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S4 Statistical distributions — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in S4, read straight through: 3 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 9MA0 section S4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

S4.1 · Understand and use simple, discrete probability distributions (mean and variance of discrete random variables excluded), including the binomial distribution as a model; calculate probabilities using the binomial distribution.

Definition. Specification S4.1: understand and use simple, discrete probability distributions (calculation of the mean and variance of discrete random variables is excluded), including the binomial distribution as a model; calculate probabilities using the binomial distribution.

  • A discrete distribution lists the values with probabilities that are each between 00 and 11 and add to 11; a discrete uniform distribution gives every value the same probability.
  • X∼B(n,p)X\sim\mathrm B(n,p) models the number of successes in nn trials when the trials are independent, each has two outcomes and the probability of success pp is constant; those conditions are the answer to “explain why the binomial is suitable”.
  • Write the model down, X∼B(10,16)X\sim\mathrm B(10,\tfrac16), before any calculation.
  • Point probabilities use P(X=r)=(nr)pr(1−p)n−rP(X=r)=\binom{n}{r}p^r(1-p)^{n-r} and cumulative probabilities come from the calculator, but the inequality must be translated for an integer variable: P(X<3)=P(X≤2)P(X<3)=P(X\leq2), P(X≥3)=1−P(X≤2)P(X\geq3)=1-P(X\leq2), P(X>3)=1−P(X≤3)P(X>3)=1-P(X\leq3).
  • A probability found in one part often becomes the pp of a second binomial in the next, and npnp gives an expected count.

Worked example

1.

The random variable X∼B(14,0.4)X\sim\mathrm B(14,0.4). Find P(X<5)P(X<5), giving your answer to 4 decimal places.

(2)

(Total for Question 1 is 2 marks)

Working

  1. 1.For an integer-valued variable, P(X<5)=P(X≤4)P(X<5)=P(X\leq4).
  2. 2.From the cumulative binomial function with n=14n=14 and p=0.4p=0.4: P(X≤4)=0.27925…P(X\leq4)=0.27925\ldots.
  3. 3.To 4 decimal places, P(X<5)=0.2793P(X<5)=0.2793.

Answer: 0.27930.2793

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1.P(X≤4)P(X\leq4) with B(14,0.4)\mathrm B(14,0.4), or a sum of five point probabilitiesM1
awrt 0.27930.2793A1
Notes
M1: P(X≤5)=0.4859P(X\leq5)=0.4859 scores M0

(2 marks)

Mark-scheme wording to use

  • “M1 for sight or use of the correct distribution. Must have B, or Bin or Bpd or Bcd and the correct value for n and p, just n = 10, p = 1/6 is M0 … 1st A1 for awrt 0.155” — Pearson 9MA0, June 2024 Paper 31, Q1(a)(i)
  • “2nd A1 for awrt 0.775” — Pearson 9MA0, June 2024 Paper 31, Q1(a)(ii)
  • “M1 for selecting an appropriate binomial model T ~ B(10, p), where p is the probability of at least 15 successes from B(24, 0.55). A1 for p = 0.299126..., accepting 0.3 or better. A1 for P(T >= 5) awrt 0.149” — Pearson 9MA0, Sample Paper 31, Q5(b)

What the examiner wants. The inequality read correctly. The June 2024 Paper 31 report says almost all wrote down the right binomial model and found P(X=3)P(X=3), but for P(X<3)P(X<3) an answer of 0.9300.930 from P(X≤3)P(X\leq3) was often seen, suggesting unfamiliarity with the calculator, and others misread the inequality and calculated 1−P(X≤2)1-P(X\leq2).

Common mistakes

  • Don't use P(X<k)P(X<k) when the question asks for P(X≤k)P(X\leq k).
  • Don't calculate a single binomial probability when the event is cumulative, or omit an endpoint from the sum.

Exam tip. Translate the inequality into the exact binomial range before using cumulative probabilities and round only at the end.

S4.2 · Understand and use the Normal distribution as a model; find probabilities using the Normal distribution; link to histograms, mean, standard deviation, points of inflection and the binomial distribution.

Definition. Specification S4.2: understand and use the Normal distribution as a model; find probabilities using the Normal distribution; link to histograms, mean, standard deviation, points of inflection and the binomial distribution.

  • Write the model as X∼N(μ,σ2)X\sim\mathrm N(\mu,\sigma^2), with the variance, not the standard deviation, in the bracket.
  • The curve is symmetric about μ\mu with points of inflection at μ±σ\mu\pm\sigma, so a bell-shaped histogram and a mean close to the median support a Normal model.
  • Probabilities come from the calculator, but the working must show the standardisation z=x−μσz=\dfrac{x-\mu}{\sigma} when μ\mu or σ\sigma is unknown: an inverse-Normal value such as 1.28161.2816 for the top 10%10\% gives one linear equation in μ\mu and σ\sigma, and two such equations are solved simultaneously.
  • Keep the same units throughout.
  • A Normal variable is continuous, so there is no continuity correction; the correction, P(X>95.5)P(X>95.5) for “more than 9595”, belongs only to approximating a binomial B(n,p)\mathrm B(n,p) by N(np, np(1−p))\mathrm N(np,\,np(1-p)) when nn is large and pp is near 0.50.5.
A Normal density is symmetric about its mean; its inflection points occur at μ−σ\mu-\sigma and μ+σ\mu+\sigma, and shaded area represents probability.

Worked example

1.

The lifetime, LL hours, of a component is modelled by L∼N(72,82)L\sim\mathrm N(72,8^2). Exactly 10%10\% of components last longer than ll hours. Find the value of ll, giving your answer to 1 decimal place.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.P(L>l)=0.10P(L>l)=0.10, so P(L<l)=0.90P(L<l)=0.90 and the inverse-Normal value is z=1.2816z=1.2816.
  2. 2.Standardise: l−728=1.2816\dfrac{l-72}{8}=1.2816.
  3. 3.So l=72+8×1.2816=82.25…=82.3l=72+8\times1.2816=82.25\ldots=82.3 hours.

Answer: l=82.3l=82.3

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1.P(L<l)=0.9P(L<l)=0.9 or z=1.2816z=1.2816 seenM1
Standardises: l−728=\dfrac{l-72}{8}= their zzM1
awrt 82.382.3A1
Notes
M1: awrt 1.28
M1: Must be a zz-value between 1 and 1.5

(3 marks)

Mark-scheme wording to use

  • “1st M1 for attempting normal with mean = 100 or ft their answer to (c) … 1st A1 for correctly labelled standard deviation allow √(250/3) = √83.3... = 9.1(28….) or correctly labelled variance. Implied by N( μ, 250/3 ) or correct answer. … 2nd M1 for attempt at continuity correction i.e. sight of 95 ± 0.5 … 2nd A1 for awrt 0.689 [Answer only 4/4]” — Pearson 9MA0, June 2024 Paper 31, Q1(d)
  • “1st M1 for standardising 16.3 and setting equal to z value where 0.5 < |z| < 0.6 … 2nd M1 for standardising 29 and setting equal to z value where 1 < |z| < 1.5 … 3rd M1 dep on 1st or 2nd M1 for solving their two linear eq'ns – reach an eq'n in one variable … 1st A1 for σ = awrt 7.03 (but see 3rd case below)” — Pearson 9MA0, June 2024 Paper 31, Q5(d)
  • “B1 for using the correct model to obtain z awrt 0.842. M1 for standardising to an appropriate z-value between 0.5 and 1. A1 for sigma awrt 23.8 hours” — Pearson 9MA0, Specimen Paper 31, Q5(b)

What the examiner wants. A Normal used as a continuous model in the right units. The June 2024 Paper 31 report says errors in finding P(T<300)P(T<300) came from applying a continuity correction, P(T<299.5)P(T<299.5), or from not converting minutes to seconds, while the strongest answers to the unknown-parameter part wrote down a pair of linear equations in μ\mu and σ\sigma with suitable zz-values.

Common mistakes

  • Don't use the variance as the denominator when standardising instead of the standard deviation.
  • Don't use the lower-tail quantile for a value exceeded by a stated percentage.

Exam tip. Convert an exceedance percentage to the corresponding lower-tail probability before standardising or using an inverse Normal calculation.

S4.3 · Select an appropriate probability distribution for a context, with appropriate reasoning, including recognising when the binomial or Normal model may not be appropriate.

  • Select a distribution by matching its assumptions to the variable and data-generating process, not merely because its parameters can be estimated.
  • A binomial model needs a fixed number of trials, two outcomes per trial, independence and a constant success probability.
  • A Normal model is continuous and symmetric with unbounded tails, so it can be unsuitable for strongly skewed, bounded or discrete data.
  • Support a choice with contextual evidence such as histogram shape, stability over time and dependence; state how a failed assumption could affect predictions.

Worked example

The masses of loaves from a stable production line form a roughly symmetric, single-peaked histogram with no clear outliers. Explain why a Normal model may be suitable and why a binomial model is not.

  1. 1.Match the continuous measurement and bell-shaped empirical pattern to the features of a Normal distribution.
  2. 2.Reject the binomial model because the response is a measured mass, not a discrete success count.

Answer: Mass is continuous and the observed distribution is approximately symmetric and unimodal, supporting a Normal model.; A binomial model counts successes in a fixed number of two-outcome trials, so it does not model individual loaf masses.

Common mistakes

  • Don't use a binomial model when the success probability changes between trials.
  • Don't choose a distribution from the graph's shape alone and ignore the type of variable and trial assumptions.

Exam tip. Justify a model using the variable type, distribution shape and process assumptions, and name any condition that may fail.