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OCR A-level Mathematics A (H240) ยท 2.04 Statistical Distributions
Mini-Lesson

Statistical distributions

OCR section 2.04 gives you two models to choose between: the binomial B(n, p) for counting successes, and the Normal N(μ, σ²) for continuous measurements. You need to select the right one, use it, and know its limitations.

Binomial B(n, p) Normal N(ฮผ, ฯƒยฒ) Choosing a model the modelling decision is worth as many marks as the calculation

Every probability below is checked against the standard tables. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.

Distributions · binomial

The binomial distribution

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 just two outcomes (success / failure).
  • The trials are independent.
  • The probability of success, p, is constant.
P(X = r) = nCr pr (1 − p)n−rE(X) = np  ยท  Var(X) = np(1 โˆ’ p)
Worked example โ€” X ~ B(20, 0.3)

E(X) = np = 20 × 0.3 = 6.

Var(X) = np(1 − p) = 20 × 0.3 × 0.7 = 4.2.

Standard deviation = √4.2 = 2.05 (3 s.f.).

Where the model breaks: drawing counters without replacement breaks independence and makes p change โ€” so it is not binomial. Sampling from a very large population is close enough to be modelled as binomial anyway.

Calculate

Your turn โ€” expected value

1X ~ B(20, 0.3). Find E(X).
Hint: E(X) = np = 20 × 0.3.
Calculate

Your turn โ€” variance

2X ~ B(20, 0.3). Find Var(X).
Hint: Var(X) = np(1 − p) = 20 × 0.3 × 0.7.
Calculate

Your turn โ€” a binomial probability

3X ~ B(10, 0.2). Find P(X = 3), to 4 decimal places.
Hint: 10C3 (0.2)³ (0.8)⁷ = 120 × 0.008 × 0.2097152.
Quick check

Is it binomial?

?Which of these is not well modelled by a binomial distribution?
Distributions · normal

The Normal distribution

X ~ N(μ, σ²) models a continuous variable that clusters symmetrically about a mean โ€” heights, masses, measurement errors.

  • The curve is symmetric and bell-shaped, centred on μ (which is also the median and mode).
  • The points of inflection lie at μ ± σ โ€” that is what σ means visually.
  • Roughly 68% lies within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ.
  • The total area under the curve is 1.
z = (x − μ) ÷ σstandardising converts any Normal variable to Z ~ N(0, 1), the standard Normal

Careful with the notation: N(μ, σ²) quotes the variance, not the standard deviation. N(50, 64) means σ = 8, not 64.

Distributions · z-scores

Standardising โ€” worked

Worked example โ€” X ~ N(50, 8²). Find P(X > 62).

Standardise: z = (62 − 50) / 8 = 12/8 = 1.5.

So P(X > 62) = P(Z > 1.5) = 1 − Φ(1.5).

From tables, Φ(1.5) = 0.93319.

∴ P(X > 62) = 1 − 0.93319 = 0.0668 (3 s.f.).

Sanity check: 62 is 1.5 standard deviations above the mean, so only a small tail should be beyond it. About 6.7% is entirely reasonable. ✓

The inverse Normal runs it backwards: given a probability, find the value.

Worked example โ€” find a such that P(X < a) = 0.9

From the inverse Normal, the z-value with 0.9 below it is z = 1.2816.

Un-standardise: a = μ + zσ = 50 + 1.2816(8) = 50 + 10.25 = 60.3 (3 s.f.).

Always sketch the bell curve and shade. It is the only reliable way to know whether you want Φ(z), 1 − Φ(z), or a difference of two.

Calculate

Your turn โ€” standardise

4X ~ N(50, 8²). Find the z-value corresponding to x = 62.
z =
Hint: z = (x − μ) ÷ σ = (62 − 50) ÷ 8.
Calculate

Your turn โ€” the probability

5X ~ N(50, 8²). Given Φ(1.5) = 0.93319, find P(X > 62), to 4 decimal places.
Hint: P(X > 62) = P(Z > 1.5) = 1 − Φ(1.5) = 1 − 0.93319.
Quick check

Reading N(ฮผ, ฯƒยฒ)

?A variable is modelled as X ~ N(50, 64). What is its standard deviation?
Distributions · approximation

The Normal approximation to the binomial

When n is large, a binomial distribution starts to look like a Normal one โ€” the bar chart of B(n, p) becomes an unmistakable bell.

B(n, p) ≈ N( np , np(1 − p) )match the mean to np and the variance to np(1 โˆ’ p) โ€” the approximation is best when p is near 0.5
Worked example โ€” identify the approximating distribution

Y ~ B(100, 0.4) with n = 100, which is large.

Mean = np = 100 × 0.4 = 40.

Variance = np(1 − p) = 100 × 0.4 × 0.6 = 24.

∴ Y is approximately N(40, 24), so the standard deviation is √24 = 4.90 (3 s.f.).

An important OCR spec detail. The specification requires you to understand that a binomial with large n can be approximated by a Normal distribution โ€” but it states that questions explicitly requiring calculations using the Normal approximation to the binomial are excluded. So learn the idea and the parameters; you will not be asked to grind out a probability with it.

Quick check

The approximating distribution

?Y ~ B(100, 0.4). Which Normal distribution approximates it?
Quick check

Which model?

?The masses of apples from an orchard cluster symmetrically around 120 g. Which model is appropriate?
Sort it

Binomial, Normal or neither?

Tap a scenario, then tap the model that fits best.

๐ŸŽฏ Binomial

๐Ÿ”” Normal

โ“ Neither

Match it

Distribution facts

Tap a description on the left, then the correct expression.

Description
Expression
Recap

The big ideas to know

Binomial: fixed n, two outcomes, independent trials, constant p

Binomial formulae: P(X = r) = nCrpr(1−p)n−r; E(X) = np; Var(X) = np(1 − p)

Normal: symmetric bell; points of inflection at μ ± σ; about 95% within 2σ

Notation: N(μ, σ²) quotes the variance โ€” N(50, 64) means σ = 8

Standardise: z = (x − μ)/σ; then use Φ(z) โ€” always sketch and shade

Inverse Normal: given the probability, find z, then x = μ + zσ

Approximation: B(n, p) ≈ N(np, np(1−p)) for large n โ€” but OCR excludes calculations with it

That is OCR 2.04 โ€” and these two models are what hypothesis testing tests. Press Finish to see your score.

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