L51 — Statistics, Probability and Mixed Problem Solving — Solutions

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1. Q1 — Scatter plots and correlation

   Fluency
   • (a) x̅ = (2+4+5+7+8+10)/6 = 36/6 = 6. y̅ = (55+63+70+78+85+90)/6 = 441/6 = 73.5.
   • (b) As hours of study increases, score increases consistently. Strong positive correlation.
   • (c) Gradient of line through (2,55) and (10,90): m = (90−55)/(10−2) = 35/8 = 4.375. y−55=4.375(x−2) ⇒ y=4.375x+46.25. At x=6: y=26.25+46.25=72.5 ≈ 73.
   • (d) Not valid — 20 hours is far outside the data range (2–10 hours). Extrapolating this far assumes the linear trend continues, which is unlikely (diminishing returns, sleep, etc.).

2. Q2 — Probability basics

   Fluency
   • (a) P(red) = 5/10 = 1/2.
   • (b) P(not green) = 8/10 = 4/5.
   • (c) P(both red) = (5/10) × (4/9) = 20/90 = 2/9.
   • (d) P(red then blue) + P(blue then red) = (5/10)(3/10) + (3/10)(5/10) = 15/100 + 15/100 = 30/100 = 3/10.

3. Q3 — Index laws and exponential equations

   Fluency
   • (a) (3x²y)³ = 27x&sup6;y³. Divide by 9xy²: 27x&sup6;y³/(9xy²) = 3x&sup5;y.
   • (b) 2^x = 2&sup6; ⇒ x = 6.
   • (c) 3^2x−1 = 3³ ⇒ 2x−1 = 3 ⇒ x = 2.
   • (d) x log 5 = log 100 = 2. x = 2/log 5 = 2/0.699 ≈ x ≈ 2.86.

4. Q4 — Networks

   Fluency
   • (a) Degree: A=2 (AB,AC), B=3 (AB,BC,BD), C=4 (AC,BC,CD,CE), D=3 (BD,CD,DE), E=2 (DE,CE). A:2, B:3, C:4, D:3, E:2.
   • (b) Euler path requires exactly 0 or 2 odd-degree vertices. Odd-degree vertices: B(3) and D(3) — exactly 2. Yes, an Euler path is possible, starting at B or D.
   • (c) Sort edges: CD=2, AC=3, BC=4, AB=5, BD=6, DE=7, CE=8. Add CD(2)✓, AC(3)✓, BC(4)✓ — wait, adds cycle A–C–B–? No: A,C,D connected; B connected to C; BD(6)✓ connects B to D but D already in tree via CD — skip. DE(7)✓. MST edges: CD, AC, BC, DE (cost=2+3+4+7=16 km). Check: 4 edges for 5 vertices ✓.
   • (d) Paths A to E: A–C–D–E = 3+2+7=12. A–B–D–E = 5+6+7=18. A–C–E = 3+8=11. Shortest: A–C–E = 11 km.

5. Q5 — Line of best fit and prediction

   Understanding
   • (a) Gradient 4.5: for each extra 1°C, approximately 4.5 more ice creams are sold. y-intercept 12: at 0°C, 12 ice creams predicted per day (model estimate — may not be meaningful in context).
   • (b) y = 4.5(28) + 12 = 126 + 12 = 138 ice creams.
   • (c) Residual = actual − predicted = 142 − 138 = +4 (the model underestimates by 4).
   • (d) 5°C is outside the data range (15°C to 35°C). This is extrapolation — the prediction is unreliable as the linear relationship may not hold at low temperatures.

6. Q6 — Conditional probability and Venn diagrams

   Understanding
   • (a) C only = 120−40 = 80. B only = 80−40 = 40. Both = 40. Neither = 200−80−40−40 = 40. Venn diagram: C only=80, C∩B=40, B only=40, outside=40.
   • (b) P(C only) = 80/200 = 2/5.
   • (c) P(C|B) = P(C∩B)/P(B) = (40/200)/(80/200) = 40/80 = 1/2.
   • (d) Neither = 40. P(neither) = 40/200 = 1/5.

7. Q7 — Growth and decay

   Understanding
   • (a) V(t) = 24000 × 0.85^t.
   • (b) V(3) = 24000 × 0.85³ = 24000 × 0.6141 ≈ $14 739.
   • (c) 24000 × 0.85^t < 10000 ⇒ 0.85^t < 5/12. t log 0.85 < log(5/12). t > log(5/12)/log(0.85) ≈ −0.3802/−0.0706 ≈ 5.39. After 6 complete years.
   • (d) 1.045^t = 2 ⇒ t log 1.045 = log 2 ⇒ t = log 2/log 1.045 ≈ 0.3010/0.01912 ≈ 15.7 years ≈ 16 years.

8. Q8 — Networks — minimum spanning tree and paths

   Understanding
   • (a) A spanning tree of n vertices has n−1 edges. 6 towns ⇒ 5 edges.
   • (b) Sort edges: D–E=6, C–D=8, E–F=10, B–C=10, A–B=12, B–D=14, A–C=18, C–E=20, D–F=16. Add: DE(6)✓, CD(8)✓, EF(10)✓, BC(10)✓ — no cycle, AB(12)✓ — no cycle. MST: D–E, C–D, E–F, B–C, A–B.
   • (c) Total = 6+8+10+10+12 = 46 km.
   • (d) A–B–C–D–E–F = 12+10+8+6+10=46. A–B–D–E–F = 12+14+6+10=42. A–B–C–D–F = 12+10+8+16=46. A–B–D–F = 12+14+16=42. Shortest: A–B–D–E–F = 42 km.

9. Q9 — Mixed: statistics and probability

   Problem Solving
   • (a) F only = 60−20 = 40. B only = 55−20 = 35. Both = 20. Neither = 150−40−20−35 = 55. (Check: 40+20+35+55=150 ✓.)
   • (b) P(F∩B) = 20/150 = 2/15.
   • (c) P(F|B) = (20/150)/(55/150) = 20/55 = 4/11.
   • (d) Only one sport = F only + B only = 40+35 = 75. P = 75/150 = 1/2.

10. Q10 — Extended: mixed Year 10 problem

    Problem Solving
    • (a) V(0) = 200 × 1 = 200 visitors.
    • (b) V(10) = 200 × 1.08¹⁰ ≈ 200 × 2.159 ≈ 432 visitors.
    • (c) 200 × 1.08^t > 1000 ⇒ 1.08^t > 5 ⇒ t log 1.08 > log 5 ⇒ t > log 5/log 1.08 ≈ 0.699/0.03342 ≈ 20.9. After 21 complete days.
    • (d) 200 × 1.08^t > 5000 ⇒ 1.08^t > 25 ⇒ t > log 25/log 1.08 ≈ 1.398/0.03342 ≈ 41.8. After 42 complete days they must upgrade.