This mini-lesson covers AQA section N: the binomial distribution B(n, p) — its conditions, probabilities, mean and variance — and the Normal distribution N(μ, σ²) — standardising with z, the inverse normal, points of inflection, and the link between the two. It finishes with choosing an appropriate model.
N(μ, σ²) — bell shaped, symmetric about μ, with points of inflection at μ ± σ. About 95% of the data lies within 2σ of μ.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. AQA expects you to use your calculator's distribution functions, and to say why a model does or does not fit. Press Start when you're ready.
Section N1 · binomial
The binomial distribution B(n, p)
X ~ B(n, p) counts the number of successes in n trials. It needs all four conditions:
a fixed number of trials, n
each trial has only two outcomes (success / failure)
Hint: Use the cumulative binomial: add P(X=0) + P(X=1) + P(X=2) + P(X=3), or use your calculator's Bcd function.
Quick check
Is it binomial?
?Which of these is NOT well modelled by a binomial distribution?
Quick check
Mean and variance
?X ~ B(40, 0.15). What are the mean and variance of X?
Section N2 · normal
The Normal distribution N(μ, σ²)
X ~ N(μ, σ²) is a continuous model: bell shaped, symmetric about the mean μ, with points of inflection at μ ± σ. Total area = 1, and mean = median = mode.
Z = (X − μ) ÷ σ ~ N(0, 1)standardising: how many standard deviations is x above the mean?
Worked example — standardising
X ~ N(50, 8²). Find P(X < 62).
z = (62 − 50) ÷ 8 = 12 ÷ 8 = 1.5
P(Z < 1.5) = Φ(1.5) = 0.933 (3 s.f.)
Worked example — inverse normal
X ~ N(100, 15²). Find x such that P(X < x) = 0.90.
Inverse normal: z = 1.2816. So x = μ + zσ = 100 + 15 × 1.2816
x = 119.2 (1 d.p.)
Useful landmarks: about 68% of the distribution lies within 1σ of μ, about 95% within 2σ, and about 99.7% within 3σ. (Exactly 95% lies within 1.96σ.)
Hint: Standardise: z = (62 − 50) ÷ 8 = 1.5. Then find Φ(1.5).
Calculate
Standardising
4X ~ N(70, 5²). Calculate the z-value of x = 78.
Hint: z = (x − μ) ÷ σ = (78 − 70) ÷ 5.
Calculate
Inverse normal
5X ~ N(100, 15²). Find the value of x for which P(X < x) = 0.90, to 1 decimal place.
Hint: Inverse normal gives z = 1.2816. Then x = μ + zσ = 100 + 15 × 1.2816.
Sort it
Choose the model
Tap a situation, then tap the model that fits it best.
🎯 Binomial
🔔 Normal
❌ Neither
Section N2–N3 · linking
Linking the binomial to the Normal
Draw the bars of B(n, p) for a large n with p not too close to 0 or 1 and the histogram looks like a bell. That is the link AQA asks for between the binomial and the Normal.
If X ~ B(n, p) with n large, then X ≈ N(np, np(1 − p))same mean, same variance · use a continuity correction: P(X ≤ 45) → P(Y < 45.5)
Worked example
X ~ B(80, 0.5). Then np = 80 × 0.5 = 40 and np(1 − p) = 80 × 0.5 × 0.5 = 20.
So X can be approximated by N(40, 20), i.e. σ = √20 ≈ 4.47.
Choosing a model (N3): say why. Binomial fails when trials are dependent or p changes. Normal fails when the data are skewed, or the quantity is discrete and n is small, or values cannot be negative but μ is close to 0.
Quick check
The 95% rule
?Approximately what percentage of a Normal distribution lies within 2 standard deviations of the mean?
Match it
Formula recall
Tap a formula on the left, then what it gives you on the right.
Formula
Meaning
Quick check
Normal approximation
?X ~ B(80, 0.5) is to be approximated by a Normal distribution. Which one?
Section N1 · discrete distributions
Discrete probability distributions
A discrete random variable X takes a listed set of values, each with a probability. The whole distribution is usually given as a table — and the probabilities must sum to 1.
Σ P(X = x) = 1this is what lets you find a missing probability
Sum to 1: 0.2 + 0.3 + k + 0.1 = 1 → k = 1 − 0.6 = 0.4
Then, for example, P(X ≥ 3) = 0.4 + 0.1 = 0.5.
AQA note: calculating the mean and variance of a general discrete random variable is excluded from A-level Maths — but you do need E(X) = np and Var(X) = np(1 − p) for the binomial.
Calculate
Missing probability
6A discrete random variable X has P(X=1) = 0.2, P(X=2) = 0.3, P(X=3) = k and P(X=4) = 0.1. Find k.
Hint: All the probabilities must add to 1: k = 1 − (0.2 + 0.3 + 0.1).
Quick check
When the Normal model fails
?The masses of some components have mean 2 g and standard deviation 3 g, and mass cannot be negative. Why is N(2, 3²) a poor model?
Recap
The big ideas to know
Binomial B(n, p): fixed n · two outcomes · independent trials · constant p