Apply probability Apply probability methods, using methods, using probability extended abstract relational thinking, in methods in solving thinking, in solving solving problems. problems. problems. 1a(i) 0 < < 1.5 Correct probability z Probability = 0.43319 calculated. 1a(ii) Correct probability > 2.2 z calculated and Probability = 0.013903 answer written as 1.39 percent percentage. 1a(iii) < - 1 z Correct probability Expected number Probability = 0.15865 calculated. calculated. Expected number 1269 (or 1270) 1a(iv) = - 1.645 z Correct weight kg calculated. 25% either side of mean 1a(v) Inverse normal Limits of weights Correct IQR ie = z value of z found. found. specified. Between 1.0325 kg and 1.1 674 kg ie IQR = 0.1349 kg For 1% when mean is 1.1 = 0.86736 x 1b(ii) Found the value New mean with Complete statement x So have to increase mean to 1.23264 kg for 1% less than. insufficient (new mean OR (the difference between 1 and 0.86736) explanation / increase old mean OR justification. by 0.1326 kg) = - 2.326 and use z completely and correctly justified. Progress shown in 1b(iii) New Std dev = 0.042925 SD calculated. finding new SD. Second choice of changing the SD. 1b(iii) Statement with Complete statement choice has mean packet weight of incomplete completely and 1.23264 kg and the second choice has a justification. correctly justified. 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. 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 Achievement Evidence Merit (r) Excellence (t) TWO (u)
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)
2012_eoy_level_2_probability_exam.docx