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Secondary 3 Elementary Mathematics Statistics Probability Quiz

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Secondary 3 Elementary Mathematics Quiz - Statistics Probability — ANSWER KEY

Total Marks: 50

Section A: Data Handling and Analysis (Questions 1–10)

1. Heights: 158, 162, 165, 170, 172, 175, 178, 180, 182, 185

(a) Median height [1]

  • Data already ordered. n = 10 (even).
  • Median = (5th + 6th values) ÷ 2 = (172 + 175) ÷ 2 = 173.5 cm ✓
    Answer: 173.5 cm

(b) Interquartile range [2]

  • Q1: median of lower half (158, 162, 165, 170, 172) = 165 cm [1]
  • Q3: median of upper half (175, 178, 180, 182, 185) = 180 cm
  • IQR = Q3 − Q1 = 180 − 165 = 15 cm [1]
    Answer: 15 cm

2. Mean of 8 numbers = 15 → Sum = 8 × 15 = 120 [1]
New mean of 9 numbers = 16 → New sum = 9 × 16 = 144
Ninth number = 144 − 120 = 24 [1]
Answer: 24

3. Marks distribution:

Marks0–910–1920–2930–3940–4950–59
f2581294

(a) Modal class [1]

  • Highest frequency = 12 → Modal class is 30–39 ✓
    Answer: 30–39

(b) Estimate of mean mark [2]

  • Midpoints: 4.5, 14.5, 24.5, 34.5, 44.5, 54.5
  • Σfx = 2(4.5) + 5(14.5) + 8(24.5) + 12(34.5) + 9(44.5) + 4(54.5) [1]
    = 9 + 72.5 + 196 + 414 + 400.5 + 218 = 1310
  • Σf = 40
  • Mean = 1310 ÷ 40 = 32.75 [1]
    Answer: 32.75 (or 32.8)

4. Cumulative frequency: Q1 = 72 g, Median = 85 g, Q3 = 94 g, n = 60

(a) IQR [1]

  • IQR = 94 − 72 = 22 g ✓
    Answer: 22 g

(b) Number of small apples (mass < 72 g) [1]

  • Q1 = 72 g → 25% of apples are below Q1
  • Number = 25% × 60 = 15 apples ✓
    Answer: 15

5. Ages:

Age10–1415–1920–2425–2930–34
f61418102

(a) Estimate of mean age [2]

  • Midpoints: 12, 17, 22, 27, 32
  • Σfx = 6(12) + 14(17) + 18(22) + 10(27) + 2(32) [1]
    = 72 + 238 + 396 + 270 + 64 = 1040
  • Σf = 50
  • Mean = 1040 ÷ 50 = 20.8 years [1]
    Answer: 20.8 years

(b) Why an estimate? [1]

  • The exact ages are unknown; we use class midpoints to represent all values in each class. The actual mean may differ slightly. ✓
    Answer: Because exact ages are not known; midpoints are used as approximations.

6. Set A: mean = 45, SD = 6; Set B: mean = 45, SD = 12

(a) Which set has greater spread? [1]

  • Set B has a larger standard deviation (12 > 6), so it has greater spread. ✓
    Answer: Set B. Reason: It has a larger standard deviation.

(b) Advantage of standard deviation over range [1]

  • Standard deviation uses all data values, not just the extremes, so it is less affected by outliers. ✓
    Answer: It takes into account every data value, not just the maximum and minimum.

7. Box-and-whisker plot: min ≈ 35, Q1 ≈ 45, median ≈ 55, Q3 ≈ 65, max ≈ 85

(a) Median score [1]

  • Median is the line inside the box ≈ 55 ✓
    Answer: 55

(b) Range [1]

  • Range = max − min ≈ 85 − 35 = 50 ✓
    Answer: 50

(c) 25% scored above a certain mark [1]

  • 25% above = upper quartile ≈ 65 ✓
    Answer: 65

8. Books read: 3, 5, 2, 4, 6, 3, 5, 7, 4, 3, 5, 6

(a) Mode [1]

  • 3 appears 3 times; 5 appears 3 times → modes are 3 and 5 (bimodal) ✓
    Answer: 3 and 5

(b) Mean [2]

  • Sum = 3+5+2+4+6+3+5+7+4+3+5+6 = 53 [1]
  • n = 12
  • Mean = 53 ÷ 12 = 4.416... ≈ 4.42 (3 s.f.) [1]
    Answer: 4.42

9. Standard deviation of number of pets [3]

xffxx²fx²
04000
18818
2714428
3412936
4281632
  • Σf = 25, Σfx = 42, Σfx² = 104 [1]
  • Mean = 42 ÷ 25 = 1.68
  • Variance = (Σfx² ÷ Σf) − mean² = (104 ÷ 25) − 1.68² [1]
    = 4.16 − 2.8224 = 1.3376
  • SD = √1.3376 = 1.1565... ≈ 1.16 (3 s.f.) [1]
    Answer: 1.16

10. Cumulative frequency diagram (n = 200)

(a) Median mark [1]

  • Median corresponds to cumulative frequency = 100. Reading from graph → approximately 52 marks. ✓
    Answer: 52 (accept 50–54 depending on graph reading)

(b) Number scoring more than 65 marks [2]

  • At 65 marks, cumulative frequency ≈ 150 (from graph) [1]
  • Number above 65 = 200 − 150 = 50 students [1]
    Answer: 50 (accept 45–55 depending on graph reading)

Section B: Probability (Questions 11–20)

11. Fair six-sided die: S = {1, 2, 3, 4, 5, 6}

(a) Prime number [1]

  • Primes: {2, 3, 5} → P(prime) = 3/6 = 1/2 ✓
    Answer: 1/2

(b) Number greater than 4 [1]

  • {5, 6} → P(>4) = 2/6 = 1/3 ✓
    Answer: 1/3

12. Bag: 5 red, 3 blue, 2 green → Total = 10

(a) P(red) [1]

  • 5/10 = 1/2 ✓
    Answer: 1/2

(b) P(not green) [1]

  • Not green = red or blue = 8 balls → 8/10 = 4/5 ✓
    Answer: 4/5

13. Standard pack of 52 cards

(a) P(King) [1]

  • 4 Kings → 4/52 = 1/13 ✓
    Answer: 1/13

(b) P(red card or Queen) [2]

  • P(red) = 26/52 = 1/2; P(Queen) = 4/52 = 1/13 [1]
  • P(red ∩ Queen) = 2/52 = 1/26 (Queen of Hearts and Queen of Diamonds)
  • P(red ∪ Queen) = 1/2 + 1/13 − 1/26 = 13/26 + 2/26 − 1/26 = 14/26 = 7/13 [1]
    Answer: 7/13

14. Two fair coins: S = {HH, HT, TH, TT}

(a) Exactly one head [1]

  • {HT, TH} → 2/4 = 1/2 ✓
    Answer: 1/2

(b) At least one tail [1]

  • {HT, TH, TT} → 3/4 ✓
    Answer: 3/4

15. Box: 4 white, 6 black → Total = 10. With replacement.

(a) Tree diagram [2]

  • First draw: P(W) = 4/10 = 0.4, P(B) = 6/10 = 0.6 [1]
  • Second draw (same probabilities due to replacement):
    • W → W: 0.4 × 0.4 = 0.16
    • W → B: 0.4 × 0.6 = 0.24
    • B → W: 0.6 × 0.4 = 0.24
    • B → B: 0.6 × 0.6 = 0.36 [1]
  • Diagram should show two levels with four branches, probabilities labeled. ✓

(b) P(both black) [1]

  • 0.6 × 0.6 = 0.36 ✓
    Answer: 0.36 (or 9/25)

(c) P(exactly one white) [2]

  • (W then B) or (B then W) [1]
  • = 0.4 × 0.6 + 0.6 × 0.4 = 0.24 + 0.24 = 0.48 [1]
    Answer: 0.48 (or 12/25)

16. P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2

(a) P(A ∪ B) [1]

  • P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.5 − 0.2 = 0.7 ✓
    Answer: 0.7

(b) Mutually exclusive? [1]

  • No, because P(A ∩ B) = 0.2 ≠ 0. Mutually exclusive events have no intersection. ✓
    Answer: No. P(A ∩ B) ≠ 0, so the events can occur together.

17. Rain: P(R) = 0.3, P(Dry) = 0.7
Umbrella: P(U|R) = 0.9, P(U|D) = 0.2

(a) Tree diagram [2]

  • First level: Rain (0.3), Dry (0.7) [1]
  • Second level:
    • Rain → Umbrella (0.9), No umbrella (0.1)
    • Dry → Umbrella (0.2), No umbrella (0.8) [1]
  • Diagram should show all branches with probabilities. ✓

(b) P(brings umbrella) [2]

  • P(U) = P(R) × P(U|R) + P(D) × P(U|D) [1]
    = 0.3 × 0.9 + 0.7 × 0.2 = 0.27 + 0.14 = 0.41 [1]
    Answer: 0.41

18. Fair coin tossed three times: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (8 outcomes)

(a) P(three heads) [1]

  • {HHH} → 1/8 ✓
    Answer: 1/8

(b) P(exactly two tails) [2]

  • Outcomes with exactly two tails: {HTT, THT, TTH} → 3 outcomes [1]
  • P = 3/8 [1]
    Answer: 3/8

19. Spinner: numbers 1 to 8

(a) P(multiple of 3) [1]

  • Multiples of 3: {3, 6} → 2/8 = 1/4 ✓
    Answer: 1/4

(b) P(factor of 8) [1]

  • Factors of 8: {1, 2, 4, 8} → 4/8 = 1/2 ✓
    Answer: 1/2

20. Survey: n = 100, Math = 60, Science = 45, Both = 25

(a) Venn diagram [2]

  • Intersection: 25 [1]
  • Math only: 60 − 25 = 35
  • Science only: 45 − 25 = 20
  • Neither: 100 − (35 + 25 + 20) = 20 [1]
  • Diagram should show two overlapping circles inside a rectangle, with all four regions labeled. ✓

(b) P(Math or Science but not both) [2]

  • Math only + Science only = 35 + 20 = 55 [1]
  • P = 55/100 = 11/20 = 0.55 [1]
    Answer: 11/20 or 0.55

END OF ANSWER KEY