NCEA Level 2Normal distributionAS 91267

The normal distribution: probabilities and z-values

What this lesson covers

  • A normal distribution is symmetric about its mean, and its spread is measured by the standard deviation.
  • A z-value counts how many standard deviations a value is from the mean.
Key idea

The bell curve

  • Many measurements pile up around the mean.
  • Fewer and fewer as you go further out.
  • Apple weights: mean 150 g, sd 12 g.
The bell curve
Key idea

Shape, centre and spread

  • Symmetric about the mean.
  • The standard deviation sets the width.
  • The total area under the curve is 1.
FeatureIsApples
Shapesymmetric bellmean = median = mode
Centrethe mean μ150 g
Spreadstandard deviation σ12 g
Total area1all apples
Key idea

A z-value counts standard deviations

  • Above the mean: positive.
  • Below the mean: negative.
  1. z = (x − μ) ÷ σ
  2. z = (165 − 150) ÷ 12x = 165
  3. = 15 ÷ 12 = 1.25
  4. z = (140 − 150) ÷ 12x = 140
  5. = −10 ÷ 12 = −0.8333below: negative
Worked example

P(X < 165)

  • Shade everything to the left of 165.
  • More than half the area, so over 0.5.
  • 0.8944.
P(X < 165)
Worked example

P(140 < X < 160)

  • Shade between the two values.
  • z from −0.8333 to 0.8333.
  • 0.5953.
P(140 < X < 160)
Worked example

P(X > 175)

  • A thin tail on the right.
  • z = 2.0833: over two sd above.
  • Small, as the picture says: 0.0186.
P(X > 175)
Worked example

The working: below and between

  • Calculator: normal cdf.
  • Lower and upper bounds.
  1. z = (165 − 150) ÷ 12 = 1.25
  2. lower −1E99, upper 165μ = 150, σ = 12
  3. P(X < 165) = 0.8944
  4. z = (140 − 150) ÷ 12 = −0.8333
  5. z = (160 − 150) ÷ 12 = 0.8333
  6. lower 140, upper 160μ = 150, σ = 12
  7. P(140 < X < 160) = 0.5953
Worked example

The working: above

  • Above: upper bound 1E99.
  • Check: 1 − P(X < 175).
  1. z = (175 − 150) ÷ 12
  2. = 25 ÷ 12 = 2.0833
  3. lower 175, upper 1E99μ = 150, σ = 12
  4. P(X > 175) = 0.0186
  5. 1 − P(X < 175) = 1 − 0.9814
  6. = 0.0186
Key idea

Does it match the picture?

  • Shaded more than half? Answer over 0.5.
  • A thin tail? A small answer.
  • A mismatch means a wrong sign or side.
Worked example

Expected numbers from a normal model

  • Expected = n × probability.
  • Round to whole apples.
  1. P(X > 175) = 0.0186from before
  2. expected = 1000 × 0.0186
  3. = 18.6
  4. about 19 applesapples are whole
Key idea

Where the marks are

  • Achieved: a basic normal probability.
  • Merit: in context, with z signs right.
  • Excellence: annotated diagrams and interpretation.
Summary

What to take away

  • Draw it, shade it, then calculate it.
Draw itshade the area you want
z(x − μ) ÷ σ, with a sign
Below the meannegative z
Probabilitythe shaded area
Checkmore or less than 0.5?
Expected numbern × probability
Your turn

Your turn: four questions

  • Sketch and shade each one first.
  • Pause here and answer each one.
  • Four decimal places.
Question
DataHeights of Year 12 students: normal, mean 170 cm, sd 8 cm.
1P(a student is taller than 180 cm)?
2P(a student is shorter than 160 cm)?
3P(between 165 and 175 cm)?
4Of 500 students, how many are expected to be over 180 cm?
Show answer

1. z = 1.25, so 0.1056. 2. z = −1.25, so 0.1056 (by symmetry, the same as question 1). 3. z from −0.625 to 0.625: 0.4680. 4. 500 × 0.1056 = 52.8, about 53 students.

Worked example

The answers: questions 1 and 2

  1. 1. z = (180 − 170) ÷ 8 = 1.25
  2. lower 180, upper 1E99μ = 170, σ = 8
  3. P(X > 180) = 0.1056
  4. 2. z = (160 − 170) ÷ 8 = −1.25
  5. lower −1E99, upper 160μ = 170, σ = 8
  6. P(X < 160) = 0.1056symmetry with 1
Worked example

The answers: questions 3 and 4

  1. 3. z = (165 − 170) ÷ 8 = −0.625
  2. z = (175 − 170) ÷ 8 = 0.625
  3. lower 165, upper 175μ = 170, σ = 8
  4. P(165 < X < 175) = 0.4680
  5. 4. expected = 500 × 0.1056
  6. = 52.8
  7. about 53 studentsstudents are whole

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