IB Maths AA SL Topic 4 — Probability Distributions Paper 1 & 2 ~10 min read

Standardisation & z-Values

Imagine comparing a maths score of 85 to a French score of 78. Different tests, different averages — how do you know which is “better”? The answer is to convert both into z-values: a universal scale that tells you how many standard deviations each score is from its mean. It’s also the move that unlocks every “find the mean” or “find the SD” exam question.

📘 What you need to know

Standardisation turns any normal distribution X ∼ N(μ, σ²) into the standard normal Z ∼ N(0, 1).
The formula: Z = X − μσ (in formula booklet).
The z-value tells you how many standard deviations X is from the mean. Positive = above mean. Negative = below mean.
Going backwards: X = μ + zσ.
z-values let you compare scores from completely different distributions — same scale for everyone.
To find an unknown μ or σ: get the z-value first using Inverse Normal on N(0, 1), then substitute and solve.

What is standardisation?

Every normal distribution has its own mean and SD, so the curves all look different from each other — some shifted left or right, some narrow, some wide. Standardisation is a clever move that puts all normal distributions onto the same scale: the standard normal Z ∼ N(0, 1), which has mean 0 and SD 1.

Original

X ∼ N(μ, σ²)

any mean, any spread

standardise →

Standard

Z ∼ N(0, 1)

mean 0, SD 1

Once a value is standardised, it becomes a z-value (or “z-score”) — a single number that tells you exactly where it sits on this universal scale.

The formula and what z really means

Standardisation formula

Z = X − μσ

✓ in formula booklet

Read it like this: subtract the mean from your value, then divide by the standard deviation. The number you get tells you how many standard deviations X is away from the mean.

What each z-value means in plain English

z = −2

far below

2 standard deviations below the mean

z = 0

at the mean

exactly equal to the mean

z = 1

above

1 standard deviation above the mean

z = 2.5

unusually high

2.5 SDs above the mean — top ~1%

A z-score’s sign matters as much as its value. A z of −1.5 isn’t “less than” 1.5 — it’s a value 1.5 SDs below the mean. Whenever you calculate one, ask yourself: “Is the original value above or below the mean? Then make sure your z-score has the right sign.”

The standard normal curve

The Z ∼ N(0, 1) curve is just a normal bell curve, but centred at 0 and with the x-axis labelled in units of standard deviations. Here’s what it looks like, with the 68-95-99.7 rule already built in:

Z ∼ N(0, 1) — the standard normal

📍

Same probabilities, different numbers

An X-value 2 SDs above its mean has z = 2 — and the probability beyond it is still ≈ 2.5% (just like in the previous note). Standardising doesn’t change probabilities; it just relabels the axis in SD units so every normal distribution looks the same.

Going from X to Z and back

The same formula works in both directions — you just rearrange depending on what you know.

From X to Z (standardise)

Z = X − μσ

From Z to X (rearrange)

X = μ + zσ

If you know the z-value and want to know the original measurement, multiply z by σ to get “how far from the mean”, then add μ to get back to the original scale. The two formulas are just rearrangements of each other — don’t memorise both, derive whichever one you need.

Comparing scores from different distributions

This is one of the most useful real-world applications of z-values. Suppose two students take different tests:

Anya scored 85 on a maths test with mean 70 and SD 10.
Ben scored 78 on a French test with mean 65 and SD 6.

Whose result is more impressive? Just looking at the raw scores, Anya’s 85 looks higher. But the tests are completely different — the averages and spreads aren’t the same. Convert both into z-values to put them on the same scale:

Anya: z = (85 − 70) / 10 = 1.5 → 1.5 SDs above her mean
Ben: z = (78 − 65) / 6 ≈ 2.17 → 2.17 SDs above his mean

Ben’s score is the more impressive result, even though the raw number is lower. The bigger z-value wins.

🤔 Why does the bigger z-value mean “more impressive”?

A larger positive z means the value is further above its mean — so fewer people scored that high. A z of 2.17 puts Ben in roughly the top 1.5% of his class, while Anya’s z of 1.5 puts her in the top 7%. Same idea applies to “worse” results — a more negative z means worse-than-average performance.

🧠

The “fair scale” idea

Z-values are like converting different currencies into one shared currency. You can’t directly compare 100 yen to 100 euros, but you can convert both to dollars. Z-values do the same for normal distributions — they let you compare apples to oranges.

The main exam use: finding unknown μ or σ

This is where standardisation really earns its keep. Many Paper 2 questions give you a probability and ask you to find an unknown mean or standard deviation. The trick is to first work out the z-value, then substitute and solve.

The 3-step method

Find z using Inverse Normal on N(0, 1) with the given probability.
Substitute into Z = (X − μ) / σ with the values you know.
Solve for the unknown.

⚠️

Don’t skip Step 1 even if a z-value seems obvious

Always run Inverse Normal with the probability — don’t try to guess the z. The exact value matters: a probability of 0.0228 gives z ≈ −2 exactly, but 0.025 gives z ≈ −1.96. Both look “close to 2” but each gives a different final answer.

The harder case: both μ and σ unknown

If both the mean and standard deviation are missing, the question will give you two probabilities. Use the 3-step method twice to get two z-values and two equations, then solve them simultaneously. We’ll work through this in WE 5.

Worked examples

WE 1

Find a z-value

The heights of plants in a greenhouse are normally distributed with mean 25 cm and standard deviation 4 cm. Find the z-value of a plant that is 31 cm tall.

Use Z = (X − μ)/σ. Watch the sign — 31 is above 25, so z is positive. Identify: X = 31, μ = 25, σ = 4 Substitute: z = (31 − 25)/4 = 6/4 z = 1.5 z = 1.5 31 is 1.5 SDs above the mean — that plant is in the upper portion of the curve!

WE 2

Find X given a z-value

Exam marks are normally distributed with mean 65 and standard deviation 12. A student has a z-value of −0.75. What was their mark?

Going backwards → use X = μ + zσ. Identify: μ = 65, σ = 12, z = −0.75 Substitute: X = 65 + (−0.75)(12) = 65 − 9 = 56 mark = 56 negative z → score below mean. always check the sign matches the context!

WE 3

Compare two scores using z

In Maths, scores are N(70, 100). In English, scores are N(60, 64). Layla scored 82 in Maths and 76 in English. In which subject did she perform better, relative to her peers?

Different distributions → can’t compare raw marks. Convert both to z.maths z μ = 70, σ = √100 = 10 z = (82 − 70)/10 = 1.2english z μ = 60, σ = √64 = 8 z = (76 − 60)/8 = 2.0Bigger z → better relative performance. English (z = 2.0) looking at raw marks, 82 > 76 — but z-values reveal the real story!

WE 4

Find an unknown standard deviation

The lifetime of a battery is normally distributed with mean 200 hours. The probability that a battery lasts more than 230 hours is 0.10. Find the standard deviation, to the nearest hour.

3-step method: find z, substitute, solve.step 1 — find z P(X > 230) = 0.10 → P(X < 230) = 0.90 Inverse Normal on N(0, 1), area = 0.90: z = 1.2816…step 2 — substitute X = 230, μ = 200, z = 1.2816 1.2816 = (230 − 200)/σ = 30/σstep 3 — solve σ = 30/1.2816 = 23.41… σ ≈ 23 hours always run Inverse Normal — never guess z from memory!

WE 5

Find both μ and σ (simultaneous equations)

The mass of a particular fruit is normally distributed. Given that P(X < 80) = 0.10 and P(X > 120) = 0.05, find the values of μ and σ.

Two unknowns → need two equations. Get two z-values, build two equations, solve.step 1 — find both z-values For P(X < 80) = 0.10: Inverse Normal, area = 0.10: z₁ = −1.2816… For P(X > 120) = 0.05 → P(X < 120) = 0.95: Inverse Normal, area = 0.95: z₂ = 1.6449…step 2 — set up equations −1.2816 = (80 − μ)/σ ⇒ 80 − μ = −1.2816σ ① 1.6449 = (120 − μ)/σ ⇒ 120 − μ = 1.6449σ ② step 3 — subtract ① from ② (120 − μ) − (80 − μ) = 1.6449σ − (−1.2816σ) 40 = 2.9265σ σ = 40/2.9265 = 13.668… σ ≈ 13.7step 4 — find μ Sub σ into ①: 80 − μ = −1.2816(13.668) = −17.518… μ = 80 + 17.518 = 97.518… μ ≈ 97.5 subtracting equations kills μ — leaves you a single equation in σ!

💡 Top tips

Get the sign right. If X is below the mean, z is negative. If above, positive. The minus sign carries through every calculation.
For comparing two scores, the bigger z is the more impressive result — regardless of what the raw numbers look like.
For finding unknown μ or σ, always run Inverse Normal first to get z. Don’t try to guess from memory.
Convert “above” probabilities before using Inverse Normal: P(X > a) = 0.10 → use 0.90 for the area on the left.
If both μ and σ are unknown, you’ll need two probabilities. Subtract one equation from the other to eliminate μ, then solve for σ first.
Type σ, not σ² into the calculator — same trap as before.
Keep full decimals on z-values until the very end. Rounding early causes huge errors.
Sketch the curve and mark where your z sits — it tells you instantly if your sign is correct.

⚠ Common mistakes

Forgetting the negative sign when X is below the mean. A z of −1.5 is very different from +1.5.
Using the wrong tail. “Above 230” with probability 0.10 means area on the LEFT is 0.90 — not 0.10.
Mixing up X and z in the formula. X is the original measurement; z is the standardised version.
Treating the bigger raw score as automatically better. Without z-values, you can’t compare across different distributions.
Skipping Inverse Normal and trying to guess z from the empirical rule. The 68-95-99.7 rule gives rough percentages — exact problems need exact z-values.
Using σ² in the formula. The denominator is the standard deviation σ, not the variance σ².
Rounding z too early. Always carry at least 4 decimal places of z through your working.
Solving simultaneous equations the slow way. Subtraction kills μ in one step — use that, don’t expand and rearrange forever.

🎉 You’ve got the full toolkit now: Normal Cdf for probabilities, Inverse Normal for values, and standardisation for everything else. Together these solve every normal distribution question on Paper 1 and Paper 2 — including the trickier ones where the question hides the mean or SD and asks you to dig it out.

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