S4 Statistical distributions — revision question pack

3 specification points · notes, questions, answers and worked methods

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.

Explanation

A discrete probability distribution lists possible values with probabilities between $0$ and $1$ whose total is $1$ ; a discrete uniform distribution assigns equal probability to each value. Use $X\sim\operatorname{B}(n,p)$ for a fixed number $n$ of independent trials, each with two outcomes and constant success probability $p$ .
For a binomial variable, $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ ; cumulative probabilities are often most efficiently found with a calculator.
Translate inequalities carefully: for integer-valued $X$ , $P(X<r)=P(X\leq r-1)$ and $P(X\geq r)=1-P(X\leq r-1)$ .
Use the cumulative binomial probability $P(X\leq4)=\sum_{r=0}^{4}\binom{14}{r}(0.4)^r(0.6)^{14-r}=0.279256\ldots$ .
Therefore $P(X\leq4)=0.2793$ .

Worked example

Let $X\sim\operatorname{B}(14,0.4)$ . Find $P(X\leq4)$ to $4$ decimal places.

1.Use the cumulative binomial probability $P(X\leq4)=\sum_{r=0}^{4}\binom{14}{r}(0.4)^r(0.6)^{14-r}=0.279256\ldots$ .
2.Therefore $P(X\leq4)=0.2793$ .

Answer: $0.2793$

Common mistakes

Don't use $P(X<k)$ when the question asks for $P(X\leq k)$ .
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.

Tier 1 · Easy

1.
A discrete random variable $X$ takes values $0$ , $1$ and $2$ with probabilities $k$ , $3k$ and $4k$ respectively. Find $k$ and $P(X\geq1)$ .
(3)
(Total for Question 1 is 3 marks)
2.
A discrete random variable $Y$ takes values $-3$ , $-1$ , $2$ and $5$ with probabilities $0.10$ , $0.25$ , $0.40$ and $0.25$ respectively. Find $P(Y>0)$ and $P(|Y|\leq2)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The random variable $X$ has a discrete uniform distribution on $\{-2,0,2,4,6\}$ . Write down its probability distribution, then find $P(X<3)$ and $P(|X|\geq4)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Independent trials each have success probability $0.20$ . Find the smallest number of trials needed for the probability of at least one success to exceed $0.95$ . State this probability for your minimum number of trials to $4$ decimal places.
(4)
(Total for Question 2 is 4 marks)
3.
Let $X\sim\operatorname{B}(20,0.35)$ . Given $P(X\leq3)=0.0443756$ , $P(X\leq4)=0.1181966$ and $P(X\leq8)=0.7623776$ , find $P(4\leq X\leq8)$ and $P(4<X\leq8)$ . Give each answer to $4$ decimal places.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A binomial random variable $X$ has $10$ trials and satisfies $P(X=0)=0.8^{10}$ . Find its success probability $p$ , then calculate $P(X\geq3)$ to $4$ decimal places.
(5)
(Total for Question 1 is 5 marks)
2.
Let $X\sim\operatorname{B}(12,0.25)$ . Given that at least one success occurs, find the probability that at least three successes occur. Give your answer to $4$ decimal places, using unrounded values in your working.
(5)
(Total for Question 2 is 5 marks)
3.
Eight independent trials share a constant success probability $p$ , where $0<p<1$ , and $X$ denotes their total number of successes. Given that $P(X=2)=3P(X=1)$ , find $p$ . Hence find $P(X\geq2)$ to $4$ decimal places.
(6)
(Total for Question 3 is 6 marks)
4.
A binomial random variable $X\sim\operatorname{B}(n,p)$ satisfies $P(X=1)=6P(X=0)$ and $P(X=2)=2P(X=1)$ , where $n\geq2$ and $0<p<1$ . Find $n$ and $p$ . Hence find the exact value of $P(X\geq2)$ .
(6)
(Total for Question 4 is 6 marks)
5.
Machine A produces $8$ independently inspected components and machine B produces $7$ . Every component, regardless of machine, has the same probability $0.40$ of being defective, and all inspections are independent. Let $T$ be the total number of defective components. State the distribution of $T$ and calculate $P(T\geq8)$ to $4$ decimal places. Explain why the same binomial distribution for $T$ would not be valid if machine B instead had defect probability $0.20$ .
(6)
(Total for Question 5 is 6 marks)

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.

Explanation

Write $X\sim\operatorname{N}(\mu,\sigma^2)$ , where $\mu$ is the mean and $\sigma$ is the standard deviation; standardise with $Z=\frac{X-\mu}{\sigma}$ .
The Normal curve is continuous, symmetric about $\mu$ , has total area $1$ , and has points of inflection at $\mu-\sigma$ and $\mu+\sigma$ .
A Normal model is plausible for a roughly symmetric, unimodal histogram with no strong outliers, but context and the variable's possible values also matter.
A binomial distribution may be approximated by a Normal distribution when $n$ is large and $p$ is close to $0.5$ ; use a continuity correction.

A Normal density is symmetric about its mean; its inflection points occur at

\mu-\sigma

and

\mu+\sigma

, and shaded area represents probability.

Worked example

The lifetime $L$ of a component, in hours, is modelled by $L\sim\operatorname{N}(72,8^2)$ . Find the lifetime exceeded by exactly $10\%$ of components, to $1$ decimal place.

1.If $P(L>l)=0.10$ , then $P(L\leq l)=0.90$ .
2.The $0.90$ standard Normal quantile is $z=1.28155\ldots$ .
3.Hence $l=72+8(1.28155\ldots)=82.252\ldots$ , so the required lifetime is $82.3$ hours.

Answer: $82.3$ hours

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.

Tier 1 · Easy

1.
A random variable has distribution $X\sim\operatorname{N}(50,6^2)$ . Find $P(X<56)$ to $4$ decimal places.
(2)
(Total for Question 1 is 2 marks)
2.
The random variable $T$ has distribution $T\sim\operatorname{N}(72,9^2)$ . State the two values of $T$ at which the Normal density curve has points of inflection.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A random variable $X$ is modelled by $X\sim\operatorname{N}(64,4^2)$ . Find $P(58<X<70)$ to $4$ decimal places.
(4)
(Total for Question 1 is 4 marks)
2.
A variable $X$ is modelled by $X\sim\operatorname{N}(\mu,5^2)$ . Given that $P(X<42)=0.10$ and that the $0.10$ standard Normal quantile is $-1.2816$ , find $\mu$ to $3$ significant figures.
(4)
(Total for Question 2 is 4 marks)
3.
A variable $X$ is modelled by $X\sim\operatorname{N}(\mu,\sigma^2)$ . Given $P(X<42)=0.20$ and $P(X<58)=0.80$ , find $\mu$ and $\sigma$ . Give $\sigma$ to $3$ significant figures. You may use the standard Normal quantiles $-0.8416$ and $0.8416$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Let $X\sim\operatorname{B}(200,0.35)$ . Use a Normal approximation with a continuity correction to estimate $P(X\geq82)$ , giving your answer to $4$ decimal places.
(5)
(Total for Question 1 is 5 marks)
2.
A variable $X$ is modelled by $X\sim\operatorname{N}(30,\sigma^2)$ . Given that $P(26<X<34)=0.80$ , find $\sigma$ to $3$ significant figures. Hence find $P(X>36\mid X>32)$ to $4$ decimal places. Use the unrounded value of $\sigma$ in your working. You may use the $0.90$ standard Normal quantile $1.2816$ .
(6)
(Total for Question 2 is 6 marks)
3.
A process measurement $X$ is modelled by a Normal distribution whose density curve has points of inflection at $492$ and $508$ . Measurements outside the interval $484\leq X\leq516$ are rejected. Find the probability that a measurement is accepted and the expected number rejected from $1200$ measurements. Given that a measurement is accepted, find the probability that it exceeds $505$ . Give probabilities to $4$ decimal places and the expected number to the nearest integer. Use unrounded values in your working.
(7)
(Total for Question 3 is 7 marks)
4.
A variable $X$ is modelled by $X\sim\operatorname{N}(\mu,\sigma^2)$ . Given that $P(X<44)=P(X>68)$ and $P(50<X<62)=0.60$ , find $\mu$ and $\sigma$ . You may use the $0.80$ standard Normal quantile $0.8416$ . Hence find the two values of $X$ at which the density curve has points of inflection, giving them to $3$ significant figures. A histogram represents $900$ such measurements, with bar area equal to frequency. Estimate the total frequency between the points of inflection, and explain why agreement with this frequency alone would not establish that a Normal model is suitable.
(7)
(Total for Question 4 is 7 marks)
5.
A variable $X$ is modelled by $X\sim\operatorname{N}(\mu,\sigma^2)$ . Given that $P(X<40)=0.10$ and $P(X<55)=0.70$ , find $\mu$ and $\sigma$ to $3$ significant figures. Hence find $P(45<X<60)$ to $4$ decimal places. Use unrounded values in your working. You may use the standard Normal quantiles $-1.2816$ and $0.5244$ .
(7)
(Total for Question 5 is 7 marks)

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.

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.

Tier 1 · Easy

1.
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)
2.
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 2 is 2 marks)

Tier 2 · Standard

1.
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 1 is 3 marks)
2.
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 2 is 4 marks)
3.
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 3 is 4 marks)

Tier 3 · Hard

1.
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 1 is 5 marks)
2.
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 2 is 5 marks)
3.
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 3 is 5 marks)
4.
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 $P(190\leq X\leq210)$ to $4$ decimal places. State one process change that would undermine both the exact binomial model and this approximation.
(6)
(Total for Question 4 is 6 marks)
5.
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 5 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

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.

Tier 1 · Easy

Mark scheme for S4.1 Tier 1 · Easy
Question	Scheme	Marks
1	$k=\frac18$ . $P(X\geq1)=\frac78$ .	3
(3 marks)		3
Notes
Probabilities sum to $1$ , so $k+3k+4k=8k=1$ and $k=1/8$ . Therefore $P(X\geq1)=3k+4k=7k=7/8$ .
2	$P(Y>0)=0.65$ $P(\|Y\|\leq2)=0.65$	2
(2 marks)		2
Notes
The positive values are $2$ and $5$ , giving $0.40+0.25=0.65$ . The values satisfying $\|Y\|\leq2$ are $-1$ and $2$ , giving $0.25+0.40=0.65$ .

Tier 2 · Standard

Mark scheme for S4.1 Tier 2 · Standard
Question	Scheme	Marks
1	$P(X=x)=\dfrac15$ for each $x\in\{-2,0,2,4,6\}$ $P(X<3)=\dfrac35$ $P(\|X\|\geq4)=\dfrac25$	3
(3 marks)		3
Notes
There are five equally likely values, so each has probability $1/5$ . The event $X<3$ contains $-2,0,2$ , giving $3/5$ . The event $\|X\|\geq4$ contains $4$ and $6$ , giving $2/5$ .
2	$14$ trials $P(X\geq1)=0.9560$ when $n=14$	4
(4 marks)		4
Notes
For $n$ trials, $P(X\geq1)=1-0.8^n$ . The requirement is $0.8^n<0.05$ , so $n>\log(0.05)/\log(0.8)=13.425\ldots$ . The smallest integer is $14$ , and $1-0.8^{14}=0.956019\ldots=0.9560$ .
3	$P(4\leq X\leq8)=0.7180$ $P(4<X\leq8)=0.6442$	4
(4 marks)		4
Notes
For the inclusive lower endpoint, subtract only values up to $3$ : $P(4\leq X\leq8)=P(X\leq8)-P(X\leq3)=0.7623776-0.0443756=0.7180020$ . For the strict lower endpoint, subtract values up to $4$ : $P(4<X\leq8)=0.7623776-0.1181966=0.6441810$ . Rounding gives $0.7180$ and $0.6442$ .

Tier 3 · Hard

Mark scheme for S4.1 Tier 3 · Hard
Question	Scheme	Marks
1	$p=0.2$ . $P(X\geq3)=0.3222$ .	5
(5 marks)		5
Notes
For $X\sim\operatorname{B}(10,p)$ , $P(X=0)=(1-p)^{10}$ . Hence $(1-p)^{10}=0.8^{10}$ , so $p=0.2$ . Then $P(X\geq3)=1-P(X\leq2)=1-[0.8^{10}+10(0.2)(0.8)^9+\binom{10}{2}(0.2)^2(0.8)^8]=0.322200\ldots$ , giving $0.3222$ .
2	$0.6293$	5
(5 marks)		5
Notes
Since $X\geq3$ implies $X\geq1$ , the required conditional probability is $P(X\geq3)/P(X\geq1)$ . Now $P(X\geq3)=1-P(X\leq2)=0.609324\ldots$ and $P(X\geq1)=1-P(X=0)=1-0.75^{12}=0.968323\ldots$ . Their ratio is $0.629257\ldots$ , giving $0.6293$ .
3	$p=\dfrac6{13}$ $P(X\geq2)=0.9445$	6
(6 marks)		6
Notes
For $X\sim\operatorname{B}(8,p)$ , $P(X=2)=28p^2(1-p)^6$ and $P(X=1)=8p(1-p)^7$ . Since $0<p<1$ , division gives $\frac{7p}{2(1-p)}=3$ , so $7p=6(1-p)$ and $p=6/13$ . Therefore $P(X\geq2)=1-P(X=0)-P(X=1)=1-(7/13)^8-8(6/13)(7/13)^7=0.944473\ldots=0.9445$ .
4	$n=3$ and $p=\dfrac23$ $P(X\geq2)=\dfrac{20}{27}$	6
(6 marks)		6
Notes
Write $q=1-p$ . The first ratio gives $np/q=6$ . Also $P(X=2)/P(X=1)=(n-1)p/(2q)=2$ , so $(n-1)p/q=4$ . Subtracting gives $p/q=2$ , hence $p=2/3$ and then $n=3$ . Therefore $P(X\geq2)=1-P(X=0)-P(X=1)=1-(1/3)^3-3(2/3)(1/3)^2=20/27$ .
5	$T\sim\operatorname{B}(15,0.40)$ $P(T\geq8)=0.2131$ If machine B had probability $0.20$ , the $15$ trials would not share one constant success probability, so their total would not have a binomial distribution with one parameter $p$ .	6
(6 marks)		6
Notes
The two independent counts have the same success probability, so pooling their $8+7=15$ trials gives $T\sim\operatorname{B}(15,0.40)$ . Thus $P(T\geq8)=\sum_{r=8}^{15}\binom{15}{r}(0.4)^r(0.6)^{15-r}=0.213103\ldots=0.2131$ . Different machine probabilities would violate the constant- $p$ condition even if all component outcomes remained independent.

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.

Tier 1 · Easy

Mark scheme for S4.2 Tier 1 · Easy
Question	Scheme	Marks
1	$0.8413$	2
(2 marks)		2
Notes
Standardise: $z=(56-50)/6=1$ . Therefore $P(X<56)=P(Z<1)=0.841344\ldots$ , which is $0.8413$ .
2	$63$ and $81$	2
(2 marks)		2
Notes
A Normal density has points of inflection at $\mu-\sigma$ and $\mu+\sigma$ . Here these are $72-9=63$ and $72+9=81$ .

Tier 2 · Standard

Mark scheme for S4.2 Tier 2 · Standard
Question	Scheme	Marks
1	$0.8664$	4
(4 marks)		4
Notes
Standardising gives $z=(58-64)/4=-1.5$ and $z=(70-64)/4=1.5$ . Hence $P(58<X<70)=P(-1.5<Z<1.5)=\Phi(1.5)-\Phi(-1.5)=0.93319\ldots-0.06681\ldots=0.8664$ .
2	$\mu=48.4$ to $3$ significant figures	4
(4 marks)		4
Notes
Standardising the tenth percentile gives $(42-\mu)/5=-1.2816$ . Hence $42-\mu=-6.408$ , so $\mu=48.408$ , which is $48.4$ to $3$ significant figures.
3	$\mu=50$ $\sigma=9.51$ to $3$ significant figures	5
(5 marks)		5
Notes
Standardising the two percentiles gives $(42-\mu)/\sigma=-0.8416$ and $(58-\mu)/\sigma=0.8416$ . Adding the equations, or using symmetry, gives $\mu=50$ . Then $8/\sigma=0.8416$ , so $\sigma=8/0.8416=9.50570\ldots$ , giving $9.51$ to $3$ significant figures.

Tier 3 · Hard

Mark scheme for S4.2 Tier 3 · Hard
Question	Scheme	Marks
1	$0.0441$	5
(5 marks)		5
Notes
Use $Y\sim\operatorname{N}(np,np(1-p))=\operatorname{N}(70,45.5)$ . The continuity correction gives $P(X\geq82)\approx P(Y>81.5)$ . Thus $z=(81.5-70)/\sqrt{45.5}=1.7049\ldots$ , so the upper-tail probability is $1-\Phi(1.7049\ldots)=0.0441$ .
2	$\sigma=3.12$ $P(X>36\mid X>32)=0.1046$	6
(6 marks)		6
Notes
By symmetry, $P(X<34)=0.90$ , so $4/\sigma=1.2816$ and $\sigma=3.1211\ldots$ . Since $X>36$ implies $X>32$ , the conditional probability is $P(X>36)/P(X>32)$ . The corresponding standardised values are $1.9224$ and $0.6408$ , giving tail probabilities $0.0272777\ldots$ and $0.260826\ldots$ . Their ratio is $0.104582\ldots$ , giving $0.1046$ .
3	$\mu=500$ and $\sigma=8$ $P(484\leq X\leq516)=0.9545$ Expected number rejected $=55$ $P(X>505\mid 484\leq X\leq516)=0.2548$	7
(7 marks)		7
Notes
The inflection points are $\mu-\sigma$ and $\mu+\sigma$ , so their midpoint gives $\mu=500$ and half their separation gives $\sigma=8$ . The acceptance bounds standardise to $-2$ and $2$ , hence the acceptance probability is $\Phi(2)-\Phi(-2)=0.954499\ldots$ . The expected rejected count is $1200[1-0.954499\ldots]=54.6003\ldots$ , giving $55$ . Within the acceptance interval, exceeding $505$ means $505<X\leq516$ ; the lower standardised value is $0.625$ . Thus the conditional probability is $[\Phi(2)-\Phi(0.625)]/[\Phi(2)-\Phi(-2)]=0.254830\ldots=0.2548$ .
4	$\mu=56$ and $\sigma=7.13$ to $3$ significant figures The points of inflection are at $X=48.9$ and $X=63.1$ to $3$ significant figures. Estimated frequency between the points of inflection $=614$ to the nearest whole number. Matching one central histogram area does not check the model's symmetry, unimodality, tails or possible outliers, so the full histogram and context must also be considered.	7
(7 marks)		7
Notes
Equal opposite-tail probabilities in a Normal distribution place the mean halfway between $44$ and $68$ , so $\mu=56$ . The interval $50<X<62$ is therefore $\mu-6<X<\mu+6$ . Its central probability is $0.60$ , leaving $0.20$ in each tail, so $6/\sigma=0.8416$ and $\sigma=6/0.8416=7.12928\ldots$ . The points of inflection are $\mu\pm\sigma=48.8707\ldots$ and $63.1293\ldots$ . The probability between them is $P(-1<Z<1)=0.682689\ldots$ , so the histogram area represents an estimated frequency $900(0.682689\ldots)=614.420\ldots$ , giving $614$ to the nearest whole number. Agreement over only this interval is not a complete model check.
5	$\mu=50.6$ and $\sigma=8.31$ to $3$ significant figures $P(45<X<60)=0.6216$	7
(7 marks)		7
Notes
Standardising gives $(40-\mu)/\sigma=-1.2816$ and $(55-\mu)/\sigma=0.5244$ . Subtraction gives $15/\sigma=1.8060$ , so $\sigma=8.305647\ldots$ and $\mu=40+1.2816\sigma=50.644518\ldots$ . Therefore $P(45<X<60)=\Phi((60-\mu)/\sigma)-\Phi((45-\mu)/\sigma)=0.621622\ldots=0.6216$ .

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.

Tier 1 · Easy

Mark scheme for S4.3 Tier 1 · Easy
Question	Scheme	Marks
1	$F\sim\operatorname{B}(12,0.04)$	2
(2 marks)		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	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
(2 marks)		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.

Tier 2 · Standard

Mark scheme for S4.3 Tier 2 · Standard
Question	Scheme	Marks
1	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
(3 marks)		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.
2	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
(4 marks)		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.
3	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
(4 marks)		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.

Tier 3 · Hard

Mark scheme for S4.3 Tier 3 · Hard
Question	Scheme	Marks
1	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
(5 marks)		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 $p$ .
2	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
(5 marks)		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.
3	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
(5 marks)		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.
4	$X\sim\operatorname{B}(400,0.50)$ 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
(6 marks)		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 $\Phi(1.05)-\Phi(-1.05)=0.706282\ldots=0.7063$ .
5	$P(T<0)=P(Z<-2.1)=0.0179$ . 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
(5 marks)		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.