 (a) (ii) X = 24.961 ≈ $25.00 The customers would be expected to spend $25 and over to get into the “Super High” category. Accept any correct rounding in the final answer. (b) Normal distribution is an appropriate model because: Shape is approximately symmetrical Most values in the centre, with fewer towards the edges. The variable being measured (amount spent per customer) is continuous The histogram of the customer spend provides evidence the target has been met at the peak of the graph is at approximately $22.50 which is over the target. Given the pre-campaign mean was at $15.67, the new approximate mean is now 1 standard deviation (σ = 7.25) higher. The shape of the graph is a bit asymmetrical, which suggests that the mean may be slightly less than $22.50 so target may not have been met, but mean will be close to target, so success of new campaign has been demonstrated. As the customers were randomly selected and the sample size is also adequate (n = 300), the findings from this sample can be seen as reliable. The target appears to have been met given the preliminary data. AS91586 Kohia 2014 3 One normal distribution features correctly identified Two normal distribution features correctly identified Two normal distribution features correctly identified AND AND AND new higher mean identified as source of success new higher mean identified as source of success with statistical justification. new higher mean identified as source of success with statistical justification Accept any sensible explanations AND Reference is made to 2 of: sampling method/sample size/sampling variability. The assumption is that the histogram follows a Normal distribution. Students may assume the mean is $22.50 and SD is still $7.25 or estimate mean and SD from the graph [Actual mean = $20.50, SD $6.00] Accept any sensible estimate. Standardising the lower end of category, $10.04: Z= 𝑥− µ σ = 10.04−15.67 7.25 Assumption of Normal distribution is made with normal distribution parameters identified. AND = -0.77655 Sensible attempt made to calculate new range for Medium spenders Standardising the upper end of category, $20.45: Z= (b) (i) 𝑥− µ σ = 20.45−15.67 7.25 = 0.6593 With the new distribution and solving for x, the lower end will be: 𝑥− 22.50 7.25 Assumption of Normal distribution is made with normal distribution parameters identified = -0.77655 => x = $18.61 The upper end will be: 𝑥− 22.50 7.25 = 0.6593 => x = $25.80 The new category for a Medium spender after the marketing campaign will lie between $18.61 and 25.80. Allow for rounding. N N1 N2 A3 A4 M5 M6 E7 E8 No relevant evidence. Making progress. 1 of u 2 of u 3 of u 1 of r 2 of r 1 of t 2 of t AS91586 Kohia 2014 4 Assumption of Normal distribution is made with normal distribution parameters identified AND Correct calculation of new range for Medium spenders Achievement Probability 7 Expected Coverage 1 THREE Merit Correct graph sketched Correct graph sketched OR AND correct distribution identified. correct distribution identified. (a) (i) 7 days The probability distribution is modelled by a Uniform Distribution. If time measured in hours vertical height is 1/168 3 (a) (ii) Correct probability identified. P( Inspector arrives between 2pm and 5pm) = 168 CAO ID card in original drawer 0 1 2 3 P( X = x) 2 6 3 6 0 1 6 E(X) = 1 and SD (X) = 1 (b) (i) Total number of ways to place ID hooks= 6 Correct probability distribution table drawn OR AND correct E(X) or SD(X). correct E(X) and SD(X). CAO Let the drawers be 1, 2 and 3 No original drawer = 2,3,1 and 3,1,2 One original drawer = 1,3,2 or 2,1,3 or 3,2,1 If two in original then will be three in original drawer. Three original drawer – 1,2,3 AS91586 Kohia 2014 Correct probability distribution table drawn 5 Excellence Assumption is the independence of the rankings between the inspectors. There should be no ‘rivalry’ nor collusion and they should not be aware of the rankings of the each other. Expected value correctly calculated. CAO Let the first inspector have distribution X and the second have distribution Y. Average of the two inspectors expected rankings = = (b) (ii) 4 + 3.5 2 E(X+ Y) 2 = E(X) + E(Y) 2 Expected value correctly calculated AND Expected value and Standard deviation correctly calculated EITHER AND Standard deviation correctly calculated assumption of independence stated in context OR = 3.75 assumption of independence stated in context As SD(X) = 1.2 and SD(Y) = 2.3: VAR(X) = 1.22 = 1.44 AND VAR(Y) = 2.32 = 5.29 Average of the two inspectors standard deviations rankings: 1 2 1 4 VAR( (X + Y) ) = [ VAR(X) + VAR(Y)] = 1 1 4 1 4 [1.44 + 5.29] = [6.73] 1 SD(½ (X + Y)) = √4 [6.73] = 2 x √6.73 = 1.297 = 1.3 The average ranking of the two inspectors is 5.29 with a SD of 1.3. N N1 N2 A3 A4 M5 M6 E7 E8 No relevant evidence. Making progress. 1 of u 2 of u 3 of u 1 of r 2 of r 1 t with minor errors 1 of t Judgement Statement Score range AS91586 Kohia 2014 Not Achieved Achievement Achievement with Merit Achievement with Excellence 0–7 8 – 12 13 – 18 19 – 24 6