Key Idea: This topic links a value to its probability in a normal distribution: z-values measure how far a value sits from the mean, and the inverse normal turns a probability back into a value — a heavy Paper-2 GDC skill.
📏 z-values: standardising
- the value you are standardising
- the mean of the distribution
- the standard deviation
- number of standard deviations from the mean (sign = side)
Positive z → above the mean; negative z → below. z is unitless (a count of standard deviations), so it lets you compare results from different distributions — the larger z is the better relative result. Standardising turns any X ~ N(μ, σ²) into the standard normal Z ~ N(0, 1).
🔁 Inverse normal: area → value
Important: invNorm(area, μ, σ) takes the lower / left-tail probability P(X < x) and returns the value x. Convert anything else first: • P(X > x) = p → use the left area 1 − p. • Middle proportion → split the leftover equally into two tails.
| Asked for | Left area to enter | GDC call |
|---|---|---|
| P(X < x) = 0.90 | 0.90 | invNorm(0.90, μ, σ) |
| P(X > x) = 0.10 (top 10%) | 1 − 0.10 = 0.90 | invNorm(0.90, μ, σ) |
| Central 90% | 0.05 and 0.95 | invNorm(0.05,…) & invNorm(0.95,…) |
| Unknown σ or μ | use the given p | z = invNorm(p, 0, 1) |
Tip: Find the standardised z = invNorm(p, 0, 1), then rearrange z = (x − μ)/σ: σ = (x − μ)/z or μ = x − zσ. You cannot feed the unknown σ straight into invNorm.
✏️ IB-style worked examples
IB-style question — find a z-value
The heights of seedlings are X ~ N(60, 8²) mm. Find the z-value of a seedling that is 72 mm tall.
Step by step:
Substitute into z = (x − μ)/σ.
Work it out.
z = 1.5 — the seedling is 1.5 standard deviations above the mean.
IB-style question — compare across distributions
In Maths (mean 70, sd 5) Leah scores 80. In History (mean 60, sd 10) she scores 75. Relative to each class, which result is stronger?
Step by step:
Standardise each score.
The larger z is further above its own mean.
Maths is the stronger result (z = 2 beats z = 1.5), even though 75 < 80 in raw marks.
IB-style question — inverse normal (left tail)
The mass of an apple is X ~ N(50, 10²) g. Find the value x such that P(X < x) = 0.9.
Step by step:
P(X < x) is already a left area — use invNorm.
Read the GDC result.
x ≈ 62.8 g.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
IB-style question — 'greater than' probability
For the same apples X ~ N(50, 10²) g, find x such that P(X > x) = 0.05 (the heaviest 5%).
Step by step:
invNorm needs the LEFT area — convert.
Apply invNorm with 0.95.
x ≈ 66.4 g.
IB-style question — find an unknown σ
X ~ N(100, σ²) and P(X < 118) = 0.8. Find the standard deviation σ.
Step by step:
Get the standardised z for a left area of 0.8.
Rearrange z = (x − μ)/σ to σ = (x − μ)/z.
σ ≈ 21.4.
IB-style question — find an unknown μ
X ~ N(μ, 5²) and P(X < 20) = 0.1. Find the mean μ.
Step by step:
Get z for a left area of 0.1 (it is negative).
Rearrange to μ = x − zσ.
μ ≈ 26.4 — since 20 lies in the lower tail, the mean is above it.
Important: invNorm always works on the lower / left-tail probability. 'Top 10%' is NOT area 0.10 — it is P(X > x) = 0.10, so the left area is 1 − 0.10 = 0.90. Convert every 'greater than' or 'top %' to a left area before you type it in.
Tap each card to reveal the answer.
X ~ N(60, 8²). z-value of x = 50? z = −1.25 — (50 − 60)/8; the minus sign means below the mean.
What is z = 0? Exactly the mean — standardising sends μ to 0 in N(0, 1).
Which GDC area for P(X > x) = 0.25? Left area 0.75 — use invNorm(0.75, μ, σ); 1 − 0.25 = 0.75.
Left areas for the central 80%? 0.10 and 0.90 — each tail is (1 − 0.80)/2 = 0.10.
Unknown σ: which invNorm gives z? invNorm(p, 0, 1) — use the standard normal, then σ = (x − μ)/z.
P(X < 20) = 0.1 — is z positive or negative? Negative (z ≈ −1.28) — any left area below 0.5 gives a negative z.
Exam Tips
- z = (x − μ)/σ: positive is above the mean, negative below; z has no units.
- Compare results from different distributions by z, never by raw scores.
- invNorm(area, μ, σ) uses the LEFT-tail area — convert 'greater than' to 1 − p.
- For an unknown σ or μ, get z = invNorm(p, 0, 1), then use z = (x − μ)/σ.
- A left area below 0.5 returns a negative z — keep the sign in your rearrangement.