A-Level CIE Mathematics: Normal Distribution Essentials | 正态分布考点精讲

📚 A-Level CIE Mathematics: Normal Distribution Essentials | 正态分布考点精讲

The normal distribution is one of the most important continuous probability distributions in CIE A-Level Mathematics. It appears in a wide range of exam questions, from direct probability calculations to approximating binomial distributions and analysing sample means. Mastering its concepts and techniques is vital for success in Paper 5 and Paper 6.

正态分布是 CIE A-Level 数学中最重要的连续概率分布之一。从直接概率计算到二项分布的近似以及样本均值的分析，它在各类考题中频繁出现。掌握其概念和技巧对在 Paper 5 和 Paper 6 中取得成功至关重要。

1. Introduction to Normal Distribution | 正态分布简介

A normal distribution is a symmetric, bell-shaped continuous distribution characterised by two parameters: the mean μ and the variance σ². The random variable X is written as X ~ N(μ, σ²). The curve extends infinitely in both directions, and the total area under the curve equals 1.

正态分布是一种对称的钟形连续分布，由两个参数描述：均值 μ 和方差 σ²。随机变量 X 记为 X ~ N(μ, σ²)。曲线向两端无限延伸，曲线下的总面积等于 1。

2. Properties of the Normal Curve | 正态曲线的性质

The normal curve is symmetric about the mean, so the mean, median and mode are equal. The points of inflection occur at μ ± σ. Approximately 68% of data lies within 1 standard deviation of the mean, 95% within 2σ, and 99.7% within 3σ. The area under the curve between μ ± 1.96σ is exactly 0.95.

正态曲线关于均值对称，因此均值、中位数和众数相等。拐点出现在 μ ± σ 处。大约 68% 的数据落在均值上下一个标准差内，95% 在 2σ 内，99.7% 在 3σ 内。曲线下 μ ± 1.96σ 的面积恰好为 0.95。

3. The Standard Normal Distribution | 标准正态分布

The standard normal distribution has a mean of 0 and a variance of 1. It is denoted by Z ~ N(0, 1). Any normal variable can be transformed to a standard normal variable using the z-score. The CIE formula booklet provides tables for the cumulative probability Φ(z) = P(Z < z) for z ≥ 0.

标准正态分布的均值为 0，方差为 1，记作 Z ~ N(0, 1)。任何正态变量都可以通过 z 分数转换为标准正态变量。CIE 公式手册提供了 z ≥ 0 时累积概率 Φ(z) = P(Z < z) 的表格。

4. Standardising: The Z-Score | 标准化：Z 分数

To convert a value x from a normal distribution N(μ, σ²) to the standard normal scale, we use the formula below. The result z measures how many standard deviations x is away from the mean.

为了将来自正态分布 N(μ, σ²) 的值 x 转换到标准正态尺度，我们使用以下公式。结果 z 衡量 x 距离均值有多少个标准差。

z = (x − μ) / σ

After standardising, probabilities can be found using the standard normal table. Always remember that the denominator is the standard deviation σ, not the variance σ².

标准化后，可以使用标准正态表求概率。务必记住分母是标准差 σ，而非方差 σ²。

5. Using Normal Distribution Tables | 使用正态分布表

The CIE table gives Φ(z) = P(Z < z) for z from 0.00 up to about 3.49. For z values with two decimal places, read the intersection of the row and column. For negative z-values, use symmetry: P(Z < −z) = 1 − Φ(z). The table only provides lower tail probabilities; you must sketch the curve to handle other intervals.

CIE 的表格给出了 z 从 0.00 到约 3.49 的 Φ(z) = P(Z < z)。对于两位小数的 z 值，读取行与列的交点。对于负 z 值，利用对称性：P(Z < −z) = 1 − Φ(z)。表格只提供左尾概率；你必须先画图来处理其他区间。

6. Finding Probabilities: Forward Problems | 求概率：正向问题

When X ~ N(μ, σ²) and you need P(X < a), first standardise to find z = (a − μ) / σ. Then look up Φ(z). For P(X > a), use 1 − Φ(z). For P(a < X < b), compute Φ(z₂) − Φ(z₁). Always convert the boundary values to z-scores before using the table.

当 X ~ N(μ, σ²) 且需要求 P(X < a) 时，先标准化得到 z = (a − μ) / σ，然后查阅 Φ(z)。对于 P(X > a)，用 1 − Φ(z)。对于 P(a < X < b)，计算 Φ(z₂) − Φ(z₁)。务必先将边界值转化为 z 分数再查表。

7. Finding Values: Inverse Normal Problems | 求数值：反向正态问题

Given a probability, find the corresponding x-value. First, locate the probability in the body of the table and read the z-value. For an unknown mean or standard deviation, set up an equation using z = (x − μ) / σ and solve simultaneously. If the table gives a left-tail area p, then x = μ + zσ where z satisfies Φ(z) = p.

给定概率，求对应的 x 值。首先在表格主体中找到概率，读出 z 值。如果均值或标准差未知，则用 z = (x − μ) / σ 建立方程并同时求解。若表格给出的左尾面积为 p，则 x = μ + zσ，其中 z 满足 Φ(z) = p。

8. The Normal Approximation to the Binomial Distribution | 二项分布的正态近似

When X ~ B(n, p) with n large enough such that np > 5 and nq > 5 (where q = 1 − p), X can be approximated by a normal distribution with μ = np and σ² = npq. That is, X ≈ N(np, npq). This is extremely useful when calculating binomial probabilities with large n becomes tedious.

当 X ~ B(n, p) 且 n 足够大使得 np > 5 且 nq > 5（q = 1 − p），X 可用均值为 μ = np、方差为 σ² = npq 的正态分布近似，即 X ≈ N(np, npq)。这在大 n 下计算二项概率变得繁琐时极为有用。

9. Continuity Correction | 连续性校正

Because the binomial is discrete and the normal is continuous, a continuity correction must be applied. Replace a discrete x with an interval of ± 0.5. Key rules: 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). Never forget the correction; marks are frequently lost here.

由于二项分布是离散的而正态分布是连续的，必须应用连续性校正。将离散的 x 替换为一个 ± 0.5 的区间。关键规则：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)。切莫忘记校正；此处经常失分。

10. Using the Normal Distribution for Sample Means | 样本均值的正态分布

If a population follows N(μ, σ²), then the sample mean X̄ from a random sample of size n is distributed as X̄ ~ N(μ, σ²/n). The standard deviation of the sample mean is σ/√n, known as the standard error. Even if the population is not normal, by the Central Limit Theorem, X̄ is approximately normal for large n (usually n ≥ 30).

若总体服从 N(μ, σ²)，则来自容量为 n 的随机样本的样本均值 X̄ 服从 X̄ ~ N(μ, σ²/n)。样本均值的标准差为 σ/√n，称为标准误。即使总体不服从正态分布，根据中心极限定理，当 n 较大（通常 n ≥ 30）时，X̄ 近似服从正态分布。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

Common pitfalls: using variance instead of standard deviation in the z-formula, forgetting continuity correction, misreading probability tables, not sketching the curve, and confusing P(Z < z) with P(Z > z). Always draw a diagram, label the mean and boundaries, and shade the required area. Show your standardisation steps clearly.

常见陷阱：在 z 公式中使用方差而非标准差，忘记连续性校正，读错概率表，不画草图，以及混淆 P(Z < z) 与 P(Z > z)。务必画图，标出均值和边界，并涂满所求区域。清晰展示标准化步骤。

12. Practice Question Examples | 练习题示例

Example 1 (Forward): X ~ N(50, 25). Find P(X < 53). Standardise: z = (53 − 50) / √25 = 3/5 = 0.6. Φ(0.6) = 0.7257, so P(X < 53) ≈ 0.726. Example 2 (Inverse): X ~ N(100, 144). Find the value of a such that P(X < a) = 0.90. From tables, z = 1.282. Then a = 100 + 1.282 × 12 = 115.38.

示例 1（正向）：X ~ N(50, 25)。求 P(X < 53)。标准化：z = (53 − 50) / √25 = 3/5 = 0.6。Φ(0.6) = 0.7257，因此 P(X < 53) ≈ 0.726。示例 2（反向）：X ~ N(100, 144)。求满足 P(X < a) = 0.90 的 a。查表得 z = 1.282。则 a = 100 + 1.282 × 12 = 115.38。

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A-Level CIE Mathematics: Normal Distribution Essentials | 正态分布考点精讲