📚 A-Level Further Mathematics: Applications of the Central Limit Theorem to Other Distributions | A-Level进阶数学：中心极限定理在其他分布中的应用
The Central Limit Theorem (CLT) is a cornerstone of statistical theory. It explains why the normal distribution appears so often in nature and in exam questions, and it underpins many approximate methods used across the A-Level Further Mathematics syllabus.
中心极限定理是统计理论的基石。它解释了为什么正态分布在自然界和考试题目中如此常见，也是进阶数学课程中许多近似方法的理论基础。
In this article, we will explore the theorem itself and then examine how it justifies the normal approximations to the binomial and Poisson distributions, as well as how it applies to the sample mean and the sum of independent random variables. We will also provide worked examples and highlight common exam pitfalls.
本文将首先介绍中心极限定理本身，然后讨论它如何为二项分布和泊松分布的正态近似提供理论依据，并研究它在样本均值和独立随机变量之和中的应用。我们还会给出典型例题并指出常见的考试陷阱。
1. The Statement of the Central Limit Theorem | 中心极限定理的陈述
Let X₁, X₂, …, Xₙ be a random sample of size n drawn from any population with mean μ and finite variance σ². When n is sufficiently large, the sample mean X̄ is approximately normally distributed with mean μ and variance σ²/n.
设 X₁, X₂, …, Xₙ 是来自任意总体的一份样本量为 n 的随机样本，该总体的均值为 μ，方差 σ² 有限。当 n 足够大时，样本均值 X̄ 近似服从均值为 μ、方差为 σ²/n 的正态分布。
In standardised form, we write:
标准化后可写成：
Z = (X̄ − μ) / (σ / √n) → N(0, 1) as n → ∞
The remarkable feature of this theorem is that it holds for almost every underlying distribution: uniform, skewed, triangular, exponential, and many others. The only requirements are independence, a finite mean, and a finite variance.
这个定理最惊人之处在于，它对几乎所有总体分布都成立：均匀分布、偏态分布、三角形分布、指数分布等等。唯一的要求是变量独立、均值有限且方差有限。
2. Key Conditions for the CLT | 使用中心极限定理的关键条件
Before applying the theorem in an exam, you must check that the conditions are satisfied.
在考试中使用中心极限定理之前，必须先检查条件是否成立。
- The observations must be independent.
观测值必须相互独立。 - The population must have a finite variance σ².
总体方差 σ² 必须有限。 - The sample size n must be large enough, usually n ≥ 30 for most distributions.
样本量 n 必须足够大，对于大多数分布通常要求 n ≥ 30。 - If the population is severely skewed, a larger sample size may be needed.
如果总体严重偏斜，可能需要更大的样本量。
In Edexcel Further Mathematics questions, you are often told to use the CLT without calculating exact probabilities. You then standardise using the population mean and variance, even if the original distribution is not normal.
在爱德思进阶数学题目中，题目通常直接要求使用中心极限定理而不需要计算精确概率。此时即使原始分布不是正态分布，你也应使用总体均值和方差进行标准化。
3. Normal Approximation to the Binomial Distribution | 二项分布的正态近似
A binomial random variable X ~ B(n, p) can be written as the sum of n independent Bernoulli random variables, each with mean p and variance p(1 − p).
二项随机变量 X ~ B(n, p) 可以写成 n 个独立伯努利随机变量之和，每个伯努利变量的均值为 p，方差为 p(1 − p)。
Applying the CLT to this sum gives:
对上述求和运用中心极限定理可得：
X ≈ N(np, np(1 − p))
This approximation is reliable when np > 5 and n(1 − p) > 5. If p is very close to 0 or 1, the binomial distribution is too skewed for the normal approximation to work well, and the Poisson approximation should be considered instead.
当 np > 5 且 n(1 − p) > 5 时，这个近似是可靠的。如果 p 非常接近 0 或 1，二项分布过于偏斜，正态近似效果不佳，此时应考虑使用泊松近似。
The approximation is especially useful when n is large and the exact binomial calculation would require many terms. For example, X ~ B(100, 0.3) is approximated by Y ~ N(30, 21).
当 n 较大且精确计算二项概率需要很多项时，这个近似尤其有用。例如，X ~ B(100, 0.3) 可以用 Y ~ N(30, 21) 来近似。
4. Normal Approximation to the Poisson Distribution | 泊松分布的正态近似
Similarly, a Poisson random variable X ~ Po(λ) can be regarded as the sum of many independent Poisson variables, each with a small parameter. As λ increases, the Poisson distribution becomes increasingly symmetric and tends towards a normal distribution.
类似地，泊松随机变量 X ~ Po(λ) 可以被看作许多参数很小的独立泊松变量之和。随着 λ 增大，泊松分布逐渐变得对称，并趋于正态分布。
X ≈ N(λ, λ) for λ > 10
The rule of thumb is that the normal approximation to the Poisson distribution is acceptable when λ is greater than about 10. For instance, a Poisson distribution with λ = 15 has mean 15 and variance 15, so we use Y ~ N(15, 15).
经验法则是：当 λ 大于 10 左右时，泊松分布的正态近似可以接受。例如，λ = 15 的泊松分布均值为 15，方差也为 15，因此我们使用 Y ~ N(15, 15)。
Note that the Poisson approximation to the binomial is different: when n is large and p is small, B(n, p) is approximated by Po(np). The normal approximation to the Poisson is only used when λ itself is large.
注意，泊松近似二项分布与这里是两回事：当 n 大且 p 小时，B(n, p) 用 Po(np) 近似。而正态近似泊松分布只在 λ 本身较大时使用。
5. Continuity Corrections | 连续性修正
Both the binomial and Poisson distributions are discrete, while the normal distribution is continuous. To improve accuracy, we apply a continuity correction when using a normal approximation.
二项分布和泊松分布都是离散分布，而正态分布是连续分布。为了提高精确度，使用正态近似时应进行连续性修正。
The table below shows the standard correction rules, where X is the discrete variable and Y is its normal approximation.
下表给出了标准的修正规则，其中 X 是离散变量，Y 是它的正态近似。
| Probability Required 需要计算的概率 |
Normal Approximation Used 使用的正态近似 |
| P(X ≤ a) | P(Y < a + 0.5) |
| P(X ≥ a) | P(Y > a − 0.5) |
| P(X < a) | P(Y < a − 0.5) |
| P(X > a) | P(Y > a + 0.5) |
| P(X = a) | P(a − 0.5 < Y < a + 0.5) |
For example, to approximate P(X ≤ 25) for X ~ B(100, 0.3), we calculate P(Y < 25.5) where Y ~ N(30, 21). If you forget the correction, your answer can be noticeably different, so examiners reward candidates who apply it consistently.
例如，要近似计算 X ~ B(100, 0.3) 中的 P(X ≤ 25)，我们计算 Y ~ N(30, 21) 下的 P(Y < 25.5)。如果忘记连续性修正，答案会有明显偏差，因此考官会赞赏始终使用修正的考生。
6. The CLT for the Sample Mean | 样本均值的中心极限定理
In many questions, you are not told that the original population is normal. Instead, you are given a distribution such as the uniform distribution on [0, 1] or an exponential distribution, and asked about the mean of a large sample.
许多题目并不会告诉你原始总体是正态分布。相反，题目会给出诸如 [0, 1] 上的均匀分布或指数分布，然后询问大样本均值的情况。
By the CLT, for a sample of size n from any distribution with mean μ and variance σ², we have:
根据中心极限定理，对于来自均值为 μ、方差为 σ² 的任意分布的大小为 n 的样本，有：
X̄ ≈ N(μ, σ² / n)
Then the standardised variable is:
标准化后的变量为：
Z = (X̄ − μ) / (σ / √n) ≈ N(0, 1)
If the population variance is unknown, but the sample size is large, you may use the sample variance as an estimate. In Edexcel Further Mathematics, however, the population variance is usually provided.
如果总体方差未知，但在样本量较大时，可以用样本方差作为估计值。不过在爱德思进阶数学中，通常会直接给出总体方差。
7. The CLT for Sums of Independent Random Variables | 独立随机变量之和的中心极限定理
Sometimes the question concerns the total, rather than the mean. If S = X₁ + X₂ + … + Xₙ is the sum of n independent observations, each with mean μ and variance σ², then for large n:
有时问题涉及总和而不是均值。设 S = X₁ + X₂ + … + Xₙ 是 n 个独立观测值之和，每个观测值均值为 μ、方差为 σ²，那么当 n 较大时：
S ≈ N(nμ, nσ²)
In standardised form:
标准化形式为：
Z = (S − nμ) / (σ√n) ≈ N(0, 1)
This version of the theorem is useful when the question gives the weight, length, or time of individual items and asks about the total of many items.
当题目给出单个物品的重量、长度或时间，并询问大量物品的总量时，这一定理形式非常有用。
For example, if the weights of 50 apples are independent random variables, each with mean 120 g and standard deviation 10 g, then the total weight T is approximately normal with mean 6000 g and variance 50 × 100 = 5000 g².
例如，如果 50 个苹果的重量是独立随机变量，每个均值为 120 克、标准差为 10 克，那么总重量 T 近似服从均值为 6000 克、方差为 50 × 100 = 5000 克² 的正态分布。
8. Worked Examples | 典型例题详解
Example 1: Binomial approximation
例 1：二项分布的正态近似
Let X ~ B(100, 0.3). Estimate P(X ≤ 25) using a continuity correction.
设 X ~ B(100, 0.3)。使用连续性修正估计 P(X ≤ 25)。
Solution: Here np = 30 and n(1 − p) = 70, both greater than 5, so the normal approximation is valid. We take Y ~ N(30, 21), so σY = √21.
解：这里 np = 30，n(1 − p) = 70，均大于 5，因此正态近似有效。取 Y ~ N(30, 21)，所以 σY = √21。
Using the continuity correction, P(X ≤ 25) ≈ P(Y < 25.5). Therefore:
使用连续性修正，P(X ≤ 25) ≈ P(Y < 25.5)。因此：
Z = (25.5 − 30) / √21 = −4.5 / 4.583 ≈ −0.982
P(X ≤ 25) ≈ Φ(−0.982) ≈ 0.163
Example 2: Poisson
Published by TutorHao | A-Level 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导