The normal distribution — every problem type, on one page.
Properties of the bell curve, z-scores, the 68-95-99.7 rule, calculator commands, and worked exam-style problems. Built for your upcoming Chapter 21 test.
A normal distribution is a continuous probability distribution shaped like a bell. It's the most important distribution in statistics because it shows up everywhere — heights, test scores, measurement errors, IQ — anywhere lots of small random factors add up.
If a random variable X is normally distributed with mean μ and standard deviation σ, we write:
NotationX ~ N(μ, σ²)
⚠️ Watch the notation: the second parameter is the variance (σ²), not the standard deviation. So if a problem says "X ~ N(50, 4)", that means μ = 50 and σ = √4 = 2.
Five properties to memorize
Bell-shaped & symmetric about the mean μ.
Mean = median = mode = μ. The peak of the curve is at x = μ.
Total area under the curve = 1. (It's a probability distribution.)
The standard deviation σ controls the spread. Bigger σ → flatter, wider curve.
The curve has asymptotes — it approaches the x-axis but never touches it, no matter how far you go.
Figure 1. Anatomy of a normal curve. Symmetric about the mean μ, with standard deviation σ controlling the spread. Total area under the curve = 1.
Key terms
Normal distributionA continuous, bell-shaped, symmetric probability distribution.
Mean (μ)The centre of the curve. Greek letter mu.
Standard deviation (σ)The spread. Larger σ = wider, flatter curve. Greek letter sigma.
Variance (σ²)Square of the standard deviation. Watch out — N(μ, σ²) uses variance!
X ~ N(μ, σ²)"X is normally distributed with mean μ and variance σ²." The tilde reads as "is distributed as".
ContinuousX can take any real value (not just integers). Probabilities are areas, not heights.
21.2Key Rule
The 68-95-99.7 rule
For any normal distribution, regardless of μ and σ, these three percentages always hold:
In words: about 68% of the data falls within one standard deviation of the mean, 95% within two, and almost everything (99.7%) within three.
Figure 2. The empirical (68-95-99.7) rule. The percentages between each pair of σ-markers add up to the totals — e.g. each of the two 13.5% slices plus the central 68% gives 95%.
Why it's useful: the empirical rule lets you estimate probabilities without a calculator for any cutoff that lands at ±σ, ±2σ, or ±3σ. Examiners love testing this on short-answer questions.
Quick example
The heights of IB students are normally distributed with mean 170 cm and standard deviation 8 cm. What percentage of students are between 162 cm and 178 cm tall?
Solution
162 cm = 170 − 8 = μ − σ
178 cm = 170 + 8 = μ + σ
So we want P(μ − σ < X < μ + σ) — exactly the 68% range.
≈ 68% of students.
21.3Standardization
z-scores & standardization
A z-score measures how many standard deviations a value is away from the mean. It lets you convert any normal distribution into the standard normal distribution, Z ~ N(0, 1), which has μ = 0 and σ = 1.
z-score formulaz = (x − μ) / σ
Reading the formula:
z = 0: the value is exactly at the mean.
z = 1: one standard deviation above the mean.
z = −2: two standard deviations below the mean.
z = 2.5: 2.5 standard deviations above (pretty extreme — only 0.62% of data is here or further).
Why we standardize
Two main reasons:
Comparison. Is scoring 82 on a math test better than scoring 75 on a chemistry test? Convert both to z-scores using each test's mean and σ, and you can compare directly.
Look-up tables. Before calculators, statisticians used z-tables for the standard normal. You'd standardize and then look up the probability. The IB still uses the concept conceptually even though we now use GDC calculator commands.
Worked example
Test scores are N(72, 64). What is the z-score of a student who scored 88?
Solution
μ = 72, σ² = 64, so σ = √64 = 8.
z = (88 − 72) / 8 = 16 / 8 = 2.
This student is 2 standard deviations above the mean — top ~2.5% of the class.
z = 2
Comparison example
Anna scored 78 on a Math test where μ = 70, σ = 5. Ben scored 84 on a Bio test where μ = 75, σ = 7. Who did better relative to their class?
Solution
Anna's z = (78 − 70) / 5 = 1.6
Ben's z = (84 − 75) / 7 = 1.29 (to 3 s.f.)
Anna's z is higher, so she did better relative to her class.
Anna (z = 1.6 vs Ben's z = 1.29)
21.4Calculator
Probability with normalcdf
To find a probability like P(a < X < b) when X is normally distributed, you use the normalcdf function on your GDC. (cdf = cumulative distribution function — the area under the curve.)
Calculator commands
What you want
TI-84 / TI-Nspire
Casio fx-CG50
P(a < X < b)
normalcdf(a, b, μ, σ)
NormCD: Lower=a, Upper=b, σ, μ
P(X < a)
normalcdf(-1E99, a, μ, σ)
Lower = −9999999, Upper = a
P(X > a)
normalcdf(a, 1E99, μ, σ)
Lower = a, Upper = 9999999
Calculator path on TI-84: Press 2nd → VARS (DISTR menu) → choose 2: normalcdf(. Then type lower, upper, μ, σ separated by commas, close the bracket, press ENTER.
Calculator path on Casio fx-CG50: from Main Menu → STAT → DIST (F5) → NORM (F1) → Ncd (F2). Fill in Lower, Upper, σ, μ. Press EXE.
Figure 3. A probability under a normal curve is the area between two vertical lines. normalcdf computes this area for you given the bounds, the mean, and the standard deviation.
Worked example
X ~ N(50, 64). Find P(45 < X < 60).
Solution
μ = 50, σ = √64 = 8.
On TI-84: normalcdf(45, 60, 50, 8)
Calculator returns ≈ 0.6285
P(45 < X < 60) ≈ 0.629 (to 3 s.f.)
"Greater than" example
X ~ N(100, 225). Find P(X > 120).
Solution
μ = 100, σ = √225 = 15.
P(X > 120) = normalcdf(120, 1E99, 100, 15)
Calculator returns ≈ 0.0912
P(X > 120) ≈ 0.0912 (about 9.1%)
Common mistake: Forgetting that the second parameter in N(μ, σ²) is variance, not standard deviation. Always check: do I have σ or σ²? Take the square root if needed.
21.5Inverse
Inverse normal — finding a value from a probability
What if you know the probability and need to find the cutoff value x? For example: "the top 10% of test scores are above what value?" You use the inverse normal function: invNorm.
Inverse normalIf P(X < x) = p, then x = invNorm(p, μ, σ)
Calculator commands
What you want
TI-84 / TI-Nspire
Casio fx-CG50
x such that P(X < x) = p
invNorm(p, μ, σ)
InvN: Area = p, σ, μ, tail=Left
x such that P(X > x) = p
invNorm(1 − p, μ, σ)
InvN: Area = p, tail=Right
⚠️ Critical: on TI-84 and most calculators, invNorm uses the area to the LEFT. If a problem says "top 10%", that's the area to the RIGHT. Either flip with 1 − p or change the tail setting on Casio.
Worked example
Heights are N(170, 64). The shortest 5% of people are below what height?
Solution
μ = 170, σ = √64 = 8.
"Shortest 5%" means we want x such that P(X < x) = 0.05.
On TI-84: invNorm(0.05, 170, 8)
Calculator returns ≈ 156.84
≈ 156.84 cm (3 s.f.)
"Top X%" example
Test scores are N(72, 100). The top 15% of students receive a distinction. What is the minimum score for a distinction?
Solution
μ = 72, σ = 10.
Top 15% = area to the right = 0.15. So area to the left = 1 − 0.15 = 0.85.
On TI-84: invNorm(0.85, 72, 10)
Calculator returns ≈ 82.36
Minimum distinction score ≈ 82.4 (or 83 if rounded up to whole marks)
21.6Advanced
Finding μ or σ from a given probability
Sometimes a problem tells you a probability and asks you to find the parameters — μ or σ — given everything else. This is the hardest problem type on this topic, and it usually appears as one of the longer questions on Paper 2.
Strategy
Standardize. Write the z-score equation: z = (x − μ) / σ.
Use invNorm to find the z-value that corresponds to the given probability — using N(0, 1).
Solve algebraically for μ or σ.
Find μ (given σ)
X is normally distributed with σ = 5. Given that P(X < 30) = 0.84, find μ.
Solution
Find the z-score: z = invNorm(0.84, 0, 1) ≈ 0.9945
Apply z formula: 0.9945 = (30 − μ) / 5
Solve: 30 − μ = 4.97, so μ = 30 − 4.97 = 25.03
μ ≈ 25.0 (3 s.f.)
Find σ (given μ)
X is normally distributed with μ = 100. Given P(X > 120) = 0.05, find σ.
Solution
P(X > 120) = 0.05 → P(X < 120) = 0.95.
z = invNorm(0.95, 0, 1) ≈ 1.6449
z = (120 − 100) / σ → 1.6449 = 20 / σ
σ = 20 / 1.6449 ≈ 12.16
σ ≈ 12.2 (3 s.f.)
Find both μ and σ (hardest)
X is normally distributed. Given P(X < 50) = 0.1 and P(X < 80) = 0.9, find μ and σ.
Solution
Find z₁: invNorm(0.1) ≈ −1.2816
Find z₂: invNorm(0.9) ≈ 1.2816
Set up two equations: −1.2816 = (50 − μ) / σ → 50 − μ = −1.2816σ 1.2816 = (80 − μ) / σ → 80 − μ = 1.2816σ
Everything you need for the test, on one screen. Print this before the exam.
Notation
X ~ N(μ, σ²)
Second parameter is the variance. Take √ to get σ.
z-score
z = (x − μ) / σ
Standardizes any normal to N(0, 1).
Empirical rule
68% / 95% / 99.7%
Within ±σ, ±2σ, ±3σ of the mean.
Probability (TI-84)
normalcdf(a, b, μ, σ)
For P(X < a): use −1E99 as lower. For P(X > a): use 1E99 as upper.
Inverse normal (TI-84)
invNorm(p, μ, σ)
Uses area to the LEFT. For top X%, use 1 − p.
Common z-values
90% → z = 1.28 95% → z = 1.645 97.5% → z = 1.96 99% → z = 2.33
These are cutoffs for the top X% (one-tailed).
Practice
Practice problems
Try each before clicking to reveal the solution. Aim to do them on paper, not just in your head — the calculator setup itself is half the skill.
Quick warm-ups
1. X ~ N(50, 25). Find P(X < 55).
σ = 5. normalcdf(-1E99, 55, 50, 5) ≈ 0.8413 (or use z = 1, look up Φ(1) ≈ 0.8413).
2. X ~ N(20, 16). Find P(X > 26).
σ = 4. normalcdf(26, 1E99, 20, 4) ≈ 0.0668.
3. X ~ N(100, 144). Find P(85 < X < 115).
σ = 12. normalcdf(85, 115, 100, 12) ≈ 0.7888.
4. What is the z-score of x = 72 if μ = 60 and σ = 6?
z = (72 − 60)/6 = 2.
Empirical rule
5. IQ scores are N(100, 225). Between what two values do 95% of people fall?
σ = 15. 95% lies within μ ± 2σ = 100 ± 30 → between 70 and 130.
6. The lifetime of a light bulb is N(1200, 10000) hours. What % of bulbs last between 1100 and 1400 hours?
σ = 100. 1100 = μ − σ, 1400 = μ + 2σ. From the empirical rule: between μ − σ and μ + σ is 68%; the extra slice from μ + σ to μ + 2σ is 13.5%. Total ≈ 81.5%. (Calculator check: normalcdf(1100, 1400, 1200, 100) ≈ 0.8186.)
Inverse normal
7. Test scores N(70, 100). What is the cutoff for the top 25% of students?
σ = 10. Top 25% = area to right = 0.25, so area to left = 0.75. invNorm(0.75, 70, 10) ≈ 76.7.
8. X ~ N(50, 100). Find x such that P(X < x) = 0.30.
σ = 10. invNorm(0.30, 50, 10) ≈ 44.76.
Finding μ or σ
9. X is normally distributed with σ = 4. Given P(X < 18) = 0.95, find μ.
z = invNorm(0.95) ≈ 1.6449
1.6449 = (18 − μ) / 4
18 − μ = 6.58 → μ ≈ 11.4
10. X is normally distributed with μ = 50. Given P(X > 65) = 0.10, find σ.
P(X < 65) = 0.90 → z = invNorm(0.90) ≈ 1.2816
1.2816 = (65 − 50) / σ → σ = 15 / 1.2816 ≈ 11.7
Exam-style longer questions
11. Bolts are produced with diameters N(8.0, 0.01) mm. Bolts are rejected if their diameter is below 7.85 mm or above 8.15 mm. (a) What % are rejected? (b) If 10000 bolts are made, how many do you expect to reject?
12. The height of students at a school is N(165, 49) cm. (a) Find P(160 < X < 175). (b) The tallest 10% of students are above what height? (c) If there are 500 students, how many are taller than 175 cm?
1. σ vs σ². The notation N(μ, σ²) uses variance. If the question says X ~ N(10, 16), then σ = 4, not 16. Always check.
2. "Top X%" with invNorm. invNorm uses area to the left. "Top 10%" means area to the right = 0.10, so the left-tail input is 0.90. Don't plug in 0.10 by accident.
3. P(X = a) for continuous distributions. For a continuous variable, P(X = a) = 0. The probability is the area at a single point, which has zero width. Don't try to compute it.
4. Forgetting to standardize. When you're working with z-scores in finding μ or σ, the z comes from N(0, 1), not from the original distribution. Use invNorm(p, 0, 1) — or just invNorm(p) on TI.
5. Rounding too early. Carry intermediate calculations to 4+ s.f. and only round at the very end. Compounding rounding errors will cost you the final mark.
What's likely on the test
Based on standard Math AA SL papers, you'll almost certainly see:
One "find the probability" question (normalcdf, straightforward) — should be quick marks.
One "find the value" question (invNorm, watch for top/bottom %).
One longer word problem — heights, lifetimes, test scores, manufacturing — testing all of the above.
Often one "find μ or σ" question — uses z-score algebra. This is the discriminator.
Manage your time: the first two should take ~3 min each, the longer problem ~8 min, the find-μ/σ question ~5 min. If you're stuck on the last one, leave it and come back.