Normal Approximation to the Binomial Distribution | 用正态分布近似二项分布

📚 Normal Approximation to the Binomial Distribution | 用正态分布近似二项分布

When the number of trials in a binomial experiment becomes large, calculating probabilities directly using the binomial formula grows increasingly tedious and computationally heavy. Fortunately, the normal distribution can serve as a remarkably accurate stand-in under certain conditions. This article explores precisely when and how to use the normal distribution to approximate the binomial distribution, a technique that appears regularly in A-Level Mathematics and Statistics examinations.

当二项试验的试验次数变得很大时，直接用二项分布公式计算概率会越来越繁琐、计算量也越来越大。幸运的是，在特定条件下，正态分布可以作为一个非常精确的替代工具。本文详细探讨在什么条件下、以及如何用正态分布近似二项分布——这是A-Level数学与统计学考试中的高频考点。

1. The Binomial Distribution and Its Shape | 二项分布及其形状

Recall that a binomial random variable X ~ B(n, p) counts the number of successes in n independent trials, each with success probability p. Its mean is μ = np and its variance is σ² = np(1 – p). The probability of exactly k successes is given by the binomial probability formula: P(X = k) = ⁿCₖ · pᵏ · (1 – p)ⁿ⁻ᵏ.

回顾一下，二项随机变量 X ~ B(n, p) 表示在 n 次独立试验中成功的次数，每次试验的成功概率为 p。其均值为 μ = np，方差为 σ² = np(1 – p)。恰好成功 k 次的概率由二项概率公式给出：P(X = k) = ⁿCₖ · pᵏ · (1 – p)ⁿ⁻ᵏ。

The shape of the binomial distribution depends heavily on n and p. When p is close to 0.5, the distribution is symmetric. As p drifts toward 0 or 1, the distribution becomes skewed. However, as n grows, the binomial distribution becomes increasingly bell-shaped and symmetric, regardless of p — provided p is not too extreme. This bell-shaped appearance is what allows us to approximate it with a normal curve.

二项分布的形状在很大程度上取决于 n 和 p。当 p 接近 0.5 时，分布是对称的。当 p 偏向 0 或 1 时，分布会变得偏斜。然而，随着 n 增大，二项分布会越来越呈现钟形且对称——只要 p 不过于极端。这种钟形外观正是我们可以用正态曲线来近似它的原因。

2. The Conditions for a Valid Approximation | 近似的适用条件

Not every binomial distribution can be safely approximated by a normal distribution. The most widely accepted rule — and the one used in every major examination syllabus — is that both expected counts must be sufficiently large:

并非所有的二项分布都能安全地用正态分布近似。最广泛接受的规则——也是各大考试大纲中使用的规则——是两个期望计数都必须足够大：

np ≥ 5 and n(1 – p) ≥ 5
Some textbooks and boards (including certain A-Level papers) adopt the stricter criterion: np ≥ 10 and n(1 – p) ≥ 10.

The reasoning is simple: np is the expected number of successes, and n(1 – p) is the expected number of failures. If either is too small, the binomial distribution is too skewed or too “discrete-looking” for a smooth normal curve to fit reliably.

理由很简单：np 是成功的期望次数，n(1 – p) 是失败的期望次数。如果其中任何一个太小，二项分布就过于偏斜或过于”离散”，光滑的正态曲线无法可靠地拟合。

As a rule of thumb, the closer p is to 0.5, the better the approximation tends to be at smaller values of n. Conversely, when p is very small or very large, you need a substantially larger n for the normal approximation to work well. When in doubt, apply the stricter criterion np ≥ 10 and n(1 – p) ≥ 10.

经验法则是：p 越接近 0.5，在较小 n 时近似效果往往越好。反之，当 p 很小时或很大时，需要相当大的 n 才能使正态近似表现良好。如有疑问，应采用更严格的标准 np ≥ 10 且 n(1 – p) ≥ 10。

3. Why the Normal Approximation Works | 正态近似为何有效

The theoretical justification comes from the Central Limit Theorem (CLT) and the De Moivre–Laplace theorem. A binomial random variable can be viewed as the sum of n independent Bernoulli random variables, each with mean p and variance p(1 – p). The CLT tells us that the sum of many independent, identically distributed random variables is approximately normally distributed, regardless of the underlying distribution.

其理论依据来自中心极限定理（CLT）和棣莫弗–拉普拉斯定理。二项随机变量可以看作 n 个独立伯努利随机变量的和，每个伯努利变量的均值为 p，方差为 p(1 – p)。中心极限定理告诉我们，大量独立同分布随机变量之和近似服从正态分布，无论其原始分布是什么。

Furthermore, the De Moivre–Laplace theorem specifically states that the binomial distribution B(n, p) converges to a normal distribution with mean np and variance np(1 – p) as n tends to infinity. This convergence is uniform and remarkably rapid when p is near 0.5, which is why the approximation performs so well in practice.

此外，棣莫弗–拉普拉斯定理专门指出，当 n 趋向无穷大时，二项分布 B(n, p) 收敛于均值为 np、方差为 np(1 – p) 的正态分布。这种收敛是一致收敛，且当 p 接近 0.5 时收敛速度非常快，这就是该近似在实践中表现如此出色的原因。

4. The Continuity Correction — Concept | 连续性修正——概念

The binomial distribution is discrete: its probabilities are concentrated at whole-number values 0, 1, 2, …, n. The normal distribution, by contrast, is continuous: probability is spread over intervals. If we naively replace P(X ≤ k) with P(Y ≤ k), where Y ~ N(np, np(1 – p)), we systematically underestimate the binomial probability because the entire “block” of probability at X = k is omitted.

二项分布是离散的：其概率集中在整数值 0, 1, 2, …, n 上。相比之下，正态分布是连续的：概率分布在区间上。如果我们天真地将 P(X ≤ k) 替换为 P(Y ≤ k)（其中 Y ~ N(np, np(1 – p))），就会系统地低估二项概率，因为 X = k 处的整块概率被遗漏了。

The solution is the continuity correction. Think of each integer k as occupying an interval from k – 0.5 to k + 0.5. The probability P(X = k) in the binomial setting then corresponds to the area under the normal curve between these two boundaries. This half-unit adjustment bridges the gap between the discrete and continuous worlds.

解决方法就是连续性修正。将每个整数 k 视为占据从 k – 0.5 到 k + 0.5 的区间。在二项分布中，概率 P(X = k) 对应正态曲线下这两个边界之间的面积。这种半单位调整弥合了离散世界与连续世界之间的鸿沟。

5. The Continuity Correction — Rules | 连续性修正——具体规则

The following table summarises how to apply the continuity correction. Let X ~ B(n, p) and let Y ~ N(np, np(1 – p)) be the approximating normal variable.

下表总结了如何应用连续性修正。设 X ~ B(n, p)，令 Y ~ N(np, np(1 – p)) 为近似的正态变量。

Binomial statement	Normal approximation with correction
P(X = k)	P(k – 0.5 ≤ Y ≤ k + 0.5)
P(X ≤ k)	P(Y ≤ k + 0.5)
P(X < k)	P(Y ≤ k – 0.5)
P(X ≥ k)	P(Y ≥ k – 0.5)
P(X > k)	P(Y ≥ k + 0.5)
P(a ≤ X ≤ b)	P(a – 0.5 ≤ Y ≤ b + 0.5)

A reliable memory aid: when the inequality involves “≤”, extend the boundary outward by 0.5 (add 0.5 to the upper limit); when it involves “≥”, extend outward by subtracting 0.5 from the lower limit. For strict inequalities, shift inward by 0.5.

一个可靠的记忆技巧：当不等式含”≤”时，将边界向外扩展 0.5（在上限加 0.5）；当含”≥”时，从下限减去 0.5 来向外扩展。对于严格不等式，则向内移动 0.5。

6. Step-by-Step Procedure | 分步计算方法

The full procedure can be condensed into five reliable steps. Each step is essential; skipping the continuity correction is the single most common error students make.

完整流程可以概括为五个可靠的步骤。每一步都至关重要；跳过连续性修正是学生最常犯的错误。

Step 1 — Verify conditions: Check that np ≥ 5 and n(1 – p) ≥ 5 (ideally both ≥ 10). If the condition fails, the normal approximation is not appropriate.
Step 2 — Identify parameters: Compute the mean μ = np and the standard deviation σ = √(np(1 – p)) for the approximating normal distribution.
Step 3 — Apply the continuity correction: Rewrite the desired binomial probability in terms of an interval or inequality involving whole numbers, then shift the boundary by 0.5 as appropriate.
Step 4 — Standardise: Convert the shifted boundary to a z-score using z = (x – μ) / σ.
Step 5 — Use the standard normal table: Look up the probability Φ(z) from the normal distribution table and compute the final answer.

第一步——验证条件：检查 np ≥ 5 且 n(1 – p) ≥ 5（理想情况下两者均 ≥ 10）。若条件不满足，则不适合使用正态近似。
第二步——确定参数：计算近似的正态分布的均值 μ = np 和标准差 σ = √(np(1 – p))。
第三步——应用连续性修正：将所需的二项概率改写为涉及整数的区间或不等式，然后按规则将边界移动 0.5。
第四步——标准化：使用 z = (x – μ) / σ 将修正后的边界转换为 z 分数。
第五步——查标准正态表：从正态分布表中查概率 Φ(z)，并计算最终答案。

7. Worked Example — P(X ≤ k) | 例题——计算 P(X ≤ k)

Problem: A fair coin is tossed 200 times. Estimate the probability that the number of heads is at most 90.

题目：一枚均匀硬币被抛掷 200 次。估计正面朝上的次数至多为 90 的概率。

Solution: Here X ~ B(200, 0.5). First verify conditions: np = 200 × 0.5 = 100 ≥ 5 and n(1 – p) = 100 ≥ 5, so the normal approximation is valid. The approximating normal variable is Y ~ N(100, 50), since the variance is np(1 – p) = 200 × 0.5 × 0.5 = 50, giving σ = √50 ≈ 7.071.

解答：此处 X ~ B(200, 0.5)。首先验证条件：np = 200 × 0.5 = 100 ≥ 5 且 n(1 – p) = 100 ≥ 5，因此正态近似有效。近似正态变量为 Y ~ N(100, 50)，因为方差为 np(1 – p) = 200 × 0.5 × 0.5 = 50，所以 σ = √50 ≈ 7.071。

We require P(X ≤ 90). Applying the continuity correction, this becomes P(Y ≤ 90.5). Standardising:

我们需要 P(X ≤ 90)。应用连续性修正，这变为 P(Y ≤ 90.5)。标准化：

z = (90.5 – 100) / √50 = -9.5 / 7.071 ≈ -1.34

From the standard normal table, Φ(-1.34) = 1 – Φ(1.34) = 1 – 0.9099 = 0.0901. Thus P(X ≤ 90) ≈ 0.0901.

查标准正态表，Φ(-1.34) = 1 – Φ(1.34) = 1 – 0.9099 = 0.0901。因此 P(X ≤ 90) ≈ 0.0901。

For comparison, the exact binomial probability (computed via software) is approximately 0.0918. The normal approximation gives an answer within 0.002 of the exact value — an error of roughly 2%, which is highly acceptable for examination purposes.

作为对比，精确二项概率（通过软件计算）约为 0.0918。正态近似给出的答案与精确值的差异在 0.002 以内——误差约为 2%，在考试中完全可以接受。

8. Worked Example — P(X ≥ k) and P(a ≤ X ≤ b) | 例题——计算 P(X ≥ k) 与 P(a ≤ X ≤ b)

Problem: A manufacturing process produces components with a 5% defect rate. A random sample of 300 components is inspected. Estimate the probability that at least 20 are defective.

题目：某制造工艺生产的零件缺陷率为 5%。随机抽取 300 个零件进行检查。估计至少有 20 个为缺陷品的概率。

Solution: Let X be the number of defective components. Then X ~ B(300, 0.05). Check conditions: np = 300 × 0.05 = 15 ≥ 5, n(1 – p) = 300 × 0.95 = 285 ≥ 5. Both are satisfied. The mean is μ = 15 and the standard deviation is σ = √(300 × 0.05 × 0.95) = √14.25 ≈ 3.775.

解答：设 X 为缺陷零件数。则 X ~ B(300, 0.05)。检查条件：np = 300 × 0.05 = 15 ≥ 5，n(1 – p) = 300 × 0.95 = 285 ≥ 5，均满足。均值为 μ = 15，标准差为 σ = √(300 × 0.05 × 0.95) = √14.25 ≈ 3.775。

We need P(X ≥ 20). With the continuity correction, this becomes P(Y ≥ 19.5). Standardising:

我们需要 P(X ≥ 20)。应用连续性修正，变为 P(Y ≥ 19.5)。标准化：

z = (19.5 – 15) / 3.775 ≈ 1.19

Thus P(X ≥ 20) ≈ P(Z ≥ 1.19) = 1 – Φ(1.19) = 1 – 0.8830 = 0.1170.

因此 P(X ≥ 20) ≈ P(Z ≥ 1.19) = 1 – Φ(1.19) = 1 – 0.8830 = 0.1170。

Extension: Suppose instead we wanted P(10 ≤ X ≤ 25). The continuity correction converts this to P(9.5 ≤ Y ≤ 25.5). Standardising both endpoints gives z₁ = (9.5 – 15)/3.775 ≈ -1.46 and z₂ = (25.5 – 15)/3.775 ≈ 2.78. Hence P(10 ≤ X ≤ 25) ≈ Φ(2.78) – Φ(-1.46) = 0.9973 – 0.0721 = 0.9252.

延伸：假如我们想计算 P(10 ≤ X ≤ 25)。连续性修正将其转换为 P(9.5 ≤ Y ≤ 25.5)。将两个端点标准化得 z₁ = (9.5 – 15)/3.775 ≈ -1.46，z₂ = (25.5 – 15)/3.775 ≈ 2.78。因此 P(10 ≤ X ≤ 25) ≈ Φ(2.78) – Φ(-1.46) = 0.9973 – 0.0721 = 0.9252。

9. When the Correction Is Omitted | 若不做连续性修正会怎样

It is instructive to see how much error is introduced by omitting the continuity correction. In the first worked example, computing P(Y ≤ 90) directly without correction gives z = (90 – 100)/7.071 ≈ -1.41, so the result is approximately 0.0793 — compared to the corrected value of 0.0901 and the exact binomial value of 0.0918.

了解省略连续性修正会引入多大误差是很有价值的。在第一个例题中，不做修正直接计算 P(Y ≤ 90) 得到 z = (90 – 100)/7.071 ≈ -1.41，结果约为 0.0793——而修正后的值为 0.0901，精确二项值为 0.0918。

The uncorrected estimate misses the exact probability by over 0.012, roughly a 13% relative error, while the corrected estimate misses by only 0.002. This dramatic difference explains why the continuity correction is non-negotiable in examination settings: marks are routinely deducted when it is omitted.

未修正的估计与精确概率相差超过 0.012，相对误差约为 13%，而修正后的估计仅相差 0.002。这一显著差异解释了为什么连续性修正在考试中是不可妥协的：省略它通常会被扣分。

10. Common Mistakes and Pitfalls | 常见错误与陷阱

Several recurring errors account for the vast majority of lost marks on this topic. Recognising them in advance is the best defence.

几个反复出现的错误导致了该主题上绝大部分的失分。提前识别它们是做好防御的最佳方式。

Forgetting the continuity correction: Always ask whether the binomial statement involves discrete values; if so, shift by 0.5.
Applying the correction in the wrong direction: For P(X ≥ k), use k – 0.5, not k + 0.5. For P(X ≤ k), use k + 0.5, not k – 0.5.
Using the binomial variance instead of np(1 – p): The normal approximation must use the correct variance; confusing it with other formulas leads to wildly wrong results.
Checking conditions incorrectly: Verify both np and n(1 – p), not just np.
Misreading the normal table: Ensure you subtract from 1 when dealing with negative z-values or upper-tail probabilities.
Rounding too early: Keep sufficient precision in σ before computing z to avoid rounding errors.

忘记连续性修正：始终问自己二项条件是否涉及离散值；如果是，则移动 0.5。
修正方向错误：对于 P(X ≥ k)，使用 k – 0.5，而不是 k + 0.5。对于 P(X ≤ k)，使用 k + 0.5，而不是 k – 0.5。
误用方差而非 np(1 – p)：正态近似必须使用正确的方差；与其他公式混淆会导致结果严重偏差。
条件检查不正确：必须同时验证 np 和 n(1 – p)，而非仅验证 np。
误读正态分布表：处理负 z 值或上尾概率时，务必记得用 1 减去表中数值。
过早四舍五入：在计算 z 分数之前，σ 需保留足够的精度，以避免舍入误差。

11. Summary of Key Formulas | 关键公式总结

The table below consolidates the essential information from this article into a single reference frame. Commit these to memory before entering the examination hall.

下表将本文的核心信息整合为一个参考框架。进入考场前请务必牢记这些内容。

Quantity	Formula
Mean of approximating normal	μ = np
Variance of approximating normal	σ² = np(1 – p)
Standard deviation	σ = √(np(1 – p))
Standardisation	z = (x – μ) / σ
Conditions	np ≥ 5 and n(1 – p) ≥ 5 (preferably ≥ 10)

In essence, the normal approximation to the binomial is a three-part recipe: check that both expected counts are large enough, replace the binomial variable with a normal variable sharing the same mean and variance, and apply the continuity correction to every probability statement before standardising. Master these three components, and you will handle any question on this topic with confidence.

总而言之，用正态近似二项分布是一个三步配方：检查两个期望计数是否足够大，用具有相同均值和方差的正态变量替换二项变量，并在标准化之前对每个概率陈述应用连续性修正。掌握这三个要素，你就能自信地应对该主题的任何题目。

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Normal Approximation to the Binomial Distribution | 用正态分布近似二项分布