IB Maths AA HL Topic 4 — Statistics & Probability Paper 2 ~7 min read

Finding Unknown Parameters

Sometimes you’re given probabilities and asked to find μ or σ — the reverse of the usual question. The trick is standardisation: convert each given probability to a z-value using InvNorm on the standard normal, then solve a linear equation in μ and σ. With one unknown, one probability is enough; with both unknown, two probabilities give two equations to solve simultaneously.

📘 What you need to know

Standardisation formula: z = x − μσ — given in the booklet.
Rearranged form: x = μ + σz — much cleaner for simultaneous equations.
One unknown: one probability gives one equation; solve directly.
Two unknowns: need two probabilities at two different x-values; solve as simultaneous equations.
Convert before InvNorm: P(X > x) = p → P(X < x) = 1 − p.
InvNorm input: use the standard normal Z ~ N(0, 1) — InvNorm(p, 0, 1) gives the z-value.
Sign of z matches the sign relative to mean: x below μ → negative z; above → positive z.
Carry plenty of decimal places (4+) for the z-value to avoid rounding errors compounding.

One unknown — find μ or σ

If only one of μ or σ is unknown, you only need one probability to pin it down. The recipe is always the same: convert the probability to a z-value, then plug into the rearranged formula.

Standardisation, in two useful forms z = x − μσ ⇔ x = μ + σz

σ unknown, μ given

σ = x − μz

divide by the z-value

μ unknown, σ given

μ = x − σz

subtract σ times z from x

Both unknown — simultaneous equations

If both μ and σ are unknown, you’ll be given two probabilities at two different x-values. Each gives one equation in the form x_i = μ + σz_i; solve simultaneously.

Simultaneous equations setup x₁ = μ + σz₁ and x₂ = μ + σz₂

Subtract to find σ: (x₂ − x₁) = σ(z₂ − z₁) → σ = (x₂ − x₁) / (z₂ − z₁). Then back-substitute either equation for μ.

🧭 Recipe — find an unknown parameter

Sketch the curve and shade the given probability — tells you the SIGN of z.
Convert to lower-tail form: P(X > x) = p → P(X < x) = 1 − p.
Find each z-value using InvNorm with mean 0 and SD 1; carry 4+ decimal places.
Substitute into x = μ + σz for each given pair.
Solve: directly for one unknown, simultaneously for two; sanity-check σ > 0.

Worked examples

WE 1

Find σ from an upper-tail probability

The random variable X ~ N(50, σ²). Given that P(X > 60) = 0.15, find σ.

Convert to lower-tail form P(X < 60) = 1 − 0.15 = 0.85 Find z-value via InvNorm z = InvNorm(0.85, 0, 1) = 1.0364… Apply z = (x − μ)/σ → σ = (x − μ)/z σ = (60 − 50) / 1.0364 = 10 / 1.0364 = 9.6485… σ ≈ 9.65 (3 sf) 60 is above the mean, so z is positive — sign check passes

WE 2

Find μ from a lower-tail probability

The heights of plants follow H ~ N(μ, 7²) cm. Given that P(H < 160) = 0.20, find μ.

Lower-tail probability is given directly z = InvNorm(0.20, 0, 1) = −0.8416… 160 is below the mean → z negative ✓ Apply x = μ + σz → μ = x − σz μ = 160 − 7(−0.8416) = 160 + 5.891 = 165.89… μ ≈ 166 cm (3 sf) double negative: subtracting a negative becomes adding — easy slip if you rush

WE 3

Find σ and Var from a lower-tail probability

Battery lifetimes follow T ~ N(40, σ²) hours. Given that P(T < 35) = 0.10, find σ and the variance of T.

Find z-value z = InvNorm(0.10, 0, 1) = −1.2816… Apply σ = (x − μ)/z σ = (35 − 40) / (−1.2816) = −5 / −1.2816 = 3.9015… Variance is σ² Var(T) = 3.9015² = 15.222… σ ≈ 3.90 hours; Var(T) ≈ 15.2 both x − μ and z are negative → σ comes out positive, as it must

WE 4

Find μ from a small upper-tail probability

The random variable X ~ N(μ, 4²). Given that P(X > 50) = 0.025, find μ.

Convert to lower-tail P(X < 50) = 1 − 0.025 = 0.975 Find z (this should look familiar — z ≈ 1.96 is the 97.5% quantile) z = InvNorm(0.975, 0, 1) = 1.9600 Apply μ = x − σz μ = 50 − 4(1.9600) = 50 − 7.840 = 42.160 μ ≈ 42.2 (3 sf) only 2.5% above 50 → 50 is well above the mean → μ < 50, as confirmed

WE 5

Find both μ and σ simultaneously

Test scores follow X ~ N(μ, σ²). Given P(X < 45) = 0.10 and P(X > 80) = 0.20, find μ and σ.

Find both z-values P(X < 45) = 0.10 → z₁ = InvNorm(0.10) = −1.2816 P(X > 80) = 0.20 → P(X < 80) = 0.80 → z₂ = InvNorm(0.80) = 0.8416 Form two equations using x = μ + σz 45 = μ − 1.2816σ … (1) 80 = μ + 0.8416σ … (2) Subtract (1) from (2) 35 = (0.8416 − (−1.2816))σ = 2.1232σ σ = 35 / 2.1232 = 16.485… Back-substitute μ = 45 + 1.2816(16.485) = 66.13 μ ≈ 66.1; σ ≈ 16.5 (3 sf) x = μ + σz form makes the algebra linear and clean — much easier than (x − μ)/σ

WE 6

Real-world: dog masses with two percentile facts

The masses of dogs at a kennel are normally distributed. The kennel’s records show that 25% of dogs weigh less than 18 kg and 10% weigh more than 32 kg. Find the mean and standard deviation of the dogs’ masses.

Translate the wording P(X < 18) = 0.25 P(X > 32) = 0.10 → P(X < 32) = 0.90 Find both z-values z₁ = InvNorm(0.25) = −0.6745 z₂ = InvNorm(0.90) = 1.2816 Form two equations 18 = μ − 0.6745σ … (1) 32 = μ + 1.2816σ … (2) Subtract 14 = (1.2816 + 0.6745)σ = 1.9561σ σ = 14 / 1.9561 = 7.157… Back-substitute μ = 18 + 0.6745(7.157) = 22.83 μ ≈ 22.8 kg; σ ≈ 7.16 kg (3 sf) μ ≈ 22.8 sits between the two cut-offs (18 and 32) — sanity check passes

💡 Top tips

Always sketch first — tells you whether each z-value should be positive or negative.
Work with x = μ + σz: it’s a clean linear form, especially for simultaneous equations.
Carry 4+ decimal places in z-values to avoid rounding errors compounding.
Use InvNorm with μ = 0, σ = 1 — you want the z-value, not an x-value.
Sanity check: σ > 0; for two-unknown problems, μ should sit between the two given x-values.

⚠ Common mistakes

Using P(X > x) directly in InvNorm — convert to P(X < x) = 1 − p first.
Wrong sign of z — if x is below the mean, z is negative; sketching catches this.
Rounding z to 2 dp early — propagates significant error into the final μ or σ.
Reporting a negative σ — always positive; if you get one, retrace the algebra.
Mixing up which z goes with which x in two-unknown problems — line up the equations carefully.

That closes the Normal Distribution sub-topic. Next up: Continuous Random Variables — the general framework where you define your own probability density function f(x), then compute probabilities and statistics by integration. The normal distribution is one specific (very famous) example, but in this last sub-topic of Topic 4 you’ll see how to handle ANY continuous distribution.

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