Definition D1
Let dᵢ = xᵢ − x̄. Then s² = (Σdᵢ²)n⁻¹ is called the variance and s = √(s²) the standard deviation.
Proof.
(1)Σaᵢ denotes a₁ + ⋯ + aₙ, u − v denotes u + (−v), and t² denotes t·t. By associativity and commutativity of addition, this sum is uniquely determined regardless of the order and grouping of terms.convention, A1, A2
(2)Since n ≥ 1, we have n ≠ 0 and n⁻¹ exists.A13, A8
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Lemma L1
Setting wᵢ = xᵢ − x̄, the data w have mean 0.
Proof.
(1)Σwᵢ = Σ(xᵢ − x̄) = 0A11
(2)n w̄ = Σwᵢ = 0 = n × 0A12, (1), A14
(3)Since n ≠ 0, cancellation gives w̄ = 0(2), A13, A9
(4)The standard deviation appears nowhere in this computation.(1), (2), (3)
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Definition D2
Suppose 0 < s. The data obtained by zᵢ = (xᵢ − x̄)s⁻¹ = dᵢ s⁻¹ are called the standardized data.
Proof.
(1)Since 0 < s, we have s ≠ 0 and s⁻¹ exists.A17, A20, A8
(2)For data whose values are all equal, every dᵢ is 0, so s = 0 and this definition cannot be applied.D1, A17, convention
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Theorem T1
The standardized data have mean 0 and standard deviation 1.
Proof.
(1)Σzᵢ = Σ(dᵢ s⁻¹) = s⁻¹ ΣdᵢD2, A10, A6
(2)Since Σdᵢ = 0, Σzᵢ = s⁻¹ × 0 = 0(1), A11, A14
(3)n z̄ = Σzᵢ = 0 = n × 0 and n ≠ 0, so by the same method as L1, z̄ = 0A12, (2), A14, A13, A9, L1
(4)Since z̄ = 0, −z̄ = 0, and hence zᵢ − z̄ = zᵢ. The deviations of the z data are the zᵢ themselves.(3), A4, A3
(5)zᵢ² = (dᵢ s⁻¹)² = dᵢ²(s⁻¹)²D2, A5, A6
(6)Σzᵢ² = (s⁻¹)² Σdᵢ²(5), A10
(7)Multiplying both sides of s² = (Σdᵢ²)n⁻¹ by n gives Σdᵢ² = n s²D1, E, A5, A6, A7, A8, A13
(8)s_z² = (Σzᵢ²)n⁻¹ = ((s⁻¹)²(n s²))n⁻¹ = (s⁻¹)²s² = (s⁻¹s)² = 1(4), (6), (7), D1, A5, A6, A7, A8
(9)Since 1 = 1² and 0 ≤ 1, s_z = √(s_z²) = √(1²) = 1(8), D1, A15, A7
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Theorem T2
Values from two data sets with different centers and scales become comparable on the same ruler once converted to z.
Proof.
(1)Data A are 54, 58, 60, 62, 66. The sum is 300 and n = 5, so x̄ = 60A12, N
(2)dᵢ = −6, −2, 0, 2, 6 and Σdᵢ² = 80, so s² = 16, s = 4(1), D1, A15, N
(3)Data B are 11, 17, 20, 23, 29. The sum is 100, so x̄ = 20A12, N
(4)dᵢ = −9, −3, 0, 3, 9 and Σdᵢ² = 180, so s² = 36, s = 6(3), D1, A15, N
(5)The value 62 in A gives z = (62 − 60)·4⁻¹ = 0.5(1), (2), D2, N
(6)The value 29 in B gives z = (29 − 20)·6⁻¹ = 1.5(3), (4), D2, N
(7)In raw values 62 exceeds 29, but in z, 0.5 is below 1.5.(5), (6), N
(8)By T1 the two z data sets share mean 0 and standard deviation 1, so they are read on the same ruler.T1
(9)Comparing raw values directly is meaningless since the two data sets differ in center and scale; comparing z is meaningful.(7), (8)
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Corollary C
But standardization does not change the shape of the distribution.
Proof.
(1)Since 0 < s, 0 < s⁻¹A17, A19
(2)If xᵢ < xⱼ, adding −x̄ to both sides gives dᵢ < dⱼ, and multiplying by s⁻¹ gives zᵢ < zⱼ(1), D2, A16, A18
(3)Therefore standardization does not change the order of the data, and each value keeps its seat.(2), A20
(4)Consider the data 0, 0, 0, 0, 10. The sum is 10, so x̄ = 2, dᵢ = −2, −2, −2, −2, 8, and Σdᵢ² = 80, so s² = 16, s = 4.A12, D1, A15, N
(5)The median of these data is 0 and the mean is 2, so the mean exceeds the median. The data are skewed toward the large side.(4), N
(6)The zᵢ are −0.5, −0.5, −0.5, −0.5, 2, and as T1 states, the mean is 0 and the standard deviation is 1.(4), D2, T1, N
(7)The median of the z data is −0.5, so the mean 0 still exceeds the median. The skew remains.(6), N
(8)The z of data A are −1.5, −0.5, 0, 0.5, 1.5, so three data points have z above −0.5, while in the data above only one does.(2), (6), D2, N
(9)Therefore the same z occupies different seats in different data sets, and a z alone does not reveal a rank.(3), (8)
(10)What standardization changes are only the center and the scale. Skewed data remain skewed after standardization.(3), (7), (9)
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