What is S4.3: Selecting a distribution?

Select a distribution by matching its assumptions to the variable and data-generating process, not merely because its parameters can be estimated.

How many marks are S4.3 questions worth?

S4.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has S4.3 appeared in?

We have not yet indexed a verified past-paper appearance for S4.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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.

Draft — not yet indexed

Selecting a distribution

Worked answers and methods for S4.3 on Edexcel A-level Maths 9MA0.

Explanation

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.Match the continuous measurement and bell-shaped empirical pattern to the features of a Normal distribution.
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.

Worked practice

Tier 1 · Easy

A manufacturer inspects

12

independently chosen switches. Each switch has probability

0.04

of being faulty. State a suitable distribution for the number

F

of faulty switches and give its parameters.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$F\sim\operatorname{B}(12,0.04)$	2
Notes
There are $12$ fixed, independent trials, each switch is faulty or not faulty, and the fault probability is constant at $0.04$ . Therefore a binomial model with $n=12$ and $p=0.04$ is suitable.

(2 marks)

Tier 2 · Standard

A quality inspector selects

20

components without replacement from a batch of only

50

and records the number that are defective. Explain why a binomial model may be inappropriate and suggest a more suitable approach.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Sampling without replacement from a small batch makes the trials dependent and changes the defect probability after each selection. If the total number of defective components in the batch is known, use a hypergeometric model; equivalently, calculate with conditional probabilities that update after each selection. A binomial model would be an adequate approximation only when the sample size is small relative to the batch size, which is not true here because $20/50=0.4$ .	3
Notes
A binomial model requires a fixed success probability and independent trials. Removing $20$ of only $50$ components changes the composition appreciably, so these conditions fail. If the batch contains a known fixed number of defectives, the exact count distribution is hypergeometric; a probability tree with updated proportions is equivalent. Binomial can approximate sampling without replacement only when the sampling fraction is small, unlike the $40\%$ fraction here.

(3 marks)

Tier 3 · Hard

A technician proposes

D\sim\operatorname{B}(500,p)

for the number of defective pixels on each screen. Defects tend to occur in neighbouring clusters, and

p

varies between production shifts. Critique the model and suggest how the data should be used before choosing a replacement.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Clustering violates independence between pixel outcomes. Shift-to-shift variation violates the constant- $p$ assumption. The binomial model is therefore likely to understate variation and tail probabilities. Analyse screens by shift and inspect the empirical count distribution or cluster structure before selecting a model.	5
Notes
Although the number of pixels is fixed and each pixel is defective or not, two essential binomial assumptions fail. Neighbouring outcomes are dependent and different shifts have different probabilities. Both mechanisms create extra variation relative to a single binomial distribution, so compare separate-shift data and observed counts with candidate models rather than forcing one common $p$ .

(5 marks)

Tier 1 · Easy

Waiting times are non-negative and their histogram is strongly right-skewed. Give two reasons why a Normal model may be unsuitable.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	A Normal distribution is symmetric, unlike the strongly right-skewed data. A Normal distribution assigns positive probability to negative waiting times, which are impossible.	2
Notes
Compare both the observed shape and the possible values with the features of a Normal distribution. The mismatch in symmetry and support makes the model doubtful.

(2 marks)

Tier 2 · Standard

A player takes

30

shots in sequence. The probability of scoring decreases as the player becomes tired. Explain why a binomial model for the total number of goals may be inappropriate and suggest a refinement.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Although there is a fixed number of two-outcome trials, the success probability is not constant. Fatigue may also make later outcomes dependent on the earlier workload or sequence. A single binomial model is therefore inappropriate. Use trial-specific or fatigue-stage scoring probabilities estimated from suitable data.	4
Notes
A binomial model requires independent trials with one constant probability of success. The stated fatigue mechanism violates the constant-probability condition and may introduce dependence. A refined model should allow the probability to change with shot number or fatigue stage.

(4 marks)

Tier 3 · Hard

A business wants to model the number of visits to its website in an hour. There is no fixed maximum number of visits, and the observed distribution is discrete, non-negative and strongly right-skewed. Explain why neither a binomial distribution nor a Normal distribution is well justified. State how the business should use its data before selecting another model.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	A binomial distribution requires a fixed number of trials with two outcomes, but the number of possible visits is not a success count from a fixed number of trials. A Normal distribution is continuous and symmetric with unbounded tails, whereas the observed counts are discrete, non-negative and strongly right-skewed. The business should examine empirical frequencies across comparable hours and check whether conditions such as time of day change the distribution. Any replacement model should be checked by comparing its predicted frequencies or tail probabilities with further observed data.	5
Notes
Match each candidate distribution to both the variable type and its generating process. The binomial trial structure is absent, while the Normal shape and support conflict with the data. Stratifying comparable hours and validating predicted against observed frequencies provides evidence for a replacement rather than choosing one only by name.

(5 marks)

Tier 2 · Standard

A system examines

50

independently selected messages, each with probability

0.08

of being spam, and records the number flagged. It also records file-download times, whose histogram is continuous, roughly symmetric and single-peaked with no clear outliers. Select a suitable distribution for each variable and justify both choices.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	For the number of spam messages, use $S\sim\operatorname{B}(50,0.08)$ because there is a fixed number of independent two-outcome trials with constant probability. A Normal model may be suitable for the download times because the variable is continuous and the supplied histogram is approximately symmetric and unimodal without clear outliers.	4
Notes
Match the count to the binomial trial conditions: $50$ fixed selections, spam or not spam, independence and constant probability $0.08$ . The time variable is measured on a continuous scale, and its stated empirical shape matches the main features of a Normal density, so a Normal model is plausible rather than guaranteed.

(4 marks)

Tier 3 · Hard

Recovery times in a study form a bimodal histogram. Every observation comes from either treatment A or treatment B, and the separate histogram for each treatment is continuous, approximately symmetric and single-peaked. Explain why one Normal distribution for all recovery times is inappropriate. Propose a more suitable modelling strategy and state two checks required before using it.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The combined bimodal shape is not well represented by one symmetric, single-peaked Normal distribution. Treatment type creates two distinct sections of the population, so combining them hides different centres or spreads. Model the two treatments separately, using a separate Normal distribution for each if its data support that choice. Check each treatment's histogram for approximate symmetry, a single peak and influential outliers. Check that observations are collected comparably and independently within each treatment, then validate each model against further data or predicted frequencies.	5
Notes
A mixture of two groups can be bimodal even when each group is individually close to Normal. One fitted Normal would place too much probability between the peaks and misrepresent both groups. Stratifying by the known treatment variable addresses the generating mechanism; the shape, support, independence and predictive fit of each proposed group model must still be checked.

(5 marks)

Tier 3 · Hard

A factory records the number

X

of faulty items among

400

independently produced items, each with constant fault probability

0.50

. Select an exact distribution for

X

and explain why a Normal approximation is reasonable. Use the approximation, with a continuity correction, to estimate

P(190\leq X\leq210)

4

decimal places. State one process change that would undermine both the exact binomial model and this approximation.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$X\sim\operatorname{B}(400,0.50)$ exactly. $np=n(1-p)=200$ , so a Normal approximation is reasonable: $Y\sim\operatorname{N}(200,100)$ . $P(190\leq X\leq210)\approx P(189.5<Y<210.5)=0.7063$ . A change causing fault probabilities to vary between items, or faults to occur in dependent clusters, would violate the binomial assumptions and invalidate the stated approximation.	6
Notes
The fixed number of independent two-outcome trials with constant $p$ gives the exact binomial model. Its mean and variance are $200$ and $100$ , and both expected outcome counts are large. Applying continuity correction gives standardised bounds $-1.05$ and $1.05$ , so the estimate is $\Phi(1.05)-\Phi(-1.05)=0.706282\ldots=0.7063$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A company proposes

T\sim\operatorname{N}(5.25,2.5^2)

for a non-negative service time

T

, measured in minutes. Calculate the probability that this model assigns to an impossible negative time and the expected number of such modelled values in

1800

services. Give the probability to

4

decimal places and the expected number to the nearest integer. Assess the model and describe what evidence should be examined before choosing a replacement.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P(T<0)=P(Z<-2.1)=0.0179$ . Expected number below zero $=1800(0.017864\ldots)=32$ to the nearest integer. Assigning about $1.8\%$ probability to impossible values is a material support mismatch, so the Normal model is doubtful unless the context or measurements have been misunderstood. Inspect the empirical distribution for skewness, bounds, outliers and changes between service conditions, then compare candidate models with fresh observed frequencies or tail probabilities.	5
Notes
Standardising zero gives $z=(0-5.25)/2.5=-2.1$ , so the lower-tail probability is $0.017864\ldots$ . Multiplying by $1800$ gives $32.1559\ldots$ , which rounds to $32$ . Model selection must consider possible values as well as centre and spread, and should be validated against observed data rather than chosen from parameters alone.

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for S4.3. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in S4 Statistical distributions

Want help turning this into marks?

Bring S4.3 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call