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 1

Compute 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 2

Recover 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 3

Compare 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 4

Standard 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 5

Standardise 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 6

Compare 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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