IB Maths AA SLTopic 4 — Probability DistributionsPaper 2 (GDC)~11 min read
Calculations with Normal Distribution
Now that you know what the bell curve looks like, it’s time to actually find probabilities from it. Two calculator functions will get you through every Paper 2 question on this topic — one finds probabilities, the other works backwards to find values. Get those down and you’ll handle anything they throw at you.
📘 What you need to know
Use Normal Cdf on your GDC to find probabilities like P(a < X < b).
Use Inverse Normal to go backwards — given a probability, find the x value.
For “less than” or “greater than”, use a very large number (like 10⁹⁹) for the missing bound.
The calculator wants standard deviation σ, not variance σ². Always type the unsquared value.
If the question gives you a probability and asks for an unknown μ or σ, use standardisation: Z = X − μσ.
Sketch the curve for every question. It takes 5 seconds and saves you from silly errors.
Two functions, opposite directions
The whole topic comes down to choosing between two GDC functions. The trick is knowing which way the question is going.
Normal Cdf · “find a probability”
x values→probability
“Given a and b, what’s P(a < X < b)?”
Inverse Normal · “find a value”
probability→x value
“P(X < x) = 0.85. What is x?”
If you’re given numbers and asked for a chance — that’s Normal Cdf. If you’re given a chance and asked for a number — that’s Inverse Normal. The direction of the question tells you the function.
Using Normal Cdf — finding a probability
Normal Cdf gives you the area under the curve between two x-values. It needs four pieces of info: the lower bound, the upper bound, the mean, and the standard deviation.
Let’s say X ∼ N(50, 64) — so μ = 50 and σ = √64 = 8. Here’s what you’d type in to find P(40 < X < 60):
Normal CdfP(40 < X < 60)
Lower (a)40
Upper (b)60
μ (mean)50
σ (SD)8
P(40 < X < 60) ≈ 0.789
Normal CdfP(X < 60) — one-sided
Lower (a)−10⁹⁹
Upper (b)60
μ (mean)50
σ (SD)8
P(X < 60) ≈ 0.894
📍
The “infinity trick” for one-sided probabilities
For P(X < a), use a really small lower bound like −10⁹⁹ or −1E99 on your calculator. For P(X > a), use a really big upper bound like 10⁹⁹ or 1E99. The calculator treats this as “all the way to infinity”.
Different calculators use different keys for this. Casio uses EXP or ×10ˣ. TI uses EE (above the comma key). If those still don’t work, just type a huge number like 9999999 — it’ll be far enough out into the tail to give the right answer.
Sketch the curve first, every time
This is the single best habit you can build for this topic. Before you touch the calculator, draw a quick bell curve, mark μ at the centre, mark the bounds, and shade the region you want. It takes 5 seconds and prevents almost every common mistake.
A 5-second sketch saves you
🤔 Why does sketching help so much?
If your shaded region looks tiny but you got an answer like 0.85, something’s wrong — go back and check. If your region covers more than half the curve but you got 0.20, also wrong. The picture acts as a built-in sanity check on your calculator answer.
Inverse Normal — going backwards
Sometimes a question gives you a probability and asks you to find the x-value that produces it. For example: “The slowest 10% of runners take longer than what time?” That’s a job for Inverse Normal.
Inverse Normal needs three things: the area to the left, the mean, and the standard deviation:
Inverse NormalP(X < x) = 0.85
Area (left)0.85
μ (mean)50
σ (SD)8
x ≈ 58.3
Inverse NormalP(X > x) = 0.10
Area (left)0.90
μ (mean)50
σ (SD)8
x ≈ 60.3
⚠️
Inverse Normal always wants the area to the LEFT
If a question gives you P(X > x) = 0.10, you can’t type 0.10 directly. Convert it to the area on the left first: 1 − 0.10 = 0.90. Forgetting this is one of the biggest mark-losers in the whole topic.
🧠
An easy way to remember it
Cdf goes x → p. Inverse goes p → x. The two functions are mirror images of each other. If a question is asking for a number, it’s Inverse. If it’s asking for a chance, it’s Cdf.
Finding an unknown μ or σ
Sometimes the exam flips the question on its head: instead of giving you the mean and SD, they give you a probability and ask you to find one of them. This is where standardisation comes in.
Standardisation converts any normal distribution to the standard normal Z ∼ N(0, 1) — a special bell curve with mean 0 and SD 1. The formula is:
Standardisation formula
Z = X − μσ
✓ in formula booklet
The Z-value (or “z-score”) tells you how many standard deviations X is from the mean. So if Z = 1.5, then X is 1.5 standard deviations above the mean.
The 3-step method for finding μ or σ
This is the move that solves nearly every “find the unknown” question:
Use Inverse Normal on Z ∼ N(0, 1) with the given probability to find the z-score.
Substitute into the formula Z = (X − μ) / σ with the values you know.
Solve for the unknown (μ or σ).
For Step 1, you’ll set μ = 0 and σ = 1 in the Inverse Normal because you’re working with the standard normal. This gives you a pure z-score that you then plug into the formula in Step 2.
Worked examples
WE 1
Find a probability over a range
The mass of a brand of apple is normally distributed with mean 150 g and standard deviation 12 g. Find the probability that a randomly chosen apple has mass between 140 g and 165 g.
Range probability → use Normal Cdf with both bounds.Identify:μ = 150, σ = 12, lower = 140, upper = 165Run Normal Cdf:P(140 < X < 165) = 0.69285…P(140 < X < 165) = 0.693always type σ (12), NOT σ² (144) into the calculator!
WE 2
One-sided probability
For the same apples (μ = 150, σ = 12), find the probability that an apple weighs more than 170 g.
“More than 170” has no upper bound → use a huge upper number.Set up:P(X > 170)Normal Cdf, lower = 170, upper = 10⁹⁹μ = 150, σ = 12P(X > 170) = 0.04779…P(X > 170) = 0.0478no upper bound → type 10⁹⁹ (or 1E99) for the upper. simple as that!
WE 3
Inverse Normal — find a value
The time taken for runners in a race is normally distributed with mean 60 minutes and standard deviation 5 minutes. The slowest 15% of runners are timed out. What time is the cut-off?
“Slowest 15%” means the upper 15% — convert to area on the left first.Slowest 15% are above the cut-off, so:P(X > cut-off) = 0.15 → P(X < cut-off) = 0.85Inverse Normal with area = 0.85, μ = 60, σ = 5x = 65.182…cut-off ≈ 65.2 minutes“slowest” / “longest” / “highest” → top tail → convert to (1 − that) for the left area!
WE 4
Find an unknown mean using standardisation
The weights of bags of sugar are normally distributed with standard deviation 25 g. The probability that a bag weighs less than 980 g is 0.0228. Find the mean weight.
Step 1: find z-score. Step 2: sub into Z = (X − μ)/σ. Step 3: solve for μ.step 1 — find zInverse Normal, area = 0.0228, μ = 0, σ = 1z = −1.9991… ≈ −2step 2 — substituteX = 980, σ = 25, Z = −2−2 = 980 − μ25step 3 — solve−50 = 980 − μμ = 980 + 50 = 1030μ = 1030 gz is NEGATIVE because 980 is BELOW the mean — keep the sign!
WE 5
Real exam-style multi-part
The heights of 16-year-olds in a town are normally distributed with mean 168 cm and standard deviation 7 cm. Let X be the height of a randomly chosen 16-year-old, in cm.
(a) Find P(X < 175). (b) Find P(160 < X < 180). (c) The tallest 5% of 16-year-olds are eligible for a basketball programme. What is the minimum height needed? (d) In a sample of 200 sixteen-year-olds, how many would you expect to be shorter than 160 cm?
μ = 168, σ = 7. Use Normal Cdf and Inverse Normal as needed.part (a) — P(X < 175)Normal Cdf, lower = −10⁹⁹, upper = 175, μ = 168, σ = 7= 0.84134…P(X < 175) = 0.841part (b) — P(160 < X < 180)Normal Cdf, lower = 160, upper = 180, μ = 168, σ = 7= 0.83205…P(160 < X < 180) = 0.832part (c) — minimum height for top 5%Top 5% → P(X > min) = 0.05 → P(X < min) = 0.95Inverse Normal, area = 0.95, μ = 168, σ = 7x = 179.51…minimum height ≈ 180 cmpart (d) — expected number shorter than 160Find P(X < 160):Normal Cdf, lower = −10⁹⁹, upper = 160P(X < 160) = 0.12655…Expected count = n × p:200 × 0.12655 = 25.31…≈ 25 students“expected number” = total × probability — the same trick as the binomial mean!
💡 Top tips
Always sketch the curve first. Even a rough drawing tells you whether your answer is sensible.
Ask yourself the direction: values → probability (Cdf) or probability → value (Inverse).
For one-sided regions, use ±10⁹⁹ as the missing bound. Some calculators accept 1E99 directly.
Inverse Normal wants area to the LEFT. If the question gives the upper tail (top 10%, slowest 5%), convert with 1 − probability first.
Type σ, not σ². The notation N(μ, σ²) uses variance; the calculator uses standard deviation.
For unknown μ or σ, standardise: Z = (X − μ)/σ. Find z first using Inverse Normal on N(0, 1), then solve.
“Expected number” out of n = n × P. Same idea as binomial expectation.
Strict and weak inequalities are equal here. P(X < a) = P(X ≤ a). No conversions needed.
⚠ Common mistakes
Typing σ² into the calculator. If the question says “variance = 64”, σ = 8, not 64. Square root it before typing.
Using Inverse Normal with the wrong area. “Top 20%” means area to the left = 0.80, not 0.20.
Forgetting the infinity trick. P(X > a) needs a huge upper bound — leaving the field blank or putting 0 will give a wrong answer.
Mixing up Cdf and Inverse. If you got the answer 0.92 when the question wanted a height in cm, you’ve used the wrong function.
Wrong sign on z. If X is below the mean, z is negative. Don’t drop the minus by mistake.
Forgetting to standardise when finding an unknown μ or σ. You can’t go straight to a calculator function — you have to use Z = (X − μ)/σ.
Rounding too early. Carry full decimals through working and only round at the end (usually 3 s.f.).
Forgetting units on the final answer. If the question is about cm, your answer is in cm.
🎉 You can now find any normal probability and any normal value — including questions where the mean or standard deviation is hidden. That’s basically all of Topic 4 covered. Next up you’ll start tying everything together with bigger statistics and probability questions where these distributions show up alongside the rest of the course.
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