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

Calculations with Normal Distribution

For a normal distribution, the GDC’s NormCdf function gives P(a < X < b) directly — input the bounds, the mean, and the standard deviation. P(X = k) is always 0 for any continuous variable, and one-sided tails like P(X < b) or P(X > a) use a very negative or very large number for the missing bound. Inverse normal goes the other way: given a probability, find the threshold value.

📘 What you need to know

Single value: P(X = k) = 0 for any continuous distribution.
Strict vs weak: P(a < X < b) = P(a ≤ X ≤ b) — both give the same area.
Range: NormCdf(lower, upper, μ, σ) → P(lower ≤ X ≤ upper).
Lower tail: for P(X < b), use lower bound = −10⁹⁹ (or any very negative number).
Upper tail: for P(X > a), use upper bound = 10⁹⁹ (or any very large number).
SD not variance: NormCdf takes σ, not σ². Square-root the second slot of N(μ, σ²) before entering.
Inverse normal: given P(X < a) = p, InvNorm(p, μ, σ) returns a.
Symmetry: P(X < μ) = P(X > μ) = 0.5 always.

Why P(X = k) = 0

For a continuous variable, the probability of getting any single exact value is zero. Picture the bell curve: a single value is a vertical line at x = k, and a line has zero width — so it encloses zero area. This is why you only ever compute probabilities for ranges.

Strict-or-weak doesn’t matter: because P(X = a) = 0, swapping < for ≤ adds nothing. P(a < X < b) and P(a ≤ X ≤ b) give identical answers — you don’t need to convert.

Range probabilities P(a < X < b)

GDC call for a finite range P(a ≤ X ≤ b) = NormCdf(a, b, μ, σ)

Different calculators name this Normal CD, NCD, or Normal Cdf. They all want four inputs in roughly the same order: lower bound, upper bound, mean, standard deviation. Some calculators ask for SD before mean — check the order.

One-sided tails P(X < b) and P(X > a)

The NormCdf function only takes a finite range, so for one-sided tails you fake the missing endpoint with a very large or very negative number. The normal density is essentially 0 beyond ±4σ, so any value far enough out works.

Lower tail P(X < b)

NormCdf(−10⁹⁹, b, μ, σ)

use a very negative lower bound

Upper tail P(X > a)

NormCdf(a, 10⁹⁹, μ, σ)

use a very large upper bound

Most calculators have a button labelled −1E99 / 1E99 or similar — that’s the conventional placeholder for “infinity”.

Inverse normal: from probability to threshold

If the question gives you a probability and asks for the value a, use the InvNorm function instead. Given P(X < a) = p, InvNorm(p, μ, σ) returns a.

Going backwards: probability → value P(X < a) = p ⇒ a = InvNorm(p, μ, σ)

If the question gives the upper-tail probability instead — i.e. P(X > a) = p — convert first: P(X < a) = 1 − p, then apply InvNorm with 1 − p.

Sanity-check trick: if p < 0.5, the answer should be below the mean; if p > 0.5, above. Sketch the curve and check before pressing buttons.

🧭 Recipe — solve any normal probability question

Identify μ and σ — square-root the variance if needed.
Sketch the curve and shade the region you want.
Translate the wording into bounds for NormCdf.
For one-sided tails, use ±10⁹⁹ as the missing bound.
For inverse questions, use InvNorm — and convert “greater than” probabilities to “less than” using 1 − p.

Worked examples

WE 1

Single value and a basic range

The random variable X ~ N(100, 144). Find: (i) P(X = 100). (ii) P(80 < X < 110).

(i) Single value on a continuous distribution P(X = 100) = 0 (ii) σ = √144 = 12 Apply NormCdf with finite bounds P(80 < X < 110) = NormCdf(80, 110, 100, 12) = 0.74988… (i) 0; (ii) 0.750 (3 sf) always square-root the variance before passing it to NormCdf

WE 2

Lower tail — banana weights

The weights of bananas at a market are modelled by X ~ N(150, 25²) grams. Find P(X < 130).

μ = 150, σ = 25 (already SD, not variance) Lower tail → use −10⁹⁹ for lower bound P(X < 130) = NormCdf(−10⁹⁹, 130, 150, 25) = 0.21185… P(X < 130) ≈ 0.212 (3 sf) 130 is below the mean (150), so the answer must be less than 0.5 — sanity check passes

WE 3

Upper tail — battery lifetimes

The lifetimes of a brand of battery are modelled by T ~ N(48, 9) hours. Find P(T > 52).

σ = √9 = 3 (variance was given, not SD) Upper tail → use 10⁹⁹ for upper bound P(T > 52) = NormCdf(52, 10⁹⁹, 48, 3) = 0.09121… P(T > 52) ≈ 0.0912 (3 sf) 52 is above the mean (48), so the upper-tail prob < 0.5 — checks out

WE 4

Mixed inclusive/strict range — coffee wait times

Coffee shop wait times are modelled by W ~ N(4, 0.81) minutes. Find P(2.5 ≤ W < 5).

σ = √0.81 = 0.9 Strict vs weak doesn’t matter for continuous → use both bounds directly P(2.5 ≤ W < 5) = NormCdf(2.5, 5, 4, 0.9) = 0.81895… P(2.5 ≤ W < 5) ≈ 0.819 (3 sf) no need to convert 2.5 ≤ W to 2.5 < W — same answer for continuous variables

WE 5

Inverse normal — bulb lifespans

The lifespans of a brand of LED bulb are modelled by X ~ N(2000, 250²) hours. The manufacturer wants to find the time a below which only 10% of bulbs fail. Find a.

Translate: P(X < a) = 0.10 Apply InvNorm directly a = InvNorm(0.10, 2000, 250) = 1679.61… Round to 3 sf a ≈ 1680 hours a ≈ 1680 hours (3 sf) p = 0.10 < 0.5, so the answer is below the mean (2000) — consistent

WE 6

Multi-part: distribution setup, inverse normal, and empirical-rule check

The masses of newborn babies at a hospital are normally distributed with mean 3.4 kg and standard deviation 0.5 kg. The hospital classifies a baby as “unusually large” if its mass exceeds the threshold m, where only 2.5% of babies exceed this mass.

(a) State the distribution of the mass X. (b) Find the value of m. (c) Approximately what percentage of babies have mass between 2.4 kg and 4.4 kg?

(a) μ = 3.4, σ = 0.5, so σ² = 0.25 X ~ N(3.4, 0.25) (b) P(X > m) = 0.025 — convert to lower-tail form P(X < m) = 1 − 0.025 = 0.975 m = InvNorm(0.975, 3.4, 0.5) = 4.37998… m ≈ 4.38 kg (c) 2.4 = 3.4 − 1.0 = μ − 2σ; 4.4 = 3.4 + 1.0 = μ + 2σ Within ±2σ → ~95% (empirical rule) (a) X ~ N(3.4, 0.25); (b) m ≈ 4.38 kg; (c) ~95% always convert “greater than” into “less than” form before InvNorm

💡 Top tips

Sketch first — knowing whether the answer should be above or below 0.5 catches sign errors.
Variance vs SD trap: NormCdf takes σ, not σ². If given variance, square-root before entering.
One-sided tails: use −10⁹⁹ or 10⁹⁹ (your calculator probably has a “1E99” button).
Strict and weak inequalities are equivalent for continuous variables — don’t waste time converting.
InvNorm wants the LOWER-tail probability. Convert “greater than” using 1 − p before entering.

⚠ Common mistakes

Inputting variance instead of SD into NormCdf — the function wants σ, not σ².
Using NormPdf instead of NormCdf — that gives the density at a point, not a probability. Ignore NormPdf for this course.
Forgetting P(X = k) = 0 — for any continuous variable, single-point probabilities are exactly zero.
Putting the bounds in the wrong order in NormCdf — lower first, upper second; reversing gives a negative answer.
Using InvNorm with an upper-tail probability directly — convert to lower-tail using 1 − p first.

Next: Standardisation of Normal Variables & z-Values. The standard normal distribution Z ~ N(0, 1) lets you compare values from completely different normal distributions (e.g. exam scores from different subjects). The z-value tells you how many standard deviations a value lies above or below the mean.

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