FS1-5 Central Limit Theorem — revision question pack

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

Checked against Edexcel 9FM0 section FS1-5. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FS1-5.1 · Applications of the Central Limit Theorem to other distributions.

Explanation

For a large independent random sample from a population with mean $\mu$ and variance $\sigma^2$ , the Central Limit Theorem gives $\overline X\mathrel{\dot\sim}N(\mu,\sigma^2/n)$ . The standard error is $\sigma/\sqrt n$ , so standardise with $Z=(\overline X-\mu)/(\sigma/\sqrt n)$ .
Applications may use any distribution studied in A-level Mathematics or Further Statistics 1; first obtain that population's mean and variance.
No proof is required.
The approximation improves with sample size and generally needs more caution for strongly skewed populations.
Examiners expect the variance of the sample mean to be divided by $n$ , approximation notation, and no continuity correction when the sample mean is treated directly.

The approximate normal distribution of a large-sample mean, centred at the population mean.

Worked example

A population has mean $30$ and variance $64$ . For a random sample of size $100$ , estimate $P(28.8<\overline X<31.6)$ .

1. $\overline X\mathrel{\dot\sim}N(30,64/100)$ , so the standard error is $0.8$ .
2.The lower standardised value is $(28.8-30)/0.8=-1.5$ .
3.The upper standardised value is $(31.6-30)/0.8=2$ .
4. $P(-1.5<Z<2)=0.9104$ to $4$ significant figures.

Answer: The estimated probability is $0.9104$ .

Common mistakes

Don't use variance $64$ for $\overline X$ instead of $64/100$ .
Don't divide the standard deviation by $n$ instead of by $\sqrt n$ .
Don't add a continuity correction to bounds already stated for the sample mean.

Exam tip

Write the approximate distribution of $\overline X$ before standardising either boundary.

Tier 1 · Easy

1.
Independent lifetimes have population mean $50$ hours and variance $100$ hours $^2$ . For a sample of $36$ lifetimes, use the Central Limit Theorem to estimate $P(\overline X<47.5)$ .
(4)
(Total for Question 1 is 4 marks)
2.
Independent observations have distribution $\operatorname{Bin}(10,0.4)$ . A sample of $60$ observations has mean $\overline X$ . Using the Central Limit Theorem, find $c$ to $3$ decimal places such that $P(\overline X<c)\approx0.90$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
A geometric population has parameter $p=0.32$ . A random sample of $45$ observations is taken. Find the value of $c$ for which $P(\overline X>c)\approx0.05$ , giving $c$ to $3$ decimal places.
(6)
(Total for Question 1 is 6 marks)
2.
Independent observations have a Poisson distribution with mean $5$ . A random sample of $n$ observations is taken and $\overline X$ denotes the sample mean. For $n=50$ , use the Central Limit Theorem to find the half-width $d$ such that $P(|\overline X-5|<d)\approx0.95$ . Give $d$ to $3$ decimal places.
(5)
(Total for Question 2 is 5 marks)
3.
The discrete random variable $X$ has distribution $\begin{array}{c|ccc}x&0&2&5\\ \hline P(X=x)&0.2&0.5&0.3\end{array}$ . A random sample of $42$ independent observations has mean $\overline X$ . Using the Central Limit Theorem, estimate $P(\overline X>2.9\mid\overline X>2.4)$ , giving your answer to $4$ decimal places.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
A population has mean $\mu$ and variance $36$ . Using the Central Limit Theorem normal approximation, estimate the least sample size $n$ for which $P(|\overline X-\mu|<1.2)\geq0.95$ .
(6)
(Total for Question 1 is 6 marks)
2.
Using a suitable approximation, find the smallest value of $n$ for which $P(\overline X<8.1)\geq0.99$ , where $\overline X$ is the mean of $n$ independent observations of a random variable $X$ . The variable $X$ counts the trial on which the third success occurs. Its success probability is unknown, but $E(X)=7.5$ .
(6)
(Total for Question 2 is 6 marks)
3.
The discrete random variable $X$ is uniformly distributed on $\{0,1,2,3\}$ , and $Y=2^X$ . A random sample of $64$ independent observations of $Y$ has mean $\overline Y$ . Use the Central Limit Theorem to estimate $P(\overline Y>4.2)$ , giving your answer to $4$ decimal places.
(7)
(Total for Question 3 is 7 marks)
4.
The score on a fair six-sided die is denoted by $X$ . A random sample of $40$ scores has mean $\overline X$ . (a) Use the Central Limit Theorem to estimate $P(\overline X<3.2)$ . Give your estimate to $4$ decimal places. (b) Explain why the Central Limit Theorem is needed here. (c) Find the corresponding estimate when the sample size is $10$ and explain the effect on the estimate.
(8)
(Total for Question 4 is 8 marks)
5.
A digital helpdesk models the number $X$ of tickets logged in a $30$ -minute interval by $X\sim\operatorname{Po}(6)$ . The mean count for $36$ selected intervals is $\overline X$ . Use the Central Limit Theorem to estimate $P(5.4<\overline X<6.5)$ , giving your answer to $4$ decimal places. State an assumption you have made. Suggest one limitation of the model.
(7)
(Total for Question 5 is 7 marks)

Answer key

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

FS1-5.1 · Applications of the Central Limit Theorem to other distributions.

Tier 1 · Easy

Mark scheme for FS1-5.1 Tier 1 · Easy
Question	Scheme	Marks
1	$P(\overline X<47.5)\approx0.0668$	4
(4 marks)		4
Notes
$\overline X\mathrel{\dot\sim}N(50,100/36)$ , so its standard error is $10/6=5/3$ . Hence $P(\overline X<47.5)\approx P(Z<(47.5-50)/(5/3))=P(Z<-1.5)=0.0668072\ldots$ .
2	$E(X)=4$ and $\operatorname{Var}(X)=2.4$ $\overline X\mathrel{\dot\sim}N(4,2.4/60)$ $z_{0.90}=1.281551\ldots$ $c=4.256$ to $3$ decimal places	4
(4 marks)		4
Notes
For $X\sim\operatorname{Bin}(10,0.4)$ , $E(X)=10(0.4)=4$ and $\operatorname{Var}(X)=10(0.4)(0.6)=2.4$ . Hence $\overline X\mathrel{\dot\sim}N(4,2.4/60)$ , with standard error $0.2$ . The $0.90$ standard-normal quantile is $1.281551565\ldots$ , so $c=4+1.281551565(0.2)=4.256310313\ldots$ . This rounds to $4.256$ . The sample mean is modelled directly, so no continuity correction is used.

Tier 2 · Standard

Mark scheme for FS1-5.1 Tier 2 · Standard
Question	Scheme	Marks
1	$E(X)=1/p=25/8$ $\operatorname{Var}(X)=(1-p)/p^2=425/64$ $\overline X\mathrel{\dot\sim}N(25/8,85/576)$ The standard error is $\sqrt{85}/24=0.384147686\ldots$ $P(\overline X>c)\approx0.05$ gives $(c-25/8)/(\sqrt{85}/24)=z_{0.95}=1.644853627\ldots$ $c=25/8+1.644853627\ldots(\sqrt{85}/24)=3.756866714\ldots=3.757$	6
(6 marks)		6
Notes
For a geometric population with $p=0.32$ , $\mu=1/p=25/8$ and $\sigma^2=(1-p)/p^2=425/64$ . Thus $\overline X\mathrel{\dot\sim}N(25/8,(425/64)/45)=N(25/8,85/576)$ , with standard error $\sqrt{85}/24=0.3841476857\ldots$ . The condition $P(\overline X>c)\approx0.05$ places $c$ at the $0.95$ standard-normal quantile, so $(c-25/8)/(\sqrt{85}/24)=z_{0.95}=1.6448536269\ldots$ . Hence $c=25/8+1.6448536269\ldots(\sqrt{85}/24)=3.7568667141\ldots$ , which is $3.757$ to $3$ decimal places. The sample mean is modelled directly, so no continuity correction is used.
2	$E(X)=\operatorname{Var}(X)=5$ $\overline X\mathrel{\dot\sim}N(5,5/50)$ $P(\|Z\|<1.959963984\ldots)=0.95$ $d=1.959963984\sqrt{5/50}=0.619795032\ldots$ $d=0.620$ to $3$ decimal places	5
(5 marks)		5
Notes
For a $\operatorname{Po}(5)$ population, $E(X)=\operatorname{Var}(X)=5$ , so with $n=50$ , $\overline X\mathrel{\dot\sim}N(5,5/50)$ . A central probability of $0.95$ leaves $0.025$ in each tail, so $d/\sqrt{5/50}=z_{0.975}=1.959963984\ldots$ . Hence $d=1.959963984\sqrt{5/50}=0.619795032\ldots$ , which is $0.620$ to $3$ decimal places.
3	$E(X)=5/2$ $E(X^2)=19/2$ $\operatorname{Var}(X)=19/2-(5/2)^2=13/4$ $\overline X\mathrel{\dot\sim}N(5/2,13/168)$ The upper-tail probabilities at $2.9$ and $2.4$ are $0.075224510\ldots$ and $0.640384530\ldots$ $P(\overline X>2.9\mid\overline X>2.4)\approx0.1175$	6
(6 marks)		6
Notes
From the probability table, $E(X)=0(0.2)+2(0.5)+5(0.3)=5/2$ and $E(X^2)=0^2(0.2)+2^2(0.5)+5^2(0.3)=19/2$ . Hence $\operatorname{Var}(X)=19/2-(5/2)^2=13/4$ . By the Central Limit Theorem, $\overline X\mathrel{\dot\sim}N(5/2,(13/4)/42)=N(5/2,13/168)$ , with standard error $\sqrt{13/168}=0.2781743201\ldots$ . The standardised values at $2.9$ and $2.4$ are $1.4379472548\ldots$ and $-0.3594868137\ldots$ , giving upper-tail probabilities $0.0752245102\ldots$ and $0.6403845297\ldots$ . Since $\{\overline X>2.9\}\subset\{\overline X>2.4\}$ , the conditional probability is $0.0752245102\ldots/0.6403845297\ldots=0.1174677193\ldots$ , which is $0.1175$ to $4$ decimal places. No continuity correction is used for the directly modelled sample mean.

Tier 3 · Hard

Mark scheme for FS1-5.1 Tier 3 · Hard
Question	Scheme	Marks
1	$n=97$	6
(6 marks)		6
Notes
Under the CLT normal approximation, the standard error is $6/\sqrt n$ . For central probability at least $0.95$ , require $1.2/(6/\sqrt n)\geq z_{0.975}=1.959964$ . Hence $n\geq(1.959964\times6/1.2)^2=96.0365\ldots$ . The estimated least integer value is $97$ .
2	$p=0.4$ $E(X)=7.5$ and $\operatorname{Var}(X)=11.25$ $\overline X\mathrel{\dot\sim}N(7.5,11.25/n)$ $0.6\sqrt{n/11.25}\geq z_{0.99}$ $n\geq169.121700\ldots$ The estimated least sample size is $170$	6
(6 marks)		6
Notes
For a negative-binomial trial count, $E(X)=r/p$ . Thus $3/p=7.5$ , giving $p=0.4$ , and $\operatorname{Var}(X)=r(1-p)/p^2=3(0.6)/(0.4)^2=11.25$ . Hence $\overline X\mathrel{\dot\sim}N(7.5,11.25/n)$ . The requirement is $0.6/\sqrt{11.25/n}\geq z_{0.99}=2.326347874\ldots$ , so $n\geq(2.326347874\sqrt{11.25}/0.6)^2=169.121700970\ldots$ . Therefore the estimated least sample size is $170$ .
3	$Y$ takes the values $1,2,4,8$ , each with probability $1/4$ $E(Y)=15/4$ $E(Y^2)=85/4$ $\operatorname{Var}(Y)=115/16$ $\overline Y\mathrel{\dot\sim}N(15/4,115/1024)$ The standardised value is $1.342806924\ldots$ $P(\overline Y>4.2)\approx0.0897$	7
(7 marks)		7
Notes
The transformation gives $Y=1,2,4,8$ with equal probabilities. Hence $E(Y)=(1+2+4+8)/4=15/4$ and $E(Y^2)=(1+4+16+64)/4=85/4$ , so $\operatorname{Var}(Y)=85/4-(15/4)^2=115/16$ . By the Central Limit Theorem, $\overline Y\mathrel{\dot\sim}N(15/4,(115/16)/64)=N(15/4,115/1024)$ . Thus the standardised value at $4.2$ is $(4.2-3.75)/(\sqrt{115}/32)=1.3428069239\ldots$ , giving an upper-tail probability $0.0896672504\ldots=0.0897$ to $4$ decimal places.
4	$E(X)=7/2$ $\operatorname{Var}(X)=35/12$ $\overline X\mathrel{\dot\sim}N(7/2,35/480)$ , with standard error $\sqrt{35/480}=0.2700309\ldots$ The standardised value at $3.2$ is $-1.110984\ldots$ $P(\overline X<3.2)\approx0.1333$ to $4$ decimal places The parent distribution is discrete, not Normal, so the Central Limit Theorem is used to approximate the distribution of the sample mean by a Normal distribution With $n=10$ the standard error doubles to $0.5400617\ldots$ , so the estimate rises to $0.2893$ The approximation is less reliable for the smaller sample	8
(8 marks)		8
Notes
For a fair die, $E(X)=(1+2+3+4+5+6)/6=7/2$ and $\operatorname{Var}(X)=35/12$ . Hence $\overline X\mathrel{\dot\sim}N(7/2,(35/12)/40)=N(7/2,35/480)$ , whose standard error is $\sqrt{35/480}=0.2700308624\ldots$ . The standardised value is $(3.2-3.5)/\sqrt{35/480}=-1.1109841197\ldots$ , so the estimate is $0.1332875924\ldots=0.1333$ . The die score is not Normally distributed, which is why the Central Limit Theorem supplies the approximate sampling distribution. If $n=10$ , the standard error is $\sqrt{35/120}=0.5400617249\ldots$ and the standardised value is $-0.5554920599\ldots$ , giving $0.2892790698\ldots=0.2893$ . This estimate is larger because the sampling distribution is more spread out, but the smaller sample also makes the CLT approximation less reliable.
5	$E(X)=\operatorname{Var}(X)=6$ $\overline X\mathrel{\dot\sim}N(6,6/36)=N(6,1/6)$ The standard error is $\sqrt6/6=0.408248\ldots$ The standardised bounds are $-1.469694\ldots$ and $1.224745\ldots$ $P(5.4<\overline X<6.5)\approx0.8188$ to $4$ decimal places Assumption: the counts in the selected intervals are independent and have the same Poisson distribution Limitation: the ticket rate may vary with time of day, so one constant Poisson mean may not describe every interval well	7
(7 marks)		7
Notes
A $\operatorname{Po}(6)$ variable has mean and variance $6$ . For $36$ intervals, the Central Limit Theorem gives $\overline X\mathrel{\dot\sim}N(6,6/36)$ , with standard error $\sqrt6/6$ . The standardised bounds are $(5.4-6)/(\sqrt6/6)=-1.4696938457\ldots$ and $(6.5-6)/(\sqrt6/6)=1.2247448714\ldots$ . Thus the estimated probability is $\Phi(1.2247448714\ldots)-\Phi(-1.4696938457\ldots)=0.8188419739\ldots=0.8188$ . The calculation assumes independent, identically distributed Poisson counts; a time-varying arrival rate would limit that model.