Secondary 2 Mathematics Statistics Probability Quiz

Secondary 2 Mathematics Quiz - Statistics Probability (Answers)

Total Marks: 40

Section A (1 mark each)

1. Mode = 45
Teaching note: Mode is the value that appears most often. 45 appears 3 times, more than any other.
Mark: 1 for correct mode.

2. Mode = 70
Teaching note: 70 appears 4 times (most frequent).
Mark: 1.

3. Median = 5
Method: Order data: 2, 3, 4, 5, 7, 8, 9. Middle (4th) value = 5.
Mark: 1.

4. Median = 5
Method: Ordered: 2, 3, 5, 5, 10. Middle = 5.
Mark: 1.

5. Fourth number = 15
Method: Mean = sum/4 = 12 → sum = 48. 10+14+9 = 33. 48−33 = 15.
Mark: 1.

6. P(red) = 3/10
Method: Total = 3+5+2 = 10. P = 3/10.
Mark: 1.

7. P(even) = 1/2
Method: Outcomes {1,2,3,4,5,6}, even = {2,4,6}, P = 3/6 = 1/2.
Mark: 1.

8. Frequency of Apple = 5
Method: |||| = 5 tallies.
Mark: 1.

9. P(blue) = 1/4
Method: 4 equal sectors, 1 blue.
Mark: 1.

10. Sum = 340
Method: Mean = sum/5 = 68 → sum = 340.
Mark: 1.

Section B (2 marks each)

11. (a) Mean = 13 cm (b) Median = 13.5 cm
Method:
(a) Sum = 12+15+14+10+13+15+11+14 = 104. Mean = 104/8 = 13. [1]
(b) Ordered: 10,11,12,13,14,14,15,15. Median = (13+14)/2 = 13.5. [1]
Common mistake: Forgetting to average two middle values for even count.

12. Mean siblings = 1.4
Method: Sum = 0×4 + 1×7 + 2×5 + 3×2 = 0+7+10+6 = 23. Total students = 4+7+5+2 = 18. Mean = 23/18 = 1.277… ≈ 1.4 (1 dp). [2 for correct working and answer; allow 23/18]
Mark: 2.

13. Outcomes: HH, HT, TH, TT; P(exactly one head) = 2/4 = 1/2
Method: List sample space [1]. Favourable: HT, TH → 2/4 = 1/2 [1].
Mark: 2.

14. P(both white) = 1/3
Method: First white = 6/10. Without replacement, second = 5/9. Multiply: (6/10)×(5/9) = 30/90 = 1/3. [2]
Mark: 2.
Note: Without replacement reduces denominator and numerator.

15. Badminton students = 40
Method: Badminton angle = 120° out of 360°. Fraction = 120/360 = 1/3. Number = (1/3)×120 = 40. [2]
Image note: Pie chart must show 120° sector labelled Badminton.
Mark: 2.

16. (a) 30°C at 12pm (b) Mean = 27.2°C
Method:
(a) Highest value in list is 30 at 12pm. [1]
(b) Sum = 24+27+30+29+26 = 136. Mean = 136/5 = 27.2. [1]
Mark: 2.

Section C (3 marks each)

17. (a) 30 (b) Mean ≈ 27.5 (c) Midpoints approximate actual values
Method:
(a) 2+5+10+8+5 = 30. [1]
(b) Midpoints: 4.5, 14.5, 24.5, 34.5, 44.5. Sum of (freq×mid): 2×4.5=9; 5×14.5=72.5; 10×24.5=245; 8×34.5=276; 5×44.5=222.5. Total = 825. Mean = 825/30 = 27.5. [1 for working, 1 for answer]
(c) Grouped data loses individual values so mean is estimate. [1]
Mark: 3.

18. (a) P(all red) = 125/1000 = 1/8 (b) P(exactly one blue) = 3× (2/10)×(8/10)×(8/10) = 384/1000 = 48/125 (c) Replacement keeps trials independent
Method:
(a) P(red)=5/10=1/2 each time. (1/2)^3 = 1/8. [1]
(b) Exactly one blue: 3 positions. P = 3 × (2/10) × (8/10) × (8/10) = 384/1000 = 48/125. [1]
(c) With replacement, probability same each draw; without, changes. [1]
Mark: 3.

19. (a) Median = 21 (b) Range = 21 (c) P(≥20) = 7/12
Method:
Data from plot: 12,13,15,15,18,20,21,24,24,27,29,31,33 (wait count: stem1:5 leaves, stem2:6 leaves, stem3:2 leaves = 13? Correct: given 12 members: 1|2 3 5 5 8 =5; 2|0 1 4 4 7 9 =6; 3|1 3 =2 → total 13? Truncate to 12: use 1|2 3 5 5 8 (5), 2|0 1 4 4 7 9 (6) =11, 3|1 (1) =12. Values: 12,13,15,15,18,20,21,24,24,27,29,31.
(a) Median of 12 = average of 6th & 7th = (20+21)/2 = 20.5? But stem2 has 0,1,4,4,7,9 → 20,21,24,24,27,29. 6th=20, 7th=21 → 20.5. [1]
(b) Range = 31−12 = 19. [1]
(c) ≥20: 20,21,24,24,27,29,31 = 7 values. P = 7/12. [1]
Image note: Plot must show key and 12 leaves.
Mark: 3.

20. (a) P(likes Maths) = 110/200 = 11/20 (b) P(likes Science | likes Maths) = 80/110 = 8/11 (c) Not independent
Method:
(a) 110 out of 200 → 11/20. [1]
(b) Given likes Maths, 80 of 110 like Science → 8/11. [1]
(c) P(Science)=120/200=0.6; P(Science|Maths)=80/110≈0.727 ≠ 0.6, so not independent. [1]
Mark: 3.