IB Maths AA HL
Topic 4 — Statistics & Probability
Paper 1 & 2
~6 min read
Standardisation of Normal Variables & z-Values
Standardisation converts any normal variable X into the standard normal Z by subtracting the mean and dividing by the SD: z = (x − μ) / σ. The z-value tells you how many standard deviations a value sits above (positive) or below (negative) its mean. Z’s universal scale lets you compare values from completely different distributions — like exam scores from different subjects.
📘 What you need to know
- Standard normal: Z ~ N(0, 1²) — mean 0, variance 1, standard deviation 1.
- Standardisation formula: z = x − μσ — given in the booklet.
- Rearranged: x = μ + σz — useful when going from z back to x.
- z is unitless: it tells you how many SDs a value lies from its mean.
- Sign of z: positive → above the mean; negative → below the mean.
- Φ(z) = P(Z < z) — the cumulative function of the standard normal.
- Probabilities transfer: P(a < X < b) = P((a−μ)/σ < Z < (b−μ)/σ).
- Comparison tool: bigger z = farther above mean → relatively higher value, regardless of original units.
The standard normal Z ~ N(0, 1)
The standard normal distribution is just a normal distribution with mean 0 and SD 1. Every other normal can be transformed onto Z by a horizontal shift (subtract the mean) and a horizontal stretch (divide by the SD). After standardisation, all normal probabilities live on the same universal scale.
Standardisation formula
z = x − μσ
From x to z
z = (x − μ) / σ
how many SDs above/below the mean
From z to x
x = μ + σz
recover the original value from a z-score
What a z-value tells you
| z-value | Meaning | Approx. percentile |
|---|
| z = 0 | exactly at the mean | 50th |
| z = 1 | 1 SD above mean | ~84th |
| z = −1 | 1 SD below mean | ~16th |
| z = 2 | 2 SDs above mean | ~97.5th |
| z = −2 | 2 SDs below mean | ~2.5th |
| z = 3 | 3 SDs above mean | ~99.85th |
Useful identities: P(Z < −z) = P(Z > z) by symmetry; P(Z > z) = 1 − Φ(z); and Φ(0) = 0.5 always.
Comparing values from different distributions
Raw scores from different distributions can’t be compared directly — a “78 in Maths” and an “84 in Physics” are on entirely different scales. Convert each to a z-value first; whichever has the larger z is relatively better, regardless of units or original scales.
🧭 Recipe — work with z-values
- Identify μ and σ for the distribution.
- Compute z using z = (x − μ) / σ; the sign tells you which side of the mean.
- To find x from z, rearrange to x = μ + σz.
- For probabilities, P(X < x) = P(Z < z) = Φ(z) — use the GDC.
- To compare across distributions, compute z for each value; the larger z is relatively higher.
Worked examples
WE 1Compute a z-value from x, μ, σ
A Maths test score follows X ~ N(70, 8²). Find the z-value of the score x = 82.
Identify μ = 70, σ = 8
Apply z = (x − μ)/σ
z = (82 − 70) / 8
= 12 / 8 = 1.5
z = 1.5
positive z → score is above the mean; specifically 1.5 SDs above
WE 2Recover x from z using x = μ + σz
Heights of students follow H ~ N(165, 6²) cm. A student has a z-value of −1.25. Find their height.
Use the rearranged form x = μ + σz
x = 165 + 6(−1.25)
= 165 − 7.5 = 157.5
Height = 157.5 cm
negative z → below mean; 7.5 cm shorter than the mean of 165
WE 3Compare values from different distributions
Anna scored 78 in a Maths exam where scores are distributed N(70, 5²). Ben scored 84 in a Physics exam where scores are distributed N(75, 8²). Whose performance was relatively better?
Compute z for each
Anna: z = (78 − 70) / 5 = 8/5 = 1.6
Ben: z = (84 − 75) / 8 = 9/8 = 1.125
Compare z-values
Anna’s z (1.6) > Ben’s z (1.125)
→ Anna is further above her mean (in SD units)
Anna performed relatively better
Ben’s RAW score is higher (84 > 78) but z standardises across distributions
WE 4Standard normal probabilities
For the standard normal Z ~ N(0, 1), find: (a) P(Z < 1.5); (b) P(Z > 0.8); (c) P(−1 < Z < 2).
(a) P(Z < 1.5) = Φ(1.5)
= 0.9332
(b) P(Z > 0.8) = 1 − Φ(0.8)
= 1 − 0.7881 = 0.2119
(c) P(−1 < Z < 2) = Φ(2) − Φ(−1)
= 0.9772 − 0.1587 = 0.8186
(a) 0.933; (b) 0.212; (c) 0.819 (3 sf)
all of these come straight from a GDC’s NormCdf with μ = 0, σ = 1
WE 5Standardise to find a probability
X ~ N(50, 16). Use standardisation to find P(46 < X < 58).
σ = √16 = 4
Standardise both bounds
z₁ = (46 − 50)/4 = −1
z₂ = (58 − 50)/4 = 2
Translate to a Z-probability
P(46 < X < 58) = P(−1 < Z < 2)
= Φ(2) − Φ(−1)
= 0.9772 − 0.1587 = 0.8186
P(46 < X < 58) ≈ 0.819 (3 sf)
when z values are clean (±1, ±2), standardisation is faster than a direct NormCdf
WE 6Compare two scores + find percentile
A student scores 88 on an English exam where scores follow N(72, 10²) and 65 on a Geography exam where scores follow N(50, 6²).
(a) Calculate the z-value for each score. (b) In which subject did the student perform relatively better? (c) For the better-performing subject, find the percentile rank (i.e. what percentage of students scored less).
(a) Compute z for each
English: z = (88 − 72)/10 = 1.6
Geography: z = (65 − 50)/6 = 2.5
(b) Compare
2.5 > 1.6 → Geography is relatively better
(c) Percentile = Φ(2.5) × 100%
Φ(2.5) = 0.9938
Percentile ≈ 99.4%
(a) zₑ = 1.6, z_g = 2.5; (b) Geography; (c) ~99.4th percentile
English raw score (88) is higher, but z says Geography is relatively much better
💡 Top tips
- z is unitless — drop the units when computing it; the answer is “in SDs”.
- Bigger z = farther from mean — use this to compare values across different distributions.
- For positive z, P(Z < z) > 0.5; for negative z, P(Z < z) < 0.5. Quick sanity check.
- Symmetry shortcut: P(Z < −z) = 1 − P(Z < z) = P(Z > z).
- Standardisation works in both directions: from x to z by dividing, from z to x by multiplying then adding back.
⚠ Common mistakes
- Subtracting in the wrong order — it’s x − μ, NOT μ − x; the sign of z is essential.
- Dividing by variance instead of SD — use σ, not σ².
- Comparing raw scores across distributions — that’s apples-to-oranges; standardise first.
- Confusing z with x — z has no units; x carries the original units (cm, kg, etc.).
- Forgetting Z’s parameters: Z has mean 0 and SD 1, NOT mean 0 and variance 0.
Next: Finding Unknown Parameters. So far we’ve assumed μ and σ are given — but exam questions often give a probability and ask you to deduce one (or both) of them. Standardisation is the workhorse here: convert each given probability into a z-value, then solve a simple linear equation in μ and σ.
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