Mini-Lesson

Statistical distributions

This mini-lesson covers Topic 4 of Edexcel A-level Mathematics (9MA0): the binomial distribution B(n, p) — its conditions, probability formula, mean and variance; the normal distribution N(μ, σ²) — symmetry, points of inflection, standardising with z, and the inverse normal; the normal approximation to the binomial with a continuity correction; and how to select an appropriate model — and spot when neither fits.

N(μ, σ²) is continuous, symmetric about μ, with points of inflection at μ ± σ. Total area = 1, so P(X > μ) = 0.5.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, take g = 9.8 m s⁻² unless a question says otherwise. Press Start when you are ready.

Distributions · binomial

The binomial distribution B(n, p)

X ~ B(n, p) counts the number of successes in n trials. Four conditions must all hold:

A fixed number of trials, n.
Each trial has exactly two outcomes (success / failure).
The trials are independent.
The probability of success, p, is constant for every trial.

P(X = r) = ⁿC_r p^r (1 − p)^{n − r}mean = np · variance = np(1 − p)

Worked example — X ~ B(10, 0.3), find P(X = 3)

¹⁰C₃ = 120, p³ = 0.3³ = 0.027, (1 − p)⁷ = 0.7⁷ = 0.0823543.

P(X = 3) = 120 × 0.027 × 0.0823543 = 0.2668 (4 d.p.)

Modelling check: drawing counters without replacement is not binomial — p changes between trials. Sampling with replacement (or from a very large population) is.

Calculate

Binomial probability

1X ~ B(10, 0.3). Find P(X = 3), giving your answer to 3 decimal places.

Hint: P(X = 3) = ¹⁰C₃ × 0.3³ × 0.7⁷ = 120 × 0.027 × 0.0823543.

Calculate

Mean & variance

2X ~ B(50, 0.2). Find the variance of X.

Hint: Variance = np(1 − p) = 50 × 0.2 × 0.8. (The mean would be np = 10.)

Quick check

Binomial conditions

?Which of these is NOT a condition for X ~ B(n, p)?

Distributions · normal

The normal distribution & standardising

X ~ N(μ, σ²) is continuous and its bell curve is symmetric about the mean μ. Probability is the area under the curve, so P(X = a) = 0 and P(X < a) = P(X ≤ a).

Mean = median = mode = μ. P(X > μ) = P(X < μ) = 0.5.
The points of inflection are at x = μ ± σ.
Roughly 68% of the data lies within 1σ of the mean, 95% within 2σ and 99.7% within 3σ.

z = (x − μ) ÷ σstandardising: turns any X ~ N(μ, σ²) into Z ~ N(0, 1²)

Worked example — X ~ N(70, 8²)

Find P(X < 82). Standardise: z = (82 − 70) ÷ 8 = 1.5.

P(Z < 1.5) = 0.9332 (4 d.p.), so P(X < 82) = 0.9332.

Calculate

Standardising

3X ~ N(70, 8²). Standardise the value x = 82, i.e. find z.

Hint: z = (x − μ) ÷ σ = (82 − 70) ÷ 8.

Calculate

Reading the probability

4Using z = 1.5 from the last question, find P(X < 82), i.e. P(Z < 1.5), to 4 decimal places.

Hint: Φ(1.5) = 0.9332. (This is the area to the LEFT of z = 1.5.)

Distributions · inverse normal

The inverse normal — going backwards

Sometimes you are given the probability and must find the value. This is the inverse normal: find z first, then unstandardise with x = μ + zσ.

Worked example — X ~ N(50, 4²), find k such that P(X < k) = 0.95

From the inverse normal, P(Z < z) = 0.95 gives z = 1.6449.

x = μ + zσ = 50 + 1.6449 × 4 = 50 + 6.5796 = 56.6 (3 s.f.)

The three z-values worth memorising

P(Z < z) = 0.90 → z = 1.2816 · = 0.95 → z = 1.6449 · = 0.975 → z = 1.96

By symmetry, P(Z > z) = 0.05 also gives z = 1.6449, and P(Z < z) = 0.05 gives z = −1.6449.

Simultaneous equations: if you are told P(X < a) and P(X < b) but not μ and σ, standardise both to get two equations in μ and σ, then solve.

Quick check

Inverse normal

?X ~ N(50, 4²). Find the value of k for which P(X < k) = 0.95.

Sort it

Binomial, normal, or the approximation?

Tap a statement, then tap where it belongs.

🎯 Binomial B(n, p)

🔔 Normal N(μ, σ²)

🔗 Normal approx. to binomial

Distributions · approximation

Normal approximation to the binomial

When n is large and p is close to 0.5, the binomial B(n, p) is nearly symmetric and can be approximated by:

B(n, p) ≈ N(np, np(1 − p))μ = np and σ² = np(1 − p), so σ = √(np(1 − p))

The binomial is discrete and the normal is continuous, so you must apply a continuity correction — widen the interval by 0.5 at each end:

P(X ≥ 60) → P(Y > 59.5) · P(X > 60) → P(Y > 60.5)
P(X ≤ 60) → P(Y < 60.5) · P(X = 60) → P(59.5 < Y < 60.5)

Worked example — X ~ B(100, 0.5), estimate P(X ≥ 60)

μ = np = 50, σ² = np(1 − p) = 100 × 0.5 × 0.5 = 25, so σ = 5. Use Y ~ N(50, 5²).

Continuity correction: P(X ≥ 60) → P(Y > 59.5). Standardise: z = (59.5 − 50) ÷ 5 = 1.9.

P(Z > 1.9) = 1 − 0.9713 = 0.0287 (4 d.p.)

Calculate

Continuity correction

5X ~ B(100, 0.5) is approximated by Y ~ N(50, 5²). Using a continuity correction, find the value of z needed to estimate P(X ≥ 60).

Hint: P(X ≥ 60) becomes P(Y > 59.5). Then z = (59.5 − 50) ÷ 5. (The probability itself is 0.0287.)

Quick check

When can you use it?

?When is the normal approximation to the binomial appropriate?

Match it

Expression → what it is

Tap an item on the left, then its partner on the right.

Expression

Meaning

Distributions · choosing a model

Selecting an appropriate model

Spec point 4.3: choose a distribution with reasoning, and recognise when a model is not appropriate.

Binomial — a count of successes out of a fixed n, with constant p and independent trials. (Number of faulty items in a box of 20.)
Normal — a continuous, roughly symmetric measurement clustered around a mean. (Heights, masses, times.)
Discrete uniform — every outcome equally likely. (The score on a fair die.)
Neither — strongly skewed data, or trials that are clearly not independent, or a variable that cannot be negative but has a mean close to zero (a normal model would then predict impossible negative values).

Exam phrasing: "State an assumption you have made." Answer in context — e.g. "the faults occur independently of one another" or "each component has the same probability of failing".

Quick check

Pick the model

?A quality inspector models the number of faulty bulbs in a randomly chosen box of 20 bulbs, where each bulb has a 3% chance of being faulty, independently of the others. Which model is appropriate?

Distributions · discrete uniform

The discrete uniform distribution

The spec also expects you to identify the discrete uniform distribution: a discrete random variable where every outcome is equally likely.

Example — a fair six-sided die

X = the score. P(X = x) = 1/6 for x = 1, 2, 3, 4, 5, 6.

Every value has the same probability, and they sum to 6 × (1/6) = 1 ✔

Contrast: the number of sixes in ten rolls is binomial B(10, 1/6), not uniform — 1 six is far more likely than 9. The score on one roll is uniform; a count of successes is binomial.

Calculator skill: you are expected to get individual (Bpd) and cumulative (Bcd) binomial probabilities straight from the calculator, and normal probabilities and inverse-normal values likewise. Show the distribution and the probability statement in your working so the method still earns marks.

Quick check

Spot the uniform

?Which of these is modelled by a discrete uniform distribution?

Recap

The big ideas to know

Binomial B(n, p): fixed n · two outcomes · independent · constant p

P(X = r) = ⁿCᵣ pʳ (1 − p)ⁿ⁻ʳ · mean = np · variance = np(1 − p)

Normal N(μ, σ²): continuous, symmetric, points of inflection at μ ± σ, area = probability

Standardise: z = (x − μ) ÷ σ · Inverse normal: x = μ + zσ (z = 1.6449 for 95%)

Approximation: n large, p near 0.5 ⇒ B(n, p) ≈ N(np, np(1 − p)) with a continuity correction of ±0.5

Choosing a model: justify it, and state your assumptions in context

That is the whole of 9MA0 Topic 4 — Statistical distributions. Press Finish to see your score.

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