NCEA Level 2Inverse normalAS 91267

Inverse normal: 7 exam questions solved step by step

What this lesson covers

  • An inverse normal problem starts from a probability and works back to a value.
  • The z-value for a known probability links the unknown mean or standard deviation to the data.
Key idea

7 questions, one method

  • 1. Sketch, and shade the known area.
  • 2. Area to the left gives z.
  • 3. z = (x − μ) ÷ σ: solve for the unknown.
  • 4. Check the side, then round.
Your turn

Question 1 · Achieved

  • Hauraki Gulf snapper lengths are
  • normal: mean 34 cm, sd 5 cm.
  • Find the length that 80% are shorter than.
Question 1 · Achieved
Worked example

Q1: 80% shorter than x

  • Area to the left: 0.8.
  • Then x = μ + zσ.
  1. μ = 34, σ = 5given
  2. P(X < x) = 0.880% shorter
  3. z = 0.8416area 0.8 to the left
  4. x = μ + zσfrom z = (x − μ) ÷ σ
  5. x = 34 + 0.8416 × 5substitute
  6. = 34 + 4.208
  7. = 38.208
  8. x = 38.2 cm1 d.p.
  9. 38.2 > 34above the mean
Your turn

Question 2 · Achieved

  • The same snapper: mean 34 cm, sd 5 cm.
  • Find the length that marks off
  • the shortest 25%.
Question 2 · Achieved
Worked example

Q2: the shortest 25%

  • Below the mean: z < 0.
  • Check the side.
  1. μ = 34, σ = 5given
  2. P(X < x) = 0.25shortest 25%
  3. z = −0.6745below the mean
  4. x = 34 + (−0.6745) × 5
  5. = 34 − 3.3725
  6. = 30.6275 ≈ 30.6 cm1 d.p.
  7. 30.6 < 34below the mean
  8. z = +0.6745 gives 37.4 cmwrong side
Your turn

Question 3 · Merit

  • The same snapper: mean 34 cm, sd 5 cm.
  • Find the limits of the middle 90%
  • of lengths.
Question 3 · Merit
Worked example

Q3: the middle 90%

  • Split the 10%: 5% each end.
  • Symmetric about 34.
  1. μ = 34, σ = 5given
  2. outside: 5% in each tail10% ÷ 2
  3. P(X < a) = 0.05: z = −1.6449
  4. a = 34 − 1.6449 × 5
  5. = 34 − 8.2245 = 25.7755
  6. P(X < b) = 0.95: z = 1.6449
  7. b = 34 + 8.2245 = 42.2245
  8. 25.78 cm to 42.22 cm2 d.p.
  9. 42.22 − 34 = 34 − 25.78symmetric
Your turn

Question 4 · Merit

  • The Days Bay ferry: crossing times are
  • normal with mean 25 min.
  • 15% of crossings take over 28 min.
  • Find the standard deviation.
Question 4 · Merit
Worked example

Q4: an unknown σ

  • Over 28: left area 0.85.
  • Substitute, solve for σ.
  1. μ = 25, P(X > 28) = 0.15given
  2. P(X < 28) = 0.85area to the left
  3. z = 1.036428 is above 25
  4. z = (x − μ) ÷ σ
  5. 1.0364 = (28 − 25) ÷ σsubstitute
  6. 1.0364σ = 3× σ
  7. σ = 3 ÷ 1.0364÷ 1.0364
  8. = 2.8946 ≈ 2.89 min2 d.p.
  9. σ > 0as it must be
Your turn

Question 5 · Merit, in context

  • Kiwifruit trays: weights normal, sd 150 g.
  • At most 2% may be under 3500 g.
  • The mean is set in whole grams.
  • Find the smallest mean to set.
Question 5 · Merit, in context
Worked example

Q5: an unknown μ

  • Solve for μ.
  • Then round the safe way.
  1. σ = 150, P(X < 3500) = 0.02given
  2. z = −2.05373500 is below μ
  3. −2.0537 = (3500 − μ) ÷ 150substitute
  4. −308.055 = 3500 − μ× 150
  5. μ = 3500 + 308.055rearrange
  6. = 3808.055 g
  7. 3808 g: P = 0.020020over 2%
  8. 3809 g: P = 0.019699under 2%
  9. set the mean to 3809 g
Your turn

Question 6 · Excellence

  • Rangitoto summit walk times are normal.
  • 10% take under 40 min.
  • 20% take over 65 min.
  • Find the mean and standard deviation.
Question 6 · Excellence
Worked example

Q6: two equations

  • One equation per fact.
  • Two unknowns: μ and σ.
  1. P(X < 40) = 0.1: z = −1.2816
  2. −1.2816 = (40 − μ) ÷ σsubstitute
  3. μ − 1.2816σ = 40× σ, rearrange
  4. P(X > 65) = 0.2given
  5. P(X < 65) = 0.8: z = 0.8416
  6. 0.8416 = (65 − μ) ÷ σsubstitute
  7. μ + 0.8416σ = 65× σ, rearrange
Worked example

Q6: solve them together

  • Subtract to lose μ.
  • Check in the second.
  1. μ − 1.2816σ = 40
  2. μ + 0.8416σ = 65
  3. 2.1232σ = 25second − first
  4. σ = 25 ÷ 2.1232 = 11.7747
  5. μ = 40 + 1.2816 × 11.7747
  6. = 40 + 15.0904
  7. = 55.0904
  8. μ ≈ 55.09 min, σ ≈ 11.77 min
  9. 55.09 + 0.8416 × 11.77 ≈ 65
Your turn

Question 7 · Excellence

  • Flour bags: normal, mean 1012 g, sd 8 g.
  • At most 3% may be under 1000 g.
  • Does the machine meet the rule?
  • If not, keep the mean: largest sd?
Question 7 · Excellence
Worked example

Q7: does it pass now?

  • A forward question first.
  • Compare with 3%.
  1. μ = 1012, σ = 8now
  2. z = (1000 − 1012) ÷ 8 = −1.5
  3. P(X < 1000) = 0.066807
  4. 0.0668 > 0.03over 3%
  5. No: about 6.7% are under
Worked example

Q7: the largest σ

  • Exactly 3% under 1000 g.
  • Round down: the safe side.
  1. P(X < 1000) = 0.03the limit
  2. z = −1.8808below the mean
  3. −1.8808 = (1000 − 1012) ÷ σsubstitute
  4. −1.8808σ = −12× σ
  5. σ = −12 ÷ −1.8808÷ −1.8808
  6. = 6.3803
  7. σ = 6.38 g at mostround down
  8. 6.38: P = 0.029994under 3%
Summary

The method that solved all seven

  • Area to the left, z with its sign, solve, round safely.
StepHow
Left areaturn "above" into "below"
Find znegative below the mean
A valuex = μ + zσ
μ or σsubstitute into z, then solve
Bothtwo equations, subtract
Roundto the side the rule needs

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