ASSESSMENT SCHEDULE 91267 Apply probability methods in solving problems Achievement Achievement with Merit Achievement with Excellence Apply probability methods in solving problems involves: • selecting and using methods • demonstrating knowledge of probability concepts and terms • communicating using appropriate representations. . probability methods, using relational thinking, in solving problems involves one or more of: • 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 mathematical statements. Apply probability methods, using extended abstract thinking, in solving problems involves one or more of: • devising a strategy to investigate or solve a problem • identifying relevant concepts in context • developing a chain of logical reasoning • making a statistical generalisation; and also, appropriate, using contextual knowledge to reflect on the answer. Sufficiency for each question: N0: no relevant evidence N1: Attempt at one question N2: 1 u A3: 2 u OR A3: 1 r + 1 u A4: 3 u OR A4: 1 r + 2 u M5: 2 r M6: 3 r E7: 1 t E8: 2 t Judgement Statement NOTE: This is an estimate as no examinations have yet been held. Not Achieved Achievement Achievement with Merit Achievement with Excellence Score range 0 - 8 9 - 13 14 - 18 19 -24 Question One Evidence Achievement (u) Merit (r) Excellence (t)
probability methods in solving problems. Apply probability methods, using relational thinking, in solving problems. Apply probability methods, using extended abstract thinking, in solving problems. 1a(i) 0 < z < 1.5 Probability = 0.43319 Correct probability calculated. 1a(ii) z > 2.2 Probability = 0.013903 1.39 percent Correct probability calculated and answer written as percentage. 1a(iii) z < - 1 Probability = 0.15865 Expected number 1269 (or 1270) Correct probability calculated. Expected number calculated. 1a(iv) z = - 1.645 kg Correct weight calculated. 1a(v) 25% either side of mean ie z = Between 1.0325 kg and 1.1 674 kg ie IQR = 0.1349 kg Inverse normal value of z found. Limits of weights found. Correct IQR specified. 1b(ii) For 1% when mean is 1.1 x = 0.86736 So have to increase mean to 1.23264 kg (the difference between 1 and 0.86736) OR z = - 2.326 and use Found the x value for 1% less than. New mean with insufficient explanation / justification. Complete statement (new mean OR increase old mean by 0.1326 kg) completely and correctly justified. 1b(iii) New Std dev = 0.042925 Progress shown in finding new SD. SD calculated. 1b(iii) Second choice of changing the SD. choice has mean packet weight of 1.23264 kg and the second choice has a mean packet weight of 1.1 kg. Thus the first choice would result in an increase in mean of 0.13264 kg per packet or 12 % more peas being used – not a good option for the company. Statement with incomplete justification. Complete statement completely and correctly justified. NOTE: There will be differences between probability answers quoted if students use normal distribution tables rather than GC. Either should be accepted. Values used here are all GC values Question TWO Evidence Achievement (u) Merit (r) Excellence (t)
Apply probability methods in solving problems. Apply probability methods, using relational thinking, in solving problems. Apply probability methods, using extended abstract thinking, in solving problems. 2a(i) [ 0.654] Correct probability calculated. 2a(ii) [ 0.332] Correct probability calculated. 2a(iii) [ 0.545] = 54.5% Correct percentage calculated. 2a(iv) = 1336 (or 1337) Correct probability calculated. Expected number calculated. 2a(v) Using GONE: 0.668 have improved dandruff. Using OFF : 0.635 have improved dandruff. I use GONE as the chance of improved dandruff is slightly greater. OR There is very little difference in the two results and it may depend on the price or aroma as to my choice. Comparison made between GONE and OFF. Correct conclusion AND supported by calculations. 2b(i) Prob required = 0.843 Correct table or numbers calculated. Correct probability calculated. 2b(ii) Prob (dandruff of a male’s dandruff improving using GONE) = 0.783 Prob (dandruff of a male’s dandruff improving OFF) = 0.4 So relative risk of dandruff improving using GONE is 1.96 so company claim is close to being correct. However sample using OFF was very small and may have been misleading and possibly changed a lot with a bigger sample. Perhaps further study with a larger should be undertaken before results used. A correct probability calculated. Both correct probabilities calculated. Probabilities correctly compared and a justified statement made.
2c The distribution of secondary school students is NOT close to normal. It is closer to be a uniform distribution with very much the same numbers at year 9, 10 and 11 with a small decrease in year 12 and year 13. No modal central value. An approximate sketch graph would be useful. Statement correct with some justification. Statement correct with full justification. Question Evidence Achievement (u) Merit (r) Excellence (t) Apply probability methods in solving problems. Apply probability methods, using relational thinking, in solving problems. Apply probability methods, using extended abstract thinking, in solving problems. 3a 0.55 Correct probability calculated. 3b 0 Correct probability calculated. 3c Correct probability calculated. 3d P (Zoe won in next 3 points) = = 0.686125 Reqd Prob = = 0.117 A correct probability calculated. Correct probability for Zoe winning in next 3 points calculated. Correct probability calculated. 3e P (Zoe won on first point) = .55 P (Zoe won on third point) = = 0.136125 Reqd prob = = 0.802 A correct probability calculated. Total probability calculated for a faulty part. Correct probability calculated. 3f P( BZB+ BBZ) = + = A correct probability calculated. Correct total probability calculated. 3g Correct probability written.