IB Mathematics: Core Concepts of the Normal Distribution | IB数学：正态分布核心概念解析

📚 IB Mathematics: Core Concepts of the Normal Distribution | IB数学：正态分布核心概念解析

The normal distribution is one of the most important probability distributions in IB Mathematics, appearing in both Analysis & Approaches (AA) and Applications & Interpretation (AI). It models continuous data that clusters around a central mean, such as heights, test scores, and measurement errors. Mastering its core concepts is essential for exam success.

正态分布是IB数学中最重要的概率分布之一，在分析与方法（AA）和应用与解释（AI）中都会出现。它用于建模围绕中心均值聚集的连续数据，例如身高、考试成绩和测量误差。掌握其核心概念对于考试成功至关重要。

1. Definition and Parameters | 定义与参数

A continuous random variable X follows a normal distribution with mean μ and standard deviation σ. We write X ~ N(μ, σ²). The probability density function (pdf) is symmetric about the mean and has a characteristic bell shape.

连续随机变量X服从均值为μ、标准差为σ的正态分布，记作X ~ N(μ, σ²)。其概率密度函数关于均值对称，具有典型的钟形曲线。

f(x) = 1 / (σ√(2π)) × e^(−(x−μ)² / (2σ²)), −∞ < x < ∞

The parameter μ determines the center of the distribution, while σ determines the spread. A larger σ produces a wider, flatter curve; a smaller σ produces a narrower, taller curve.

参数μ决定分布的中心位置，σ决定分布的离散程度。σ越大，曲线越宽越扁；σ越小，曲线越窄越高。

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

The curve is symmetric about x = μ, so the mean, median, and mode are all equal.

曲线关于x = μ对称，因此均值、中位数和众数三者相等。
The total area under the curve is exactly 1, representing total probability.

曲线下的总面积为1，代表总概率。
The curve approaches the horizontal axis asymptotically as x → ±∞, but never touches it.

当x → ±∞时，曲线渐近地接近横轴，但永远不会触及横轴。
Points of inflection occur at x = μ − σ and x = μ + σ.

拐点出现在x = μ − σ和x = μ + σ处。

3. The Empirical Rule (68-95-99.7 Rule) | 经验法则（68-95-99.7法则）

For any normal distribution, approximately 68% of the data lies within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

对于任何正态分布，约68%的数据落在距均值一个标准差范围内，95%落在两个标准差范围内，99.7%落在三个标准差范围内。

区间	概率
μ ± 1σ	≈ 68%
μ ± 2σ	≈ 95%
μ ± 3σ	≈ 99.7%

This rule provides a quick way to estimate probabilities without a calculator and helps identify unusual values in a dataset.

该法则提供了一种无需计算器即可快速估算概率的方法，并有助于识别数据中的异常值。

4. Standard Normal Distribution and Z-Scores | 标准正态分布与Z分数

The standard normal distribution has mean 0 and standard deviation 1, denoted Z ~ N(0, 1). Any normal variable X can be transformed into Z using the z-score formula:

标准正态分布的均值为0，标准差为1，记作Z ~ N(0, 1)。任何正态变量X都可以通过Z分数公式转换为Z：

Z = (X − μ) / σ

The z-score measures how many standard deviations a value is above or below the mean. A positive z-score indicates a value above the mean, while a negative z-score indicates a value below the mean.

Z分数衡量一个值高于或低于均值多少个标准差。正的Z分数表示该值高于均值，负的Z分数表示该值低于均值。

5. Calculating Probabilities Using the Normal Distribution | 使用正态分布计算概率

To find P(X < a) for a normal distribution, we first convert a to a z-score, then use the standard normal table or a GDC (graphical display calculator). For example, if X ~ N(100, 15²), then P(X < 120) = P(Z < (120 − 100)/15) = P(Z < 1.333).

要计算正态分布中的P(X < a)，首先将a转换为Z分数，然后使用标准正态分布表或图形计算器（GDC）。例如，若X ~ N(100, 15²)，则P(X < 120) = P(Z < (120 − 100)/15) = P(Z < 1.333)。

Common probability notation includes P(X ≤ a), P(X > a) = 1 − P(X ≤ a), and P(a < X < b) = P(X < b) − P(X < a).

常见的概率记号包括P(X ≤ a)、P(X > a) = 1 − P(X ≤ a)以及P(a < X < b) = P(X < b) − P(X < a)。

6. Inverse Normal Calculations | 逆正态计算

Inverse normal problems ask for the value of x corresponding to a given cumulative probability. For example, find the 90th percentile of X ~ N(100, 15²), i.e. find x such that P(X ≤ x) = 0.90.

逆正态问题要求给定累积概率下对应的x值。例如，求X ~ N(100, 15²)的第90百分位数，即求x使得P(X ≤ x) = 0.90。

Using the inverse normal function on a GDC, or by finding the z-score corresponding to 0.90 (z ≈ 1.2816) and solving x = μ + zσ, we get x ≈ 100 + 1.2816 × 15 ≈ 119.22.

使用GDC上的逆正态函数，或查找0.90对应的Z分数（z ≈ 1.2816）并解x = μ + zσ，可得x ≈ 100 + 1.2816 × 15 ≈ 119.22。

7. Sampling Distributions and the Central Limit Theorem | 抽样分布与中心极限定理

When taking a random sample of size n from a population with mean μ and standard deviation σ, the sample mean X̄ has approximately a normal distribution if n is large (usually n ≥ 30), regardless of the population distribution. This is the Central Limit Theorem (CLT).

当从均值为μ、标准差为σ的总体中抽取容量为n的随机样本时，如果n足够大（通常n ≥ 30），样本均值X̄近似服从正态分布，无论总体分布是什么。这就是中心极限定理（CLT）。

The sampling distribution of X̄ has mean μ and standard deviation σ/√n. This allows us to make inferences about population means from sample data.

样本均值X̄的抽样分布均值为μ，标准差为σ/√n。这使我们能够从样本数据对总体均值进行推断。

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

When n is large and p is not too close to 0 or 1, the binomial distribution B(n, p) can be approximated by a normal distribution with mean np and variance np(1 − p). A common rule is that both np and n(1 − p) should be at least 5 (or 10, depending on the textbook).

当n很大且p不太接近0或1时，二项分布B(n, p)可以用均值为np、方差为np(1 − p)的正态分布来近似。一个常用规则是np和n(1 − p)都应至少为5（或10，取决于教材）。

Because the binomial is discrete and the normal is continuous, a continuity correction is applied. For example, P(X ≤ k) is approximated by P(X ≤ k + 0.5) using the normal distribution.

由于二项分布是离散的而正态分布是连续的，需要进行连续性修正。例如，P(X ≤ k)用正态分布近似为P(X ≤ k + 0.5)。

9. Common Exam Pitfalls | 常见考试陷阱

Forgetting to use the continuity correction when approximating a binomial distribution with a normal distribution.

在使用正态分布近似二项分布时忘记进行连续性修正。
Confusing σ with σ². Always check whether the problem gives the standard deviation or the variance.

混淆σ与σ²。始终检查题目给出的是标准差还是方差。
Using P(X < a) instead of P(X ≤ a) when the question uses "at least" or "at most". Note that for continuous distributions, P(X = a) = 0, so P(X < a) = P(X ≤ a).

当问题使用”至少”或”至多”时，错误使用P(X < a)代替P(X ≤ a)。注意对于连续分布，P(X = a) = 0，因此P(X < a) = P(X ≤ a)。
Misreading the z-table: always confirm whether the table gives P(Z < z) or P(0 < Z < z).

误读Z表：始终确认表格给出的是P(Z < z)还是P(0 < Z < z)。

10. Practice Question | 练习题

The heights of students at a school are normally distributed with mean 170 cm and standard deviation 8 cm. (a) Find the probability that a randomly chosen student is taller than 180 cm. (b) Find the height that separates the top 10% of students from the rest.

某学校学生的身高服从均值170 cm、标准差8 cm的正态分布。(a) 求随机选择一名学生身高超过180 cm的概率。(b) 求将最高的10%学生与其余学生分开的身高临界值。

Solution: (a) z = (180 − 170)/8 = 1.25. P(Z > 1.25) = 1 − 0.8944 = 0.1056. (b) The z-score for the 90th percentile is 1.2816, so h = 170 + 1.2816 × 8 ≈ 180.25 cm.

解答：(a) z = (180 − 170)/8 = 1.25。P(Z > 1.25) = 1 − 0.8944 = 0.1056。(b) 第90百分位数的Z分数为1.2816，因此h = 170 + 1.2816 × 8 ≈ 180.25 cm。

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IB Mathematics: Core Concepts of the Normal Distribution | IB数学：正态分布核心概念解析