IB Maths AA HLTopic 4 — Statistics & ProbabilityPaper 1 & 2~7 min read
The Normal Distribution
A normal distribution is the bell-shaped curve described by N(μ, σ²) — symmetric about the mean μ, with spread controlled by the variance σ². Real-world variables like heights, weights and exam scores are modelled this way, provided the data is symmetrical with one mode. The 68-95-99.7 rule gives quick percentage estimates within ±1, ±2, ±3 standard deviations of the mean.
📘 What you need to know
Notation: X ~ N(μ, σ²) — μ is the mean, σ² is the variance.
Standard deviation: σ = √(σ²) — always take the square root of the variance.
Symmetry: the curve is symmetric about x = μ, so mean = median = mode.
Bell-shaped: total area under the curve is 1; tails extend to ±∞ but vanish quickly.
68-95-99.7 rule: ~68% within μ±σ; ~95% within μ±2σ; ~99.7% within μ±3σ.
Changing μ only translates the curve horizontally — shape is unchanged.
Single value: P(X = k) = 0 for any continuous variable; only ranges have nonzero probability.
The N(μ, σ²) notation
If you read X ~ N(μ, σ²), the first parameter is always the mean and the second is the variance. To get the standard deviation, take the square root of the second parameter — this is the single biggest source of slip-ups in this topic.
Reading the parametersX ~ N(μ, σ²) ⇒ mean = μ, variance = σ², SD = σ = √(σ²)
The 68-95-99.7 empirical rule
The shape of every normal curve is the same — only the location and width change. Because of this, the percentage of the population within a fixed number of standard deviations of the mean is always the same, regardless of μ or σ:
Total area under any normal curve is 1. The percentage within ±k standard deviations is the same for every N(μ, σ²).
Tail probability shortcut: the area outside ±2σ is roughly 5% (≈ 2.5% in each tail); outside ±3σ is roughly 0.3% (≈ 0.15% in each tail). Useful for “more than”/”less than” estimates without a GDC.
When to use a normal model
Scenario
Normal model?
Reason
Adult heights or weights in a country
YES
large population, symmetric, single peak
Mass of randomly selected apples
YES
biological measurement, bell-shaped
Rolling a single fair die
NO
discrete; uniform, not bell-shaped
Daily rainfall (mm)
NO
spike at zero; right-skewed
Lifespan of LED bulbs
NO
typically right-skewed (long tail)
Output of a random number generator
NO
uniform, no single mode
Two non-negotiable conditions for normal modelling: the variable should be roughly symmetric and have a single mode. A long tail or a spike at zero is a deal-breaker.
🧭 Recipe — answer empirical-rule questions
Read μ and σ² from the notation; take the square root for σ.
Sketch the curve with marks at μ, μ±σ, μ±2σ, μ±3σ.
Locate the value in question — how many σ above/below the mean is it?
Apply 68% / 95% / 99.7% for the central region; subtract from 1 for tails.
Use symmetry to split a tail in half if you only need one side.
Worked examples
WE 1
Read parameters from N(μ, σ²)
The random variable T ~ N(75, 64). State the mean, variance and standard deviation of T.
Identify each parameterFirst slot is the mean: μ = 75Second slot is the variance: σ² = 64Square-root for SDσ = √64 = 8μ = 75; σ² = 64; σ = 8never report 64 as the standard deviation — it’s the variance
WE 2
Apply the 68-95-99.7 rule
Adult male heights (in cm) in a population are modelled by H ~ N(178, 49). (a) State the standard deviation. (b) Approximately what percentage of men are between 164 cm and 192 cm tall? (c) Approximately what percentage are taller than 199 cm?
(a) σ = √49 = 7(b) Locate the bounds in σ-units from the mean164 = 178 − 14 = μ − 2σ192 = 178 + 14 = μ + 2σ→ within ±2σ → ~95%(c) 199 = 178 + 21 = μ + 3σ~99.7% within ±3σ → ~0.3% outsideBy symmetry, half of that is in the upper tail→ P(H > 199) ≈ 0.15%(a) σ = 7; (b) ~95%; (c) ~0.15%always count σ-multiples first, THEN apply 68/95/99.7
WE 3
Compare two normal curves
Two random variables are A ~ N(50, 4) and B ~ N(50, 25). (a) State the SD of each. (b) Describe how the two curves compare in shape. (c) State the mode of B.
(a) Square-root each varianceσ_A = √4 = 2; σ_B = √25 = 5(b) CompareBoth curves are centred at μ = 50 (same mean)B has the larger SD → B is wider and shorterA has the smaller SD → A is narrower and taller(c) Normal is symmetric → mode = meanMode of B = 50σ_A = 2, σ_B = 5; B is wider and shorter; mode of B = 50same mean = same horizontal position; bigger σ = squashed flatter
WE 4
Identify which scenarios suit a normal model
For each random variable, state with reason whether a normal distribution is appropriate.
(a) Daily rainfall (in mm) at a weather station.
(b) The mass of randomly selected apples from a single orchard.
(c) The score on one roll of a fair six-sided die.
(d) The lifespan of a brand of LED bulb (which has a long right tail).
(a) Rainfall — NOmany days have zero rain → spike at 0; not symmetric(b) Apple mass — YESbiological measurement: symmetric, bell-shaped, single mode(c) Die score — NOdiscrete, uniform — not bell-shaped or continuous(d) LED lifespan — NOstated long right tail → not symmetric(a) No; (b) Yes; (c) No; (d) Notwo checks: symmetric? single peak? both must pass for normal modelling
WE 5
Use symmetry to find a probability
The random variable X ~ N(60, σ²). (a) Find P(X > 60). (b) Given that P(X < 50) = 0.2, find P(50 < X < 70).
(a) Symmetry about the meanP(X > 60) = 0.5(b) 50 and 70 are symmetric about 60 (each 10 away)By symmetry: P(X > 70) = P(X < 50) = 0.2P(50 < X < 70) = 1 − P(X < 50) − P(X > 70)= 1 − 0.2 − 0.2 = 0.6(a) 0.5; (b) P(50 < X < 70) = 0.6σ wasn’t needed — symmetry alone solved it
WE 6
Real-world: egg masses + empirical rule
The mass M (in grams) of a randomly chosen egg from a particular flock is modelled by M ~ N(62, 16).
(a) State the mean and standard deviation of M. (b) State two assumptions needed to use this model. (c) Approximately what percentage of eggs have mass between 58 g and 66 g? (d) An egg is classified as “extra large” if its mass exceeds 70 g. Approximately what percentage are extra large?
(a) Read parametersμ = 62 g; σ² = 16, so σ = √16 = 4 g(b) Assumptions• distribution of egg masses is symmetrical• distribution is bell-shaped (single mode)(c) 58 = 62 − 4 = μ − σ; 66 = 62 + 4 = μ + σWithin ±σ → ~68%(d) 70 = 62 + 8 = μ + 2σ~95% within ±2σ → ~5% outsideHalf in each tail → P(M > 70) ≈ 2.5%(a) μ = 62, σ = 4; (c) ~68%; (d) ~2.5%always state assumptions explicitly — they’re often a separate mark
💡 Top tips
Variance vs SD: N(μ, σ²) gives variance. Take the square root before using it as a standard deviation.
Sketch first — it makes “how many σ from the mean” obvious.
Symmetry is free: P(X < μ) = P(X > μ) = 0.5 for any normal.
Mean = median = mode for any normal — useful sanity check.
Halve the tail: by symmetry, the area in each tail beyond μ + kσ is half the total outside ±kσ.
⚠ Common mistakes
Confusing variance with SD — the second slot in N(μ, σ²) is variance, not standard deviation.
Using normal for skewed data — income, lifespans and rainfall do NOT fit; the model needs symmetry.
Reading “68%” as “68% of values” — it’s specifically “68% within 1 SD of the mean”.
Forgetting P(X = a) = 0 — for a continuous variable, single-value probabilities are exactly 0.
Mixing up which tail when applying symmetry — sketch the curve and shade the region asked for.
Next: Calculations with Normal Distribution. Now that you can read parameters and sketch the curve, the GDC’s NormCdf and InvNorm functions will give you exact (rather than empirical-rule-approximate) probabilities for any range — even when the bounds aren’t whole multiples of σ.
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