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Edexcel A-level Further Maths revision notes

Central Limit Theorem

Section FS1-5
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
1 specification point

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section FS1-5

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.

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Explanation

  • For a large independent random sample from a population with mean μ\mu and variance σ2\sigma^2, the Central Limit Theorem gives X˙N(μ,σ2/n)\overline X\mathrel{\dot\sim}N(\mu,\sigma^2/n). The standard error is σ/n\sigma/\sqrt n, so standardise with Z=(Xμ)/(σ/n)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 nn, 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 3030 and variance 6464. For a random sample of size 100100, estimate P(28.8<X<31.6)P(28.8<\overline X<31.6).

  1. 1.X˙N(30,64/100)\overline X\mathrel{\dot\sim}N(30,64/100), so the standard error is 0.80.8.
  2. 2.The lower standardised value is (28.830)/0.8=1.5(28.8-30)/0.8=-1.5.
  3. 3.The upper standardised value is (31.630)/0.8=2(31.6-30)/0.8=2.
  4. 4.P(1.5<Z<2)=0.9104P(-1.5<Z<2)=0.9104 to 44 significant figures.

Answer: The estimated probability is 0.91040.9104.

Common mistakes

  • Don't use variance 6464 for X\overline X instead of 64/10064/100.
  • Don't divide the standard deviation by nn instead of by n\sqrt n.
  • Don't add a continuity correction to bounds already stated for the sample mean.

Exam tip

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

Tier 1 · Easy

ORIGINAL

1.

Independent lifetimes have population mean 5050 hours and variance 100100 hours2^2. For a sample of 3636 lifetimes, use the Central Limit Theorem to estimate P(X<47.5)P(\overline X<47.5).

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

A geometric population has parameter p=0.32p=0.32. A random sample of 4545 observations is taken. Find the value of cc for which P(X>c)0.05P(\overline X>c)\approx0.05, giving cc to 33 decimal places.

(6)

(Total for Question 1 is 6 marks)

Tier 3 · Hard

ORIGINAL

1.

A population has mean μ\mu and variance 3636. Using the Central Limit Theorem normal approximation, estimate the least sample size nn for which P(Xμ<1.2)0.95P(|\overline X-\mu|<1.2)\geq0.95.

(6)

(Total for Question 1 is 6 marks)

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