Using the normal distribution

Standardising: the z-score

There are infinitely many normal distributions — one per pair (μ , σ). A change of variable reduces them all to one.

The standardised variable

        X - μ
   Z = -------            then  Z ~ N(0 , 1)
          σ

Two operations: centring (subtracting μ puts the centre at 0), then scaling (dividing by σ sets the unit to one standard deviation).

X ~ N(100 ; 15²)                    Z ~ N(0 ; 1)
        __                                __
      _/  \_             ------>        _/  \_
   __/      \__                      __/      \__
  ---|---|---|---                   ---|---|---|---
    85 100 115                        -1   0   1

The resulting number z is called the z-score: the distance to the mean, measured in standard deviations.

X = 130 for N(100 ; 15²)   ->   z = (130 - 100)/15 = 2

  "130 is two standard deviations above the mean"

Why this is useful

The z-score makes measurements in different units comparable:

A pupil scores 14/20 in maths (class mean 11, standard deviation 2)
            and 15/20 in French (mean 13, standard deviation 4)

maths  : z = (14 - 11)/2 = +1.5
French : z = (15 - 13)/4 = +0.5

  ->  the better RELATIVE performance is in maths, despite the lower mark

This is the principle of all standardisation: standardised marks, test scores, paediatric growth charts.

The values worth knowing

Once in N(0,1), probabilities are read from a table or a calculator. A few landmarks suffice in practice:

z       P(Z ≤ z)         corresponding central region
-----   ---------        ----------------------------
1.00     0.841           68 %   within [-1 ; 1]
1.645    0.950           90 %   within [-1.645 ; 1.645]
1.96     0.975           95 %   within [-1.96 ; 1.96]
2.576    0.995           99 %   within [-2.576 ; 2.576]

The number 1.96 is the most used in all of statistics: it is what appears in 95 % confidence intervals.

The computational method

1. translate the wording into an inequality on X
2. standardise: subtract μ, divide by σ
3. read off the probability for Z
4. use symmetry if needed

X ~ N(100 ; 15²).  What proportion exceeds 130?

  P(X > 130) = P(Z > 2) = 1 - P(Z ≤ 2) = 1 - 0.977 = 0.023

  about 2.3 % of the population.

The two symmetry reflexes that save half the table:

P(Z ≤ -z) = 1 - P(Z ≤ z)
P(-z ≤ Z ≤ z) = 2 P(Z ≤ z) - 1

Summary