1. 2015 NZQA Exam L2 Probability
Note: make sure each student has a paper copy of the exam paper first; this PPT is designed for the purposes of demonstrating (some) solutions. The slides with poor quality ‘snapshots’ of the exam paper are for the teacher’s benefit only.

2. Qu 1 (a) i
P(0 < Z < 0.75) =

3. 𝝁=𝟑𝟒 𝝈=𝟖 P(34< X < 40) = 34 40

4. Qu 1 (a) ii

5. 𝝁=𝟑𝟒 𝝈=𝟖 Inverse Normal P(X < ????) = 0.90 X = 44.25 minutes
90% 34 X

6. Qu 1 (a) iii

7. 𝝁=𝟎 𝝈=𝟏 Standard Normal P(z < ???) = 0.95 z = 1.645
95% ??

8. 𝒛= 𝑿−𝝁 𝝈 𝟏.𝟔𝟒𝟓= 𝟒𝟎−𝟑𝟒 𝝈 𝝈= 𝟒𝟎−𝟑𝟒 𝟏.𝟔𝟒𝟓 𝝁=𝟑𝟒, 𝑿=𝟒𝟎, 𝒛=𝟏.𝟔𝟒𝟓 𝝈=𝟑.𝟔𝟒𝟕
𝝁=𝟑𝟒, 𝑿=𝟒𝟎, 𝒛=𝟏.𝟔𝟒𝟓
𝒛= 𝑿−𝝁 𝝈
𝟏.𝟔𝟒𝟓= 𝟒𝟎−𝟑𝟒 𝝈
𝝈= 𝟒𝟎−𝟑𝟒 𝟏.𝟔𝟒𝟓
𝝈=𝟑.𝟔𝟒𝟕
The new std deviation is minutes

9. Old std dev = 8 minutes
The new std deviation is minutes
Question states ‘…each doctor added reduces the stdv dev by 0.4 minutes.
The difference between 8 and is 4.353
4.353 ÷ 0.4 = 10.88
This means that approximately 11 more doctors need to added to the duty teams.

10. P(0 < X < 90) P(-3 < Z < 1.5) = 0.932
Qu 1 (b) i
P(0 < X < 90) P(-3 < Z < 1.5) = 0.932

11. 𝝁=𝟔𝟎 𝝈=𝟐𝟎 P(0< X < 90) = 60 90

12. Qu 1 (b) ii

13. Qu 1 (b) ii ANS: 60 out of 80 patients were seen in less than 90 seconds (0.75 or 75%)

14. Qu 1 (b) iii

15. Qu 1 (b) iii contd

16. Graph 2: As above but one peak.
Qu 1 (b) iii ans
Possible valid comparative statements that may relate to the points listed below.
Shape:
Graph 1: Not symmetrical; skewed to the left; two peaks; bunching of values to the right.
Graph 2: As above but one peak.

17. Qu 1 (b) iii ans
Centre:
Graph 1: Mode at 75 – 90 seconds; Median at 60 – 75 seconds; mean to the right of centre.
Graph 2: As above.

18. Graph 1: Range of about 105 seconds
Qu 1 (b) iii ans
Spread:
Graph 1: Range of about 105 seconds
Graph 2: Range of 2 minutes (120 seconds)
Proportions:
Proportions are similar except for 30 – 60 seconds.

19. Qu 2 (a) i

20. Qu 2 (a) ii
Conditional Prob : ‘Given “Failed”, what is the probability they were from Year 12?
ANS: 33 out of 300

21. Qu 2 (a) iii
The proportion who were Yr 13 students who ‘Passed’ is 𝟖𝟓𝟑 𝟏𝟓𝟎𝟎 . ( 𝟖𝟓𝟑 𝟏𝟓𝟎𝟎 ×𝟓𝟐𝟓𝟎𝟎=29855)

22. Qu 2 (a) ANS

23. Qu 2 (a) iv
Relative Risk : calculate the ordinary ‘risk’ for Yr 12 and for Yr 13 first…

24. Qu 2 (a) iv ANS

25. Qu 2 (b) i

26. 5 subjects
6 subjects
Total
Passed
1200
Failed
192
300
682
1500

27. 5 subjects
6 subjects
Total
Passed
626
574
1200
Failed
192
108
300
818
682
1500

28. Qu 2 (b) i ANS

29. Qu 2 (b) ii
Relative Risk : calculate the ordinary ‘risk’ for passing “6 Subjects” and for passing “5 Subjects” first…

30. 𝑷(𝑷𝒂𝒔𝒔𝒆𝒅 𝒊𝒏 𝟔 𝑺𝒖𝒃𝒋𝒆𝒄𝒕𝒔)= 𝟓𝟕𝟒 𝟔𝟖𝟐 =𝟎.𝟖𝟒𝟏𝟔
𝑷(𝑷𝒂𝒔𝒔𝒆𝒅 𝒊𝒏 𝟓 𝑺𝒖𝒃𝒋𝒆𝒄𝒕𝒔)= 𝟔𝟐𝟔 𝟖𝟏𝟖 =𝟎.𝟕𝟔𝟓𝟑
6 Subjects ‘relative’ to 5 Subjects:
𝟎.𝟖𝟒𝟏𝟔 𝟎.𝟕𝟔𝟓𝟑 =𝟏.𝟏𝟎 (𝟐𝒅𝒑)
“1.10 means that a ‘6 Subject’ student is 10% more likely to pass than a ‘5 Subject student”

31. “1.10 means that a ‘6 Subject’ student is 10% more likely to pass than a ‘5 Subject student”
NZQA answer also states “…10% more likely to pass if taking 6 subjects BUT this is deceptive, as candidates with more ability are likely to be taking 6 subjects.”
There could also be a comment on the representativeness of the sample. (eg. ‘Even though the sample was randomly selected, it might not be truly representative of the population of Yr 12 / 13 students sitting international exams…’)
The next slide gives an alternate answer…in terms of ‘Failing’.

32. 𝑷(𝑭𝒂𝒊𝒍𝒆𝒅 𝒊𝒏 𝟔 𝑺𝒖𝒃𝒋𝒆𝒄𝒕𝒔)= 𝟏𝟎𝟖 𝟔𝟖𝟐 =𝟎.𝟏𝟓𝟖𝟒
𝑷(𝑭𝒂𝒊𝒍𝒆𝒅 𝒊𝒏 𝟓 𝑺𝒖𝒃𝒋𝒆𝒄𝒕𝒔)= 𝟏𝟗𝟐 𝟖𝟏𝟖 =𝟎.𝟐𝟑𝟒𝟕
6 Subjects ‘relative’ to 5 Subjects:
𝟎.𝟏𝟓𝟖𝟒 𝟎.𝟐𝟑𝟒𝟕 =𝟎.𝟔𝟕𝟓 (𝟑 𝒅𝒑)
“0.675 means that a ‘6 Subject’ student has 67.5% of the risk of failing (ie. less chance of failing) compared to a ‘5 Subject student”.

33. Qu 2 (b) ii

34. Probability tree : Always draw a large diagram, clearly labelled. Why?
Qu 3 (a) i
Probability tree : Always draw a large diagram, clearly labelled. Why?

35. Sold
Not
0.7
Sold
Not
0.8 M F
0.55
Sold
Not
0.2
Sold
Not
0.35

36. M F Sold Not 0.7 Sold Not 0.8 0.55 0.3 0.2 Sold Not 0.2 0.45 Sold Not
0.35
0.8
0.65

37. Qu 3 (a)
P(Male and sold at age ‘1 month’)
0.55 × 0.7 = 0.385
P(Female and sold at age ‘3 months’)
0.45 × 0.8 × 0.35 = 0.126

38. (0.55 × 0.3 × 0.2) + (0.45 × 0.8 × 0.65) = 0.267 Qu 3 (a) iii M F 0.55
Not M F
Not
0.55
0.3
0.2
Not
0.45
Not
0.8
0.65
(0.55 × 0.3 × 0.2) + (0.45 × 0.8 × 0.65) = 0.267

39. Qu 3 (a) ANS

40. Qu 3 (a) iv

41. 0.033 × 550 calves = 18 males kept 0.033 M F Qu 3 (a) iv 0.55 0.3 Not
0.2
0.033
Not
Not
0.033 × 550 calves = 18 males kept

42. Qu 3 (a) v

43. Qu 3 (b)

44. Qu 3 (b) i
= 0.85

45. Qu 3 (b) ii

46. Qu 3 (b) ii