📚 Exercise 21C: Normal Distribution | 正态分布练习解析
Exercise 21C is a classic set of IB Mathematics problems focused on the normal distribution, often involving probability calculations, standardisation, inverse normal look‑ups, and real‑life contexts such as exam scores, heights, and weights. This article unpacks the core ideas behind Exercise 21C, walks through typical question types, and provides worked solutions and tips to help you master normal distribution problems.
练习21C是IB数学中一组典型正态分布题目，通常涉及概率计算、标准化、逆正态查找以及考试成绩、身高、体重等实际情景。本文梳理练习21C背后的核心概念，讲解常见题型，并提供详细解答与技巧，帮助你彻底掌握正态分布问题。
1. Normal Distribution Fundamentals | 正态分布基础
The normal distribution, often called the Gaussian distribution, is a continuous probability distribution that is symmetric about its mean μ. It is fully described by two parameters: the mean μ and the standard deviation σ. The graph of the probability density function is a bell‑shaped curve, where the total area under the curve equals 1.
正态分布，又称高斯分布，是一种关于均值μ对称的连续概率分布。它完全由两个参数决定：均值μ和标准差σ。其概率密度函数图形呈钟形曲线，曲线下的总面积为1。
The probability that a random variable X lies between two values a and b is given by the area under the curve between a and b. For the normal distribution, about 68% of the data lies within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ (the empirical rule). Exercise 21C heavily uses these properties together with standard normal tables.
随机变量X落在两个值a和b之间的概率等于a到b之间曲线下方的面积。正态分布大约68%的数据落在均值±1σ内，95%落在±2σ内，99.7%落在±3σ内（经验法则）。练习21C大量结合这些性质与标准正态表使用。
f(x) = 1/(σ√2π) · e^( –(x – μ)² / (2σ²) )
概率密度函数：f(x) = 1/(σ√2π) · e^( –(x – μ)² / (2σ²) )
2. Standardisation and Z‑scores | 标准化与Z值
To work with normal distribution problems, we convert a normal variable X ~ N(μ, σ²) to the standard normal variable Z ~ N(0, 1) using the z‑score formula. This step is essential because standard normal tables provide probabilities for Z only.
为了求解正态分布问题，我们通过Z值公式将正态变量X ~ N(μ, σ²)转化为标准正态变量Z ~ N(0, 1)。这一步至关重要，因为标准正态表仅提供Z值的概率。
z = (x – μ) / σ
The z‑score measures how many standard deviations an observed value x is away from the mean. A positive z indicates x is above the mean, while a negative z indicates below. Exercise 21C often begins by asking students to calculate z for given x, μ, and σ.
Z值衡量观测值x与均值相差多少个标准差。正值表示x高于均值，负值表示低于均值。练习21C常常要求先计算给定x、μ和σ下的z值。
are normally distributed with μ = 170 cm and σ = 10 cm, a height of 185 cm gives z = (185 – 170)/10 = 1.5. A height of 155 cm gives z = –1.5.
例如，若身高服从μ = 170 cm、σ = 10 cm的正态分布，身高185 cm对应的z = (185 – 170)/10 = 1.5。身高155 cm对应的z = –1.5。
3. Using the Standard Normal Table | 使用标准正态表
ables provide values for positive z from 0.00 to 3.99. For negative z, we use symmetry: Φ(–z) = 1 – Φ(z).
标准正态表给出累积概率Φ(z) = P(Z ≤ z)，即z左侧曲线下的面积。大多数表格列出0.00到3.99的正z值。对于负z，利用对称性：Φ(–z) = 1 – Φ(z)。
In Exercise 21C, you will often look up Φ(1.5) which is 0.9332, meaning 93.32% of the data lies below a z of 1.5. To find the probability that Z exceeds 1.5, compute 1 – 0.9332 = 0.0668.
在练习21C中，你经常需要查表，如Φ(1.5)=0.9332，表示93.32%的数据落在z=1.5以下。若要计算Z大于1.5的概率，则用1 – 0.9332 = 0.0668。
For an interval probability P(a < Z < b), we compute Φ(b) – Φ(a). For example, P(–1 < Z < 1.5) = Φ(1.5) – Φ(–1) = 0.9332 – (1 – 0.8413) = 0.7745.
对于区间概率P(a < Z < b)，计算Φ(b) – Φ(a)。例如，P(–1 < Z < 1.5) = Φ(1.5) – Φ(–1) = 0.9332 – (1 – 0.8413) = 0.7745。
4. Direct Probability Problems | 直接概率问题
g and σ = 20 g. Find the probability that a randomly chosen apple weighs less than 120 g.
典型的练习21C题目：苹果重量服从μ=150 g、σ=20 g的正态分布。求随机选取一个苹果重量小于120 g的概率。
Step 1: Compute z = (120 – 150)/20 = –1.5. Step 2: P(X < 120) = P(Z < –1.5) = Φ(–1.5) = 1 – Φ(1.5) = 1 – 0.9332 = 0.0668. So about 6.68% of apples weigh less than 120 g.
步骤1：计算z = (120 – 150)/20 = –1.5。步骤2：P(X < 120) = P(Z < –1.5) = Φ(–1.5) = 1 – 0.9332 = 0.0668。因此大约6.68%的苹果重量小于120 g。
e proportion of data between 130 g and 170 g. Compute z₁ = -1.0, z₂ = 1.0. Then P = Φ(1.0) – Φ(–1.0) = 0.8413 – 0.1587 = 0.6826, matching the empirical rule.
另一种常见形式：求重量在130 g到170 g之间的比例。计算z₁ = -1.0，z₂ = 1.0。P = 0.8413 – 0.1587 = 0.6826，符合经验法则。
5. Inverse Normal Calculations | 逆正态计算
Inverse normal problems start with a known probability (or percentage) and ask for the corresponding x‑value or z‑score. Exercise 21C includes these to test understanding of the cumulative distribution's reverse operation.
逆正态问题从一个已知概率（或百分比）出发，要求找到对应的x值或z值。练习21C包含此类题目以考察对累积分布反向运算的理解。
ance: In a test with μ = 65 marks and σ = 8 marks, find the mark that separates the top 10% of students. Here we need the z‑score such that Φ(z) = 0.9, because the top 10% le, z ≈ 1.2816 (interpolating as needed). Then x = μ + zσ = 65 + 1.2816 × 8 ≈ 75.25 marks.
例如：某测验μ=65分，σ=8分，求划分前10%学生的分数。这里我们需要满足Φ(z)=0.9的z值，因为前10%就是第90百分位数。查表得z ≈ 1.2816，然后x = μ + zσ = 65 + 1.2816×8 ≈ 75.25分。
Sometimes the problem gives a symmetric central proportion, e.g., the middle 80%. Then we find z such that P(– 0.9, again z ≈ 1.2816. Exercise 21C contains several such symmetrical interval questions.
有时题目给出对称中心比例，如中间80%。则找出z使P(–z < Z < z) = 0.80，每侧尾部概率0.10，同样需要Φ(z)=0.9，z ≈ 1.2816。练习21C包含多个此类对称区间问题。
6. Combining Mean and Standard Deviation Unknown | 均值和标准差均未知的情形
Advanced questions in Exercise 21C might give two probability statements and ask you to find both μ and σ. This requires setting up a system of two equations using z‑scores.
练习21C中的进阶题可能给出两个概率条件，要求同时求出μ和σ。这需要通过z值建立两个方程的方程组。
< 498 ml) = 0.01 and P(volume > 508 ml) = 0.05. Find μ and σ. From P(Z < z₁) = 0.01, z₁ ≈ –2.3263. So (498 – μ)/σ = –2.3263. From P(Z > z₂) = 0.05, i.e. P(Z < z₂) = 0.95, z₂ ≈ 1.6449. Thus (508 – μ)/σ = 1.6449. Solve simultaneously: μ ≈ 503.2 ml, σ ≈ 2.24 ml.
例如：一台灌装机。P(容量<498 ml)=0.01，P(容量>508 ml)=0.05。求μ和σ。由P(Z < z₁)=0.01得z₁≈ –2.3263，所以(498 – μ)/σ = –2.3263。由P(Z > z₂)=0.05即P(Z < z₂)=0.95，得z₂≈1.6449，所以(508 – μ)/σ = 1.6449。联立解得μ≈503.2 ml，σ≈2.24 ml。
This two‑equation approach appears frequently and requires careful table reading and algebraic manipulation, a key skill tested in Exercise 21C.
这种双方程方法很常见，需要仔细查表和代数运算，是练习21C测试的关键技能。
7. Contextual Applications and Modelling | 实际情境应用与建模
IB Mathematics places strong emphasis on applying the normal distribution to real‑world data. Exercise 21C includes problems about examination results, battery lifetimes, heights, and manufacturing tolerances. You are often asked to interpret results in context.
IB数学非常重视将正态分布应用于实际数据。练习21C包含考试成绩、电池寿命、身高和制造公差等问题，常要求联系情境解释结果。
life of 24 hours with σ = 2 hours. A customer will only accept the product if fewer than 5% of batteries last less than 21 hours. Should the customer accept the claim? Compute z = (21–24)/2 = –1.5, P(Z < –1.5) = 0.0668 = 6.68%, which exceeds 5%, so the claim is not acceptable. This illustrates how normal calculations inform decision‑making.
例如：电池厂商声称平均寿命24小时，σ=2小时。客户只接受寿命不足21小时的电池比例低于5%的产品。是否接受该声称？计算z = (21–24)/2 = –1.5，P(Z < –1.5)=0.0668=6.68%，超过5%，因此不接受。这展示了正态计算如何辅助决策。
Modelling assumptions should also be examined: Is the distribution truly normal? The data must be continuous, symmetric, and unimodal. Exercise 21C sometimes asks you to justify why a normal model is appropriate.
建模假设也需要审视：分布真的符合正态吗？数据应连续、对称且单峰。练习21C有时要求你说明为什么正态模型是合适的。
8. Typical Pitfalls and How to Avoid Them | 常见陷阱与避免方法
Confusing x and z: You must always standardise before using the table. Use z = (x – μ)/σ, not the reverse.
混淆x和z：使用表格前必须标准化。用z = (x – μ)/σ，而不是反过来。
Forgetting to subtract from 1: When finding a right‑tail probability P(Z > z) for z > 0, compute 1 – Φ(z). Many students take Φ(z) directly.
忘记从1中减去：求右尾概率P(Z > z)且z>0时，要计算1 – Φ(z)。不少学生直接取Φ(z)。
Incorrect sign for negative z: Use symmetry correctly; Φ(–z) = 1 – Φ(z). Do not just use a negative z in the table as if it were positive.
处理负z时符号错误：正确使用对称性；Φ(–z) = 1 – Φ(z)。不要像正z那样直接查表。
Rounding errors: In inverse normal problems, interpolate z‑values if necessary, and keep at least 4 significant figures during calculations to avoid final answer drift.
四舍五入误差：在逆正态问题中，必要时对z值进行插值，计算过程中保留至少四位有效数字，避免最终答案偏离。
9. Step‑by‑Step Worked Example from Exercise 21C | 练习21C逐步详解示例
Question: The marks in an exam are normally distributed with mean 62 and standard deviation 9. a) Find the probability that a student scores above 75. b) If the bottom 20% of students retake the exam, find the cut‑off mark. c) Find the interquartile range of the marks.
题目：某考试分数服从正态分布，均值62，标准差9。a) 求学生得分高于75的概率。b) 如果末尾20%的学生需要重考，找出分数线。c) 求分数的四分位距。
Solution a): z = (75 – 62)/9 ≈ 1.4444. Φ(1.4444) ≈ 0.9257 (interpolating). P(X > 75) = 1 – 0.9257 = 0.0743.
解答a)：z = (75 – 62)/9 ≈ 1.4444。Φ(1.4444)≈0.9257（插值）。P(X > 75) = 1 – 0.9257 = 0.0743。
Solution b): Need z such that Φ(z) = 0.20. Table gives z ≈ –0.8416. Then cut‑off x = 62 + (–0.8416)×9 ≈ 54.43 marks.
解答b)：需要使Φ(z)=0.20的z值。查表得z ≈ –0.8416。分数线x = 62 + (–0.8416)×9 ≈ 54.43分。
Solution c): Q1 corresponds to p=0.25, z ≈ –0.6745; Q3 to p=0.75, z ≈ 0.6745. So Q1 ≈ 62 – 0.6745×9 ≈ 55.93, Q3 ≈ 62 + 0.6745×9 ≈ 68.07. IQR ≈ 12.14 marks.
解答c)：Q1对应p=0.25，z ≈ –0.6745；Q3对应p=0.75，z ≈ 0.6745。因此Q1≈55.93，Q3≈68.07。IQR≈12.14分。
10. Practice Strategies for Exercise 21C | 练习21C的备考策略
To master Exercise 21C, first consolidate how to use the standard normal table forwards and backwards. Practice at least 20 straightforward probability questions before tackling inverse and two‑equation problems.
要掌握练习21C，先巩固正向和反向使用标准正态表。先练习至少20道直接概率题，再处理逆问题和双方程题。
Draw a clear graph for every problem: label the mean, x‑values, shaded area, and z‑scores. This visual aid significantly reduces sign errors. Also, memorise the empirical rule and key percentile z‑values (e.g., 1.645 for 95th, 1.2816 for 90th) to speed up work.
每道题画出清晰的图形：标出均值、x值、阴影区域和z值。这种视觉辅助能显著减少符号错误。同时，记住经验法则和关键百分位数的z值（如1.645对应95百分位，1.2816对应90百分位）以加快解题速度。
When checking answers, verify that a cumulative probability less than 0.5 corresponds to a negative z, and a z between –3 and 3 usually covers nearly all data. If your result looks extreme, re‑check your calculations.
检查答案时，验证累积概率小于0.5对应负z值，且z值在–3到3之间通常覆盖几乎所有数据。如果结果看起来很极端，重新检查计算。
11. Linking to IB Assessment | 与IB评估的联系
In IB Mathematics (AA and AI), normal distribution questions appear in Paper 1 (no calculator) and Paper 2 (with GDC). In calculator papers, you may use the built‑in normalcdf and invNorm functions. However, understanding the table method is still required for non‑calculator papers and conceptual questions.
在IB数学（AA和AI）中，正态分布题目出现在试卷一（无计算器）和试卷二（可使用图形计算器）。在计算器试卷中，你可以使用内置的normalcdf和invNorm功能。但无计算器试卷和概念题仍要求掌握查表方法。
Exercise 21C mirrors IB exam style, often asking you to justify your choice of distribution, state assumptions, and interpret results in context. Always include units and a concluding sentence in context to earn full marks.
练习21C模拟IB考试风格，常要求说明选择分布的理由、陈述假设并结合情境解释结果。始终包含单位并用情境总结句以获得满分。
12. Summary and Final Tips | 总结与最后建议
rdise to z → use the table for probabilities, or reverse the process for inverse problems.
正态分布是建模连续数据的强大工具。练习21C训练你计算概率、求百分位数并求解未知参数。核心流程是：确定μ和σ→标准化为z→利用表格求概率，或者反向操作解决逆问题。
Remember that behind every normal distribution problem there is a bell curve waiting to be labelled with x, μ, z, and shaded areas. Let that picture guide your reasoning, and you will find Exercise 21C not only manageable but enjoyable.
记住，每道正态分布题目背后都有一条待标注x、μ、z和阴影区域的钟形曲线。让这张图引导你的思路，你将发现练习21C不仅容易应对，而且充满乐趣。
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