IB Maths AA SL Topic 4 — Stats & Probability Paper 2 🎯 Skill ~4 min practice

AA SL Z-Values & Standardising skills

A z-value tells you how many standard deviations a data point is from the mean. Standardise to compare values from different distributions, or to find an unknown μ or σ. One formula, two directions — and the GDC handles the inverse.

The Method

z = x − μσ number of standard deviations x is from the mean · ✓ in formula booklet

Direction 1

Standardise: x → z

z = (x − μ) / σ

use when given a raw value, want z-score

Direction 2

Un-standardise: z → x

x = μ + zσ

use when given a z-score, want the raw value

Identify μ, σ, and either x or z — read the question carefully.
Pick the direction — going to z (standardise), or going from z back to x (un-standardise).
Find unknown z-values with invNorm on the GDC: invNorm(area, 0, 1) — μ=0, σ=1 for the standard normal.

Standardising shifts the mean to 0

Every normal distribution becomes the same standard normal Z ~ N(0, 1) after standardising. x = 85 on the original scale becomes z = 1.5 — meaning “1.5 standard deviations above the mean”.

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What a z-score means at a glance

z = 0 sits exactly at the mean. z = 1 is one SD above. z = −2 is two SDs below. The sign tells you direction; the magnitude tells you how far.

How to find a z-value with invNorm

TI-84 Plus

Press 2nd → VARS (DISTR menu)
Choose 3: invNorm(
Enter: area to left, μ=0, σ=1
Press ENTER → that’s your z

Casio fx-CG50

Open STAT menu → DIST (F5)
Choose NORM (F1) → InvN (F3)
Set Tail: Left, Area = your value
Set σ = 1, μ = 0 → execute

Worked examples

WE 1 EASY

A test has mean 60 and SD 8. A student scores 76. Find the student’s z-score.

step 1 — identify μ = 60, σ = 8, x = 76step 2 — apply z = (x − μ) / σ z = (76 − 60) / 8 = 16 / 8 = 2z = 2 z = 2 means the score is 2 standard deviations above the mean — top end!

WE 2 MEDIUM

A normal distribution has mean 100 and SD 15. The 80th percentile corresponds to z ≈ 0.8416. Find the value at the 80th percentile (3 sf).

step 1 — identify direction have z, want x → un-standardisestep 2 — apply x = μ + zσ x = 100 + (0.8416)(15) = 100 + 12.624 = 112.624…80th percentile ≈ 113 (3 sf) when z is given, multiply by σ then add μ — never the other way!

WE 3 HARD

Heights are normally distributed with mean 170 cm and unknown SD. The top 10% are taller than 183 cm. Find σ to 3 sf.

step 1 — find z for top 10% P(Z > z) = 0.10 → P(Z < z) = 0.90 z = invNorm(0.90, 0, 1) ≈ 1.2816step 2 — set up equation z = (x − μ) / σ 1.2816 = (183 − 170) / σstep 3 — solve for σ σ = 13 / 1.2816 ≈ 10.143…σ ≈ 10.1 cm (3 sf) find z first, then rearrange the formula to solve for the unknown!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always sketch the bell curve when finding an unknown μ or σ.

Q1 EASY A distribution has mean 50 and SD 10. Find the z-score of x = 65. Show answer ▼Hide answer ▲

z = (65 − 50) / 10 = 15/10 z = 1.5

Q2 EASY A distribution has mean 200 and SD 25. Find the z-score of x = 180. Show answer ▼Hide answer ▲

z = (180 − 200) / 25 = −20/25 z = −0.8 below the mean → negative z. always check the sign!

Q3 MEDIUM A distribution has mean 75 and SD 12. The z-value for a particular score is 1.4. Find the score (3 sf). Show answer ▼Hide answer ▲

un-standardise: x = μ + zσ x = 75 + 1.4 × 12 = 75 + 16.8 x = 91.8

Q4 MEDIUM Test scores are N(μ, 10²). The bottom 5% score below 50. Find μ (3 sf). Show answer ▼Hide answer ▲

step 1 — z for bottom 5% z = invNorm(0.05, 0, 1) ≈ −1.6449 step 2 — set up −1.6449 = (50 − μ) / 10 −16.449 = 50 − μ μ = 50 + 16.449 μ ≈ 66.4 (3 sf)

Q5 HARD In maths, Lia scores 78 (mean 70, SD 6). In English, she scores 82 (mean 75, SD 10). In which subject did she perform relatively better? Show answer ▼Hide answer ▲

standardise both scores z(maths) = (78 − 70) / 6 ≈ 1.333 z(English) = (82 − 75) / 10 = 0.7 compare z-scores 1.333 > 0.7 Lia performed relatively better in maths z-scores let you compare different distributions on the same scale — the whole point of standardising!

⚠ Common mistakes

Wrong direction. “Find the z-score for x = 80” → standardise (use z = (x − μ)/σ). “Find x given z = 1.5” → un-standardise (use x = μ + zσ).
Sign errors with negative z. If x is below the mean, z is negative. Don’t drop the minus.
Mixing up σ and σ². The formula needs σ (standard deviation), not variance.
Looking up z in old printed tables. The IB expects you to use the GDC’s invNorm — much faster and more accurate.
Confusing “top X%” with “below X%”. Top 10% means area = 0.10 in the right tail → use 0.90 area in invNorm.
Forgetting to set μ=0 and σ=1 when finding a pure z-value with invNorm. Otherwise you get an x-value, not a z.

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Want the theory?

Read the full Standardisation & Z-Values notes for why z-scores are useful, the link to percentiles, and how the standard normal Z ~ N(0, 1) underpins every normal calculation.

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