1. Sample Assessment Probability

2. Question 1a(i) Highlight important information (words) as you read

3. Question 1a(i) At the time of the 2006 Census, 65% of people aged 15 years and over were employed. Of the people employed at the time of the 2006 Census, 77% were in full-time work and the rest were in part-time work.

4. 1a(i) Of the people employed full-time: 18 % had no qualification 33% had a school qualification as their highest qualification 49% had a post-school qualification as their highest qualification.

5. Of the people employed part-time: 20% had no qualification 43% had a school qualification as their highest qualification 37% had a post-school qualification as their highest qualification. Calculate the proportion of people from the 2006 Census who were employed part-time with no qualification.

8. 65% of people (aged 15 years and over) were employed 77% were in full-time work

9. Achieved answer Assumption is that of independence

10. Merit answer Assumption is that of independence. This is reasonable as the people were selected randomly and a census involves a large number of people and hence the proportion would not change significantly when one person is taken out.

11. Question 1(b) (i)

12. 2-way table group of 120

13. 32 are at least 65 years old

14. 83 earn at least $50 000

15. 17 are under 65 years old and earn under $50 000.

16. Finish table

17. 71/120 person from this group earns at least $50 000 and is under 65 years old

18. Achieved answer For events to be mutually exclusive As this is not zero, the two events are NOT mutually exclusive

19. It was also found that of this group: 98 have a post-school qualification everyone is either at least 65 years old, or earns at least $50 000, or has a post-school qualification 19 are at least 65 years old but do not have a post-school qualification 4 are at least 65 years old, earn at least $50 000, and have a post-school qualification.

20. Calculate the probability that a randomly selected person from this group has a post- school qualification and earns at least $50 000. You should explain your reasoning in calculating your answer.

21. Venn Diagram a c db

22. 32 are at least 65 years old ≥ 65 years old 32 – (a + b+ c)

23. 83 earn at least $50 000 ≥ 65 years old ≥ $50 000 83-(a+c+d)

24. everyone is either at least 65 years old, or earns at least $50 000, or has a post-school qualification 0 ≥ 65 years old ≥ $50 000 Post-school qualification

25. 17 are under 65 years old and earn under $50 000 0 ≥ 65 years old ≥ $50 000 17

26. 98 have a post-school qualification 98 – 17 = a + b + d = 81 0 ≥ 65 years old ≥ $50 000 17

27. 19 are at least 65 years old but do not have a post-school qualification 32 – (a + b) = 19  (a + b) = 13 0 ≥ 65 years old ≥ $50 000 17 68

28. 4 are at least 65 years old, earn at least $50 000, and have a post-school qualification 0 ≥ 65 years old ≥ $50 000 17 68 4

29. 0 ≥ 65 years old ≥ $50 000 17 68 4 9

30. Adding everything = 120, so b = 8 0 ≥ 65 years old ≥ $50 000 17 68 4 9 19 - b 11 - b b

31. Adding everything = 120, so b = 8 (not needed) 0 ≥ 65 years old ≥ $50 000 17 68 4 9 11 3 8

32. Excellence

33. Merit: Candidate determines the count/probability of another event not given in the question (partial completion of Venn diagram) for part (ii)

34. Question 2(a) The data below shows internet access by household type for the West Coast of the South Island. Consider the events ‘a household of a privately owned house’ and ‘a household has access to the internet’. Explain whether these events are independent.

36. Add the totals needed

37. P(household of a privately owned house) = 0.398 P(access to internet) = 0.665 P(household of a privately owned house ∩ access to internet) = 0.261 P(household of a privately owned house) x P(access to internet) = 0.398 x 0.645 = 0.265 (3 s.f.) Therefore the events are not independent as P(A) x P(B) ≠ P(A ∩ B)

38. Is the household of a privately owned house more likely to have access to the internet than the household of a rented house? Support your answer with appropriate statistical statements.

39. P(internet/household of a privately owned house) = 0.261/0.398 = 0.656 (3 s.f.) P(internet/household of a rented house) = 0.404/0.602 = 0.671 (3 s.f.) No, the household of a privately owned house is not more likely to have access to the internet than the household of a rented house.

40. Question 2(b) The Newborn Metabolic Screening Programme screens newborn babies for different metabolic disorders. Screening attempts to identify babies who have a metabolic disorder. These babies are then referred for diagnostic testing to confirm or rule out having a metabolic disorder.

41. For a particular metabolic disorder: screening identifies about 18 in 20 000 babies as being more likely than others to have this disorder subsequent diagnostic testing (including retesting) finds around 10% of babies identified through screening actually have this disorder approximately 6 in 66 400 babies actually have this disorder.

42. Start here For a particular metabolic disorder: screening identifies about 18 in 20 000 babies as being more likely than others to have this disorder subsequent diagnostic testing (including retesting) finds around 10% of babies identified through screening actually have this disorder approximately 6 in 66 400 babies actually have this disorder.

43. A table may be useful Identified through screening Have Disorder 6/66400 Do not have disorder approximately 6 in 66 400 babies actually have this disorder

44. screening identifies about 18 in 20 000 babies as being more likely than others to have this disorder Identified through screening Have Disorder 6/66400 Do not have disorder 18/20 000

45. subsequent diagnostic testing (including retesting) finds around 10% of babies identified through screening actually have this disorder Identified through screening Have Disorder 1.8/20 0006/66400 Do not have disorder 18/20 0001

46. If a baby actually has this disorder, what is the approximate risk of not being diagnosed with this disorder through the screening process? Identified through screening Have Disorder 1.8/20 0006/66400 Do not have disorder 18/20 0001

47. If a baby actually has this disorder, what is the approximate risk of not being diagnosed with this disorder through the screening process? Identified through screening Not identified through screening Have Disorder 1.8/20 0006/66400 - 18/200000 =3/8300000 6/66400 Do not have disorder 18/20 0001

48. If a baby actually has this disorder, what is the approximate risk of not being diagnosed with this disorder through the screening process? Identified through screening Not identified through screening Have Disorder 1.8/20 0006/66400 - 18/200000 =3/8300000 6/66400 Do not have disorder 18/20 0001

49. Question 3 Emma was given a set of six different keys to her new house. One of the keys opened the dead lock on the front door, and a different key opened the door lock on the front door. Emma did not know which keys were the correct keys for each lock, and was able to open both locks after four key attempts.

50. Emma’s friend Sene thought this was a low number of key attempts, and wondered what process Emma used to find the correct keys. Sene designed and carried out a simulation to estimate how many key attempts it would take before both locks were open, using a ‘trial and error’ process, to see if this might have been the process Emma used.

51. For the ‘trial and error’ process, Sene assumed (VERY IMPORTANT): that a key was selected at random to try to open the dead lock once Emma had tried a key for the dead lock, she did not try it again once Emma found the correct key for the dead lock, she removed this from the set of keys and tried the same process with the door lock.

52. The results of Sene’s simulation are shown below:

53. Calculate the theoretical probability of a person using a 'trial and error’ process taking more than two key attempts before both locks are open. Compare this probability with the results from Sene’s simulation and discuss any differences.

54. To be successful on only 2 attempts (non-replacement):

55. Theoretical probability To be successful on only 2 attempts (non-replacement): More than 2 attempts

56. Comparing this probability Theoretical probability Experimental probability: Conclusion: The results from the simulation are similar to the theoretical value.

57. Sene used the central 90% of her simulation results to check if the number of key attempts Emma took (four) was likely if the ‘trial and error’ process was used, and concluded it was. Discuss if the results of Sene’s simulation support this conclusion.

58. The central 90% (i.e. 900 out of the 1000) of simulation results are between 3 and 10 key attempts, so 4 attempts is within this range. As the simulation results have a unimodal distribution it is appropriate to use the central 90%.

59. Calculate the theoretical probability that a person using Emma’s process takes four key attempts before both locks are open.

60. List the possibilities first 1 attempt on the dead lock and 3 attempts on the door lock 2 attempts on the dead lock and 2 attempts on the door lock 3 attempts on the dead lock and 1 attempt on the door lock

61. Calculate the probabilities 1st attempt on the dead lock followed by 3rd attempt on the door lock 2nd attempt on the dead lock followed by 2nd attempt on the door lock 3rd attempt on the dead lock followed by 1st attempt on the door lock

62. Calculate the probabilities 1st attempt on the dead lock followed by 3rd attempt on the door lock 2nd attempt on the dead lock followed by 2nd attempt on the door lock 3rd attempt on the dead lock followed by 1st attempt on the door lock