NCEA Level 3The normal distributionAS 91586

The normal distribution: forwards, backwards and conditional

What this lesson covers

  • A normal probability is an area under the curve, found from the mean and standard deviation.
  • Working backwards from a probability finds an unknown value, mean or standard deviation.
Key idea

Level 3 normal: what is new

  • Name the model every time.
  • Unknown mean or standard deviation.
  • Conditions, and proposing a σ.
Worked example

Name the variable and the model

  • X = time for a random trip, in minutes.
  • X ~ N(42, 6): mean 42, sd 6.
  • Name the model and its parameters.
This question
Random variableX = trip time (min)
Distributionnormal
Parametersμ = 42, σ = 6
WrittenX ~ N(42, 6)
Worked example

P(X > 50): sketch it first

  • Mean 42 in the middle.
  • Shade the right tail beyond 50.
  • Expect a small area.
P(X > 50): sketch it first
Worked example

P(X > 50): the working

  • Standardise: z = (x − μ)/σ.
  • Keep 4 decimal places.
  1. X ~ N(42, 6)
  2. P(X > 50)
  3. z = (50 − 42)/6 = 8/6
  4. = 1.33334 d.p.
  5. normal cdf: 50 to ∞, μ 42, σ 6
  6. P(X > 50) = 0.0912
Worked example

Backwards: the fastest 10%

  • Area to the left: 0.1.
  • Then x = μ + zσ.
  1. X ~ N(42, 6)
  2. P(X < k) = 0.1
  3. inverse normal: area 0.1
  4. z = −1.2816
  5. k = μ + zσ
  6. = 42 + (−1.2816)(6)
  7. = 42 − 7.6896 = 34.3104
  8. k = 34.31 min
Worked example

An unknown standard deviation

  • Find z first.
  • Then solve for σ.
  1. X ~ N(240, σ)
  2. P(X < 230) = 0.05
  3. inverse normal: area 0.05
  4. z = −1.6449
  5. (230 − 240)/σ = −1.6449
  6. −10/σ = −1.6449
  7. σ = 10/1.6449 = 6.0794
  8. σ = 6.08 ml
Worked example

An unknown mean

  • Find z first.
  • Then solve for μ.
  1. X ~ N(μ, 5)
  2. P(X < 230) = 0.01
  3. inverse normal: area 0.01
  4. z = −2.3263
  5. (230 − μ)/5 = −2.3263
  6. 230 − μ = −2.3263 × 5 = −11.6315
  7. μ = 230 + 11.6315 = 241.6315
  8. μ = 241.63 ml
Worked example

Where marks are lost

  • No named model.
  • z rounded early: σ comes out wrong.
  • Show the working, every time.
Loses marksKeeps them
Modelnot statedX ~ N(240, σ)
zrounded to −1.6−1.6449
σ6.25 ml6.08 ml
Workinganswer onlyz, equation, solve
Worked example

Given it has passed 45 minutes

  • The condition: X > 45.
  • Inside it, the event: X > 50.
  • Smaller area ÷ larger area.
Given it has passed 45 minutes
Worked example

P(X > 50 | X > 45)

  • Both ÷ the condition.
  • Over 50 is already over 45.
  1. P(X > 50 | X > 45)
  2. = P(X > 50 and X > 45)/P(X > 45)
  3. = P(X > 50)/P(X > 45)over 50 is over 45
  4. P(X > 50) = 0.0912from before
  5. z = (45 − 42)/6 = 0.5
  6. P(X > 45) = 0.3085
  7. = 0.0912/0.3085
  8. = 0.2956
Worked example

Propose σ: the first condition

  • Mean 242 ml.
  • At most 1% under 230.
  1. X ~ N(242, σ)
  2. P(X < 230) ≤ 0.01
  3. inverse normal: area 0.01
  4. z = −2.3263
  5. (230 − 242)/σ = −2.3263
  6. σ = 12/2.3263 = 5.1584
  7. σ ≤ 5.158round a limit down
Worked example

Propose σ: the second condition

  • At most 5% over 250.
  • The smaller limit wins.
  1. σ ≤ 5.158
  2. P(X > 250) ≤ 0.05
  3. inverse normal: area 0.95
  4. z = 1.6449
  5. (250 − 242)/σ = 1.6449
  6. σ = 8/1.6449 = 4.8635
  7. σ ≤ 4.863round a limit down
  8. both: σ ≤ 4.863the smaller wins
Worked example

Justify it: check both

  • Propose σ = 4.5 ml.
  • Check both conditions.
  1. propose σ = 4.5
  2. z = (230 − 242)/4.5 = −2.6667
  3. P(X < 230) = 0.0038 < 0.01
  4. z = (250 − 242)/4.5 = 1.7778
  5. P(X > 250) = 0.0377 < 0.05
  6. σ = 5: z = 8/5 = 1.6
  7. P(X > 250) = 0.0548 > 0.05too many spill
Key idea

Where the marks are

  • Achieved: name the model, sketch, probability.
  • Merit: unknown μ or σ, or a condition.
  • Excellence: propose and justify a σ.
Summary

What to take away

  • Name it, sketch it, then work forwards or backwards.
Name itX ~ N(μ, σ) in context
Sketchshade the area first
Unknown μ or σfind z, then solve
ConditionalP(A and B) ÷ P(B)
Propose σcheck every condition
Rounding4 dp until the end
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Name the model and sketch it first.
  • Question 3 is conditional.
GivenFind
1X ~ N(42, 6), 35 to 45probability
2σ = 6, 15% over 50μ
3N(42, 6), and X > 36P(X > 48)
4μ = 242, ≤ 2% < 235largest σ
Show answer

1. z from −1.1667 to 0.5: 0.6915 − 0.1217 = 0.5698. 2. z = 1.0364: μ = 50 − 1.0364 × 6 = 43.78 min. 3. P(X > 48 | X > 36) = 0.1587 ÷ 0.8413 = 0.1886. 4. z = −2.0537: σ ≤ 7 ÷ 2.0537 = 3.408, so propose 3.4 ml or less (P(X < 235) = 0.0198).

Worked example

The answers: questions 1 and 2

  1. 1. X ~ N(42, 6)
  2. z = (35 − 42)/6 = −1.1667
  3. z = (45 − 42)/6 = 0.5
  4. P(−1.1667 < Z < 0.5)
  5. = 0.6915 − 0.1217 = 0.5698
  6. 2. P(X > 50) = 0.15, σ = 6
  7. z = 1.0364inverse normal, 0.85
  8. (50 − μ)/6 = 1.0364
  9. μ = 50 − 6.2184 = 43.78 min
Worked example

The answers: questions 3 and 4

  1. 3. P(X > 48)/P(X > 36)over 48 is over 36
  2. z = (48 − 42)/6 = 1 → 0.1587
  3. z = (36 − 42)/6 = −1 → 0.8413
  4. = 0.1587/0.8413 = 0.1886
  5. 4. P(X < 235) ≤ 0.02, μ = 242
  6. z = −2.0537inverse normal, 0.02
  7. (235 − 242)/σ = −2.0537
  8. σ = 7/2.0537 = 3.4085
  9. σ ≤ 3.408: propose 3.4 ml

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