NCEA Level 2Inverse normalAS 91267

Inverse normal: finding a value, the mean or the standard deviation

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

Backwards: from area to value

  • So far: a value in, a probability out.
  • Inverse normal: a probability in, a value out.
  • "Top 10%", "fastest 15%", "only 2% under".
Worked example

The heaviest 10% of apples

  • Shade the top 10% first.
  • The cut-off is unknown: call it x.
  • z = 1.2816, so x = 165.38 g.
The heaviest 10% of apples
Worked example

The working

  • Area to the left: 0.9.
  • Then x = μ + zσ.
  1. P(X < x) = 1 − 0.1 = 0.9top 10%
  2. left tail, area 0.9σ = 1, μ = 0
  3. z = 1.2816
  4. z = (x − μ) ÷ σ, so x = μ + zσ
  5. x = 150 + 1.2816 × 12substitute
  6. = 150 + 15.3792
  7. = 165.38 g
Worked example

The lightest 5%: a negative z

  • Below the mean: z < 0.
  • Check: x is below 150.
  1. left tail, area 0.05σ = 1, μ = 0
  2. z = −1.6449below the mean
  3. x = μ + zσ
  4. = 150 + (−1.6449) × 12
  5. = 150 − 19.7388
  6. = 130.26 g
  7. 130.26 < 150below the mean
Worked example

An unknown standard deviation

  • Bottles: mean 500 ml, 2% under 490 ml.
  • z for 0.02 is −2.0537.
  • σ = 10 ÷ 2.0537 = 4.869 ml.
An unknown standard deviation
Worked example

The steps for σ

  • z from the probability.
  • Substitute, solve for σ.
  1. left tail, area 0.02σ = 1, μ = 0
  2. z = −2.0537490 is below 500
  3. z = (x − μ) ÷ σ
  4. −2.0537 = (490 − 500) ÷ σsubstitute
  5. −2.0537 = −10 ÷ σ
  6. −2.0537σ = −10× σ
  7. σ = −10 ÷ −2.0537÷ −2.0537
  8. = 4.869 ml
  9. σ > 0as it must be
Worked example

An unknown mean

  • σ known, μ unknown.
  • Same equation: solve for μ.
  1. σ = 4, P(X < 490) = 0.01given
  2. left tail, area 0.01σ = 1, μ = 0
  3. z = −2.3263below the mean
  4. −2.3263 = (490 − μ) ÷ 4substitute
  5. −9.3052 = 490 − μ× 4
  6. μ = 490 + 9.3052rearrange
  7. = 499.31 mlround up
Key idea

The sign of z does the work

  • A value below the mean has negative z.
  • Get the sign wrong and σ comes out negative.
  • A negative σ is impossible: a free check.
Key idea

Where the marks are

  • Achieved: find the z-value.
  • Merit: use it for a value or a missing σ.
  • Excellence: interpret the answer in context.
Summary

What to take away

  • Area first, z with its sign, then solve.
Startwith the area, not the value
Left areathe calculator's inverse normal
Sign of znegative below the mean
A valuex = μ + zσ
Unknown σ or μz = (x − μ) ÷ σ, then solve
Checkdoes the answer sit on the right side?
Your turn

Your turn: four questions

  • Sketch and shade each one first.
  • Pause here and answer each one.
  • Watch the sign of z.
Question
DataRace times: normal, mean 42 min, sd 6 min.
1Fastest 15% get a medal. What time must they beat?
2The slowest 5% take longer than what time?
3Another race: mean 60 min, 10% take over 75 min. Find σ.
4A third: σ = 3 min, and 2.5% take over 20 min. Find μ.
Show answer

1. z = −1.0364: 42 − 1.0364 × 6 = 35.78 min. 2. z = 1.6449: 42 + 1.6449 × 6 = 51.87 min. 3. 1.2816 = 15 ÷ σ, so σ = 11.70 min. 4. 1.96 = (20 − μ) ÷ 3, so μ = 20 − 5.88 = 14.12 min.

Worked example

The answers: questions 1 and 2

  1. 1. left area 0.15: z = −1.0364
  2. x = 42 − 1.0364 × 6
  3. = 42 − 6.2184 = 35.78 min
  4. 2. left area 0.95: z = 1.6449
  5. x = 42 + 1.6449 × 6
  6. = 42 + 9.8694 = 51.87 min
Worked example

The answers: questions 3 and 4

  1. 3. left area 0.9: z = 1.2816
  2. 1.2816 = (75 − 60) ÷ σsubstitute
  3. 1.2816σ = 15× σ
  4. σ = 15 ÷ 1.2816 = 11.70 min
  5. 4. left area 0.975: z = 1.96
  6. 1.96 = (20 − μ) ÷ 3substitute
  7. 5.88 = 20 − μ× 3
  8. μ = 20 − 5.88 = 14.12 min

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