Uniform and triangular: 7 exam questions solved step by step

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What this lesson covers

  • Continuous uniform and triangular probabilities are areas of rectangles and triangles.
  • A model is judged by comparing its probabilities with the data's relative frequencies.
Key idea

7 questions, one method

  • 1. Sketch the pdf, shade the area.
  • 2. Height: 1 ÷ (b − a), or 2 ÷ (b − a).
  • 3. Area of a rectangle or triangle.
  • 4. Judge the model against data.
Your turn

Question 1 · Achieved

  • Pizza delivery times are uniform
  • between 20 and 45 minutes.
  • Find P(a delivery takes under 30 min).
Question 1 · Achieved
Worked example

Q1: a rectangle

  • Flat pdf: height 1 ÷ (b − a).
  • Width × height.
  1. X ~ U(20, 45)minutes
  2. height = 1 ÷ (45 − 20) = 1/25
  3. 25 × 1/25 = 1total area 1
  4. P(X < 30) = width × height
  5. = (30 − 20) × 1/25
  6. = 10/25 = 0.4
  7. 30 × 1/25 = 1.2over 1: impossible
  8. a 40% chanceunder 30 min
Your turn

Question 2 · Achieved

  • Cross-country times are triangular:
  • a = 12, c = 15, b = 24 minutes.
  • Find the peak height and
  • P(a runner takes over 20 min).
Question 2 · Achieved
Worked example

Q2: the peak, then a triangle

  • Total area 1 fixes the peak.
  • ½ × base × height.
  1. a = 12, c = 15, b = 24
  2. peak = 2 ÷ (24 − 12) = 1/6
  3. ½ × 12 × 1/6 = 1total area 1
  4. f(20) = (4/9) × (1/6)similar triangles
  5. = 4/54 = 2/27
  6. P(X > 20) = ½ × base × height
  7. = ½ × 4 × 2/27
  8. = 4/27 = 0.14814 d.p.
Your turn

Question 3 · Merit

  • The same run: a = 12, c = 15, b = 24.
  • Find P(14 < X < 20).
  • Careful: the middle is not a triangle.
Question 3 · Merit
Worked example

Q3: 1 minus both tails

  • The tails are triangles.
  • Take them off 1.
  1. P(14 < X < 20)
  2. = 1 − P(X < 14) − P(X > 20)not a triangle
  3. f(14) = (2/3) × (1/6) = 1/9similar triangles
  4. P(X < 14) = ½ × 2 × 1/9 = 1/9
  5. P(X > 20) = 4/27from question 2
  6. 1 − 1/9 − 4/27
  7. = 27/27 − 3/27 − 4/27
  8. = 20/27 = 0.74074 d.p.
Your turn

Question 4 · Merit, in context

  • A garden centre says a delivery takes
  • at least 2 days, at most 9,
  • and most often 3. Use a triangular
  • model to find P(over 6 days).
Question 4 · Merit, in context
Worked example

Q4: choose a, c, b

  • Least, most likely, most.
  • Then the right tail.
  1. a = 2, c = 3, b = 9least, likeliest, most
  2. peak = 2 ÷ (9 − 2) = 2/7
  3. f(6) = (3/6) × (2/7) = 1/7similar triangles
  4. P(X > 6) = ½ × 3 × 1/7
  5. = 3/14 = 0.21434 d.p.
  6. about 21% take over 6 days
Your turn

Question 5 · Merit, in context

  • A learner shearer's times are uniform,
  • from 2 to 6 minutes a sheep.
  • Of 200 sheep, how many over 5 min?
  • A sheep has taken 3 min: P(over 5)?
Question 5 · Merit, in context
Worked example

Q5: count, then given

  • n × probability.
  • "Given" shrinks the total.
  1. X ~ U(2, 6), height 1/4
  2. P(X > 5) = (6 − 5) × 1/4 = 1/4
  3. 200 × 1/4 = 50 sheep
  4. P(X > 5 | X > 3)
  5. = P(X > 5) ÷ P(X > 3)over 5 is over 3
  6. P(X > 3) = (6 − 3) × 1/4 = 3/4
  7. (1/4) ÷ (3/4) = 1/3
  8. = 0.33334 d.p.
  9. about 1 in 3 of those sheep
Your turn

Question 6 · Excellence

  • The garden centre: model a = 2, c = 3, b = 9.
  • Their records for 60 deliveries are below.
  • Compare model and data at both tails.
Under 3 daysOver 6 daysTotal
Deliveries11960
Worked example

Q6: model against data

  • Model: an area.
  • Data: a relative frequency.
  1. P(X < 3) = ½ × 1 × 2/7 = 1/7
  2. = 0.1429model
  3. data: 11/60 = 0.1833
  4. P(X > 6) = 3/14 = 0.2143from question 4
  5. data: 9/60 = 0.15
  6. 0.1429 < 0.1833model too low
  7. 0.2143 > 0.15model too high
  8. too many long waits: b too big
Your turn

Question 7 · Excellence

  • Change the model to b = 8.
  • Recalculate both tails and
  • judge: is b = 8 a better model?
Question 7 · Excellence
Worked example

Q7: try b = 8

  • Smaller b: taller peak.
  • Compare both tails again.
  1. a = 2, c = 3, b = 8
  2. peak = 2 ÷ (8 − 2) = 1/3
  3. P(X < 3) = ½ × 1 × 1/3 = 1/6
  4. = 0.1667data 0.1833
  5. f(6) = (2/5) × (1/3) = 2/15similar triangles
  6. P(X > 6) = ½ × 2 × 2/15 = 2/15
  7. = 0.1333data 0.15
  8. both gaps now about 0.017were 0.04 and 0.06
  9. b = 8 is the better model
Summary

The method that solved all seven

  • Sketch, find the height, find the area, then judge the model against the data.
StepHow
Uniformheight 1 ÷ (b − a)
Triangularpeak 2 ÷ (b − a)
Heightssimilar triangles
Middle1 minus both tails
Choosea least, c most likely, b most
Judgemodel against data, each tail

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