Normal Distribution Key Points for IB & CCEA Mathematics | IB & CCEA 数学：正态分布考点精讲

📚 Normal Distribution Key Points for IB & CCEA Mathematics | IB & CCEA 数学：正态分布考点精讲

The normal distribution is the single most important probability distribution in statistics. It underpins large portions of the IB Mathematics (Analysis & Approaches, Applications & Interpretation) and CCEA A‑Level Mathematics syllabuses. A solid grasp of its properties, calculations, and applications is essential for success in exams. This article breaks down every major topic, from the bell curve equation to inverse normal and normal approximations.

正态分布是统计学中最重要的概率分布，也是 IB 数学（分析与方法、应用与解释）以及 CCEA A‑Level 数学课程的核心内容。透彻理解其性质、计算方法和应用场景，是考试取得高分的关键。本文将逐一拆解所有重要考点，从钟形曲线方程到逆正态，再到正态近似。

1. What is the Normal Distribution? | 什么是正态分布？

A continuous random variable X follows a normal distribution if its probability density curve is bell‑shaped and symmetric about the population mean μ. The total area under the curve equals 1, representing the total probability. The shape is completely determined by the mean μ and the standard deviation σ.

若连续型随机变量 X 的概率密度曲线呈钟形且关于总体均值 μ 对称，则 X 服从正态分布。曲线下的总面积等于 1，代表总概率。曲线的形状完全由均值 μ 和标准差 σ 决定。

The mean μ locates the centre of the distribution. The median and mode coincide with the mean.
均值 μ 确定了分布的中心位置。中位数和众数与均值重合。
A larger σ flattens and widens the curve; a smaller σ makes it taller and narrower.
σ 越大，曲线越扁平、越宽；σ 越小，曲线越高耸、越窄。
About 68% of data falls within μ ± 1σ, 95% within μ ± 2σ, and 99.7% within μ ± 3σ (the empirical rule).
大约 68% 的数据落在 μ ± 1σ 内，95% 落在 μ ± 2σ 内，99.7% 落在 μ ± 3σ 内（经验法则）。

2. Probability Density Function of the Normal Distribution | 正态分布的概率密度函数

The probability density function (PDF) for a normal random variable X is given by:

正态随机变量 X 的概率密度函数 (PDF) 为：

f(x) = (1/(σ√(2π))) e^{–(x–μ)²/(2σ²)}

Here π is the constant pi and e is Euler’s number. The formula is rarely used directly to calculate probabilities in exams – tables or calculators are used instead – but you must recognise that the PDF depends only on μ and σ.

其中 π 为圆周率，e 为欧拉数。考试中极少直接使用该公式计算概率，而是使用概率表或计算器，但你必须明白 PDF 只依赖于 μ 和 σ。

The curve has maximum height when x = μ, and it has points of inflection at x = μ ± σ. Because the function is symmetric, the probability P(X ≤ μ) = P(X ≥ μ) = 0.5.

当 x = μ 时曲线达到最高点，拐点位于 x = μ ± σ 处。由于函数对称，满足 P(X ≤ μ) = P(X ≥ μ) = 0.5。

3. Standard Normal Distribution and Z‑scores | 标准正态分布与 Z 分数

Any normal distribution X ~ N(μ, σ²) can be transformed to the standard normal distribution Z ~ N(0, 1²), which has mean 0 and variance 1. The transformation is called standardising:

任何一个正态分布 X ~ N(μ, σ²) 都可以通过标准化变换为标准正态分布 Z ~ N(0, 1²)，其均值为 0、方差为 1。变换公式为：

z = (x – μ) / σ

The z‑score tells you how many standard deviations a data value lies from the mean. Positive z means above the mean; negative z means below. This standardisation allows you to compare values from different normal populations and to use a single probability table for all normal calculations.

z 分数表示某个数据值距离均值有几个标准差。正 z 值表示高于均值，负 z 值表示低于均值。标准化使你能够比较来自不同正态总体的数值，并可利用同一张概率表进行所有正态计算。

4. Using the Standard Normal Table | 使用标准正态分布表

In many exam papers, a table provides cumulative probabilities Φ(z) = P(Z ≤ z) for positive z‑scores. Because of symmetry, probabilities for negative z‑scores can be deduced using Φ(–z) = 1 – Φ(z).

许多试卷会提供标准正态分布表，给出正 z 值对应的累积概率 Φ(z) = P(Z ≤ z)。利用对称性，负 z 值的概率可通过 Φ(–z) = 1 – Φ(z) 求得。

A small extract of such a table might look like:

下表为概率表的小片段：

z	0.00	0.01	0.02
0.0	0.5000	0.5040	0.5080
1.0	0.8413	0.8438	0.8461
1.5	0.9332	0.9345	0.9357

Always ensure you understand whether your table gives P(Z ≤ z) or P(0 ≤ Z ≤ z). CCEA often uses a cumulative lower‑tail table, while IB calculators return any probability directly.

务必明确所给表格提供的是 P(Z ≤ z) 还是 P(0 ≤ Z ≤ z)。CCEA 通常使用左下侧累积概率表，而 IB 计算器可直接得出任意概率值。

5. Calculating Probabilities | 计算概率

To find P(X < a), first standardise a to z = (a – μ)/σ, then look up Φ(z). If the question asks for P(X > a), use P(X > a) = 1 – P(X ≤ a). For an interval P(a < X < b), compute Φ(z_b) – Φ(z_a), where z_a and z_b are the z‑scores of a and b.

计算 P(X < a) 时，先将 a 标准化为 z = (a – μ)/σ，再查表得 Φ(z)。若求 P(X > a)，利用 P(X > a) = 1 – P(X ≤ a)。对于区间概率 P(a < X < b)，计算 Φ(z_b) – Φ(z_a)，其中 z_a、z_b 分别为 a、b 的 z 分数。

Worked example: X ~ N(100, 15²). Find P(85 < X < 115).
z₁ = (85 – 100)/15 = –1.00, z₂ = (115 – 100)/15 = 1.00.
Using table, Φ(1.00) = 0.8413, so probability = 0.8413 – (1 – 0.8413) = 0.6826 (empirical rule).

例题：X ~ N(100, 15²)，求 P(85 < X < 115)。
z₁ = (85 – 100)/15 = –1.00，z₂ = (115 – 100)/15 = 1.00。
查表得 Φ(1.00) = 0.8413，因此概率 = 0.8413 – (1 – 0.8413) = 0.6826（符合经验法则）。

6. Inverse Normal: Finding Critical Values | 逆正态：求临界值

Sometimes you are given a probability and need to find the corresponding value of x. This is the inverse normal problem. For the standard distribution, find z such that P(Z ≤ z) = p. Then apply x = μ + zσ.

有时题目给出概率，要求找出对应的 x 值，这就是逆正态问题。对标准正态分布，先找到满足 P(Z ≤ z) = p 的 z 值，再代入公式 x = μ + zσ。

Many IB and CCEA questions involve finding the value that cuts off a given upper‑tail percentage, e.g. the top 10%. If P(X > k) = 0.10, then P(Z > z) = 0.10 ⇒ Φ(z) = 0.90. Look up Φ^–1(0.90) ≈ 1.2816; then k = μ + 1.2816σ.

许多 IB 和 CCEA 试题会要求找出切去某个右侧尾部概率的分界值，如上侧 10%。若 P(X > k) = 0.10，则 P(Z > z) = 0.10 ⇒ Φ(z) = 0.90。查表得 Φ^–1(0.90) ≈ 1.2816，于是 k = μ + 1.2816σ。

Always state clearly which tail is used. Draw a sketch to avoid sign errors, especially when finding symmetrical bounds such as a central 95% interval, which requires z = ±1.96.

务必清楚表明使用的是哪个尾部。画图可以帮助避免符号错误，尤其在求对称边界时，如中间 95% 的区间，对应的 z 值为 ±1.96。

7. Finding Unknown Mean or Standard Deviation | 寻找未知的均值或标准差

In exam questions you may be given two probability statements and asked to find μ or σ. Set up a pair of simultaneous equations by standardising each given condition. For instance, if you know P(X < 20) = 0.15 and P(X > 80) = 0.05, you can write:

考试中可能给出两个概率条件，要求求解 μ 或 σ。通过标准化每个条件建立联立方程组。例如，已知 P(X < 20) = 0.15 且 P(X > 80) = 0.05，可列出：

(20 – μ)/σ = –1.0364, (80 – μ)/σ = 1.6449

Solve simultaneously to obtain μ and σ. This technique appears frequently in CCEA A‑Level papers and IB HL questions. Double‑check the sign of z: a left‑tail probability less than 0.5 gives a negative z, a right‑tail probability less than 0.5 gives a positive z.

联立求解即可得到 μ 和 σ。这种方法常见于 CCEA A‑Level 和 IB HL 试题。注意检查 z 的符号：左侧概率小于 0.5 时 z 为负，右侧概率小于 0.5 时 z 为正。

8. Distribution of Sample Means & Central Limit Theorem | 样本均值的分布与中心极限定理

When you take repeated random samples of size n from any population with mean μ and standard deviation σ, the distribution of the sample mean X̅ approaches a normal distribution as n increases. This is the Central Limit Theorem (CLT).

从均值为 μ、标准差为 σ 的任意总体中反复抽取容量为 n 的随机样本，样本均值 X̅ 的分布会随着 n 增大而趋近于正态分布，这就是中心极限定理 (CLT)。

If the population itself is normal, then X̅ ~ N(μ, σ²/n) exactly for any n. Otherwise, the rule of thumb is that n ≥ 30 is sufficient for the approximation to be valid. The standard deviation of the sample mean, σ/√n, is called the standard error.

若总体本身为正态分布，则对任意 n 均有 X̅ ~ N(μ, σ²/n)。否则，经验准则是当 n ≥ 30 时，该近似已足够准确。样本均值的标准差 σ/√n 称为标准误。

This theorem allows you to calculate probabilities involving sample means. For instance, if X ~ N(50, 10²) and you take a sample of size 25, then X̅ ~ N(50, 10²/25) i.e. N(50, 4).

这一定理使我们可以计算涉及样本均值的概率。例如，若 X ~ N(50, 10²) 且抽取容量为 25 的样本，则 X̅ ~ N(50, 10²/25)，即 N(50, 4)。

9. Normal Approximation to the Binomial | 二项分布的正态近似

When a binomial distribution X ~ B(n, p) has a large n, calculating exact probabilities becomes tedious. If both np ≥ 5 and nq ≥ 5 (with q = 1 – p), the binomial can be approximated by a normal distribution N(μ, σ²) where μ = np and σ = √(npq).

当二项分布 X ~ B(n, p) 的 n 很大时，精确计算概率将变得繁琐。若 np ≥ 5 且 nq ≥ 5（q = 1 – p），则可用正态分布 N(μ, σ²) 来近似，其中 μ = np，σ = √(npq)。

Because the binomial is discrete and the normal is continuous, a continuity correction must be applied. For P(X ≤ a) use P(X < a + 0.5); for P(X ≥ a) use P(X > a – 0.5). This adjustment significantly improves accuracy.

由于二项分布是离散的而正态分布是连续的，必须进行连续性校正。对于 P(X ≤ a)，使用 P(X < a + 0.5)；对于 P(X ≥ a)，使用 P(X > a – 0.5)。这一调整能显著提高精度。

Example: X ~ B(200, 0.4). Find P(70 ≤ X ≤ 90). Mean = 80, variance = 48, σ = √48 ≈ 6.928. With continuity correction: P(69.5 < X < 90.5). Standardise and use normal table.

例题：X ~ B(200, 0.4)，求 P(70 ≤ X ≤ 90)。均值 = 80，方差 = 48，σ = √48 ≈ 6.928。加上连续性校正：P(69.5 < X < 90.5)，标准化后查表计算。

10. Checking Normality & Exam Tips | 检验正态性与考试技巧

Before applying normal procedures, you should check that the data or model justifies normality. Look for a roughly symmetric histogram, a straight‑line pattern on a Q‑Q plot (quantile‑quantile plot), or an approximate bell shape. In exam contexts, the question will state that a variable is normally distributed, or you will be told to assume so.

在使用正态方法前，应先检验数据或模型是否满足正态性。观察直方图是否大致对称，Q‑Q 图（分位数‑分位数图）是否近似为直线，或曲线是否呈钟形。考试中，题目通常会明确变量服从正态分布，或要求你假定其正态。

Key exam advice: Always sketch a bell curve and shade the area of interest. Label the mean and the x values. This simple visualisation prevents errors with tail directions. When using a graphical calculator (allowed in IB), learn to use the normalcdf and invNorm functions efficiently. For CCEA, show your standardisation steps clearly, even if using a calculator, to gain method marks.

重要考试建议：务必画出钟形曲线草图并标出所求区域的面积，标出均值和 x 值。简单的图示可以避免尾部方向的错误。在使用图形计算器时（IB 允许使用），要熟悉 normalcdf 和 invNorm 函数的高效用法。对于 CCEA，即使使用计算器，也要清晰写出标准化步骤，以获得方法分。

Finally, always round your final answers sensibly and pay attention to units. If a question gives mean and s.d. to one decimal place, your answer should not quote four decimal places of probability without justification.

最后，合理取舍最终答案的精度并注意单位。如果题目给出的均值和标准差保留了一位小数，你的概率答案在没有特别说明的情况下也不宜给出四位小数。

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Normal Distribution Key Points for IB & CCEA Mathematics | IB & CCEA 数学：正态分布考点精讲