📚 Exercise 14B.4: Solving Normal Distribution Problems | 正态分布习题解析
Exercise 14B.4 in the IB Mathematics curriculum presents a classic normal distribution problem that tests your ability to standardise values, read probability tables and perform inverse calculations. Whether you are working from the Analysis & Approaches or Applications & Interpretation syllabus, fluency with the normal model is essential. In this article we unpack a representative problem involving apple masses, taking you through each part with clear reasoning and practical tips.
IB 数学课程中的习题14B.4是一道经典的正态分布问题，考察你对数值标准化、查阅概率表以及逆向计算的能力。无论你学习的是分析与方法（AA）还是应用与解释（AI）课程，熟练运用正态模型都是基本功。本文以一道关于苹果质量的典型题目为例，带你逐一拆解，辅以清晰的推理和实用技巧。
1. Understanding the Exercise Structure | 了解题目结构
A typical Exercise 14B.4 states: The masses of apples from an orchard are normally distributed with mean 120 g and standard deviation 15 g. An apple is chosen at random. (a) Find the probability that its mass is less than 100 g. (b) Find the probability that its mass lies between 110 g and 130 g. (c) The heaviest 5% of apples are classified as 'premium'. Determine the minimum mass required to be premium. This three-part layout progresses from direct probability to inverse normal reasoning.
典型的习题14B.4会给出：某果园苹果的质量服从正态分布，均值为120克，标准差为15克。随机抽取一个苹果。(a) 求其质量小于100克的概率。(b) 求质量介于110克和130克之间的概率。(c) 最重的5%的苹果被定为“特级”，求成为特级所需的最低质量。这种三部分的设置由直接概率计算逐步过渡到逆正态推理。
2. Reviewing the Normal Distribution | 回顾正态分布
A continuous random variable X that follows a normal distribution with mean μ and variance σ² is written as X ~ N(μ, σ²). Its bell-shaped curve is symmetric about the mean, and the total area under the curve equals 1. Probabilities correspond to areas under the curve and cannot be found by direct formula – instead we standardise to the standard normal distribution Z ~ N(0, 1²).
若连续随机变量X服从均值为μ、方差为σ²的正态分布，记作X ~ N(μ, σ²)。其钟形曲线关于均值对称，曲线下总面积为1。概率对应曲线下的面积，无法用简单公式直接求得——我们需要将其标准化为标准正态分布Z ~ N(0, 1²)后再求解。
3. Standardisation and the Z‑Score | 标准化与Z分数
The transformation that connects any normal variable X to the standard normal Z is:
z = (x − μ) / σ
Here, μ = 120 g and σ = 15 g. This formula converts a raw data value into a z‑ that tells us how many standard deviations the value lies away from the mean. A negative z‑ bles and calculator functions for the normal distribution rely on this standardised scale.
将任意正态变量X与标准正态Z联系起来的变换为：
z = (x − μ) / σ
本题中μ = 120 g，σ = 15 g。该公式将原始数据值转化为z分数，标明该值与均值相差多少个标准差。z为负表示低于均值，为正表示高于均值。所有正态分布的概率表与计算器功能都建立在这一标准化尺度之上。
4. Using the Standard Normal Table | 使用标准正态分布表
with a printed table, you look up the z value to find the area to its left. For instance, a small excerpt looks like:
| z | Φ(z) |
|---|---|
| 0.67 | 0.7486 |
| 1.00 | 0.8413 |
| 1.33 | 0.9082 |
| 1.64 | 0.9495 |
| 1.65 | 0.9505 |
ur calculator provides normalcdf or invNorm functions, you can bypass the printed table, but you must still understand the underlying logic.
标准正态分布的累积分布函数（CDF）记作Φ(z)，表示P(Z ≤ z)。使用印刷表格时，可通过查找z值获得左侧面积。例如，上表摘录了部分数据。许多表格还利用对称性：Φ(−z) = 1 − Φ(z)。如果你的计算器提供normalcdf或invNorm函数，可以省去查表步骤，但你仍然需要理解背后的逻辑。
5. Working Through Part (a): Probability Less Than a Value | 求解(a)部分：小于某值的概率
We need P(X < 100). Standardise the boundary:
z = (100 − 120) / 15 = −20 / 15 ≈ −1.33
= Φ(−1.33). Using symmetry, Φ(−1.33) = 1 − Φ(1.33) = 1 − 0.9082 = 0.0918. Thus, about 9.18% 0 g. Always round to three or four significant figures unless the question says otherwise.
我们需要求P(X < 100)。将边界值标准化：
z = (100 − 120) / 15 = −20 / 15 ≈ −1.33
因此P(X < 100) = P(Z < −1.33) = Φ(−1.33)。利用对称性，Φ(−1.33) = 1 − Φ(1.33) = 1 − 0.9082 = 0.0918。所以，大约9.18%的苹果质量低于100克。除非题目另有要求，通常保留三或四位有效数字。
6. Working Through Part (b): Probability Between Two Values | 求解(b)部分：介于两值之间的概率
For P(110 < X < 130), calculate the two z‑scores:
z₁ = (110 − 120)/15 = −10/15 ≈ −0.6667 → −0.67
z₂ = (130 − 120)/15 = +10/15 ≈ 0.6667 → 0.67
The probability is the area between these two z values:
P(110 < X < 130) = Φ(0.67) − Φ(−0.67)
symmetric about the mean and spans roughly ±0.67σ, capturing almost half the distribution.
对于P(110 < X < 130)，先计算两个z分数：
z₁ = (110 − 120)/15 = −10/15 ≈ −0.6667 → −0.67
z₂ = (130 − 120)/15 = +10/15 ≈ 0.6667 → 0.67
概率为这两个z值之间的面积：
P(110 < X < 130) = Φ(0.67) − Φ(−0.67)
Φ(0.67) = 0.7486，而Φ(−0.67) = 1 − 0.7486 = 0.2514。相减得0.7486 − 0.2514 = 0.4972。这个结果是合理的，因为区间关于均值对称且跨度约为±0.67σ，囊括了将近一半的分布。
7. Inverse Calculations for Part (c): Finding a Cut-Off | (c)部分的逆向计算：求截断值
The heaviest 5% correspond to a right‑ them; by interpolation we often use z₀ = 1.645. Then transform back to the original scale:
x₀ = μ + z₀ × σ = 120 + 1.645 × 15 = 120 + 24.675 = 144.675 g
Rounding appropriately, the minimum mass to be classed as premium is about 145 g. Always interpret the result in context: an apple must weight at least 145 g to be in the top 5%.
最重的5%对应右尾概率0.05。我们需要求x₀使得P(X > x₀) = 0.05，即P(X ≤ x₀) = 0.95。先从标准正态入手：求z₀满足Φ(z₀) = 0.95。查表得Φ(1.64) = 0.9495，Φ(1.65) = 0.9505，因此所需z介于二者之间；常用插值取z₀ = 1.645。再转换回原始尺度：
x₀ = μ + z₀ × σ = 120 + 1.645 × 15 = 120 + 24.675 = 144.675 g
适当舍入后，成为特级苹果的最低质量约为145克。务必结合情境解释结果：一个苹果至少须重145克才能跻身前5%。
8. Visualizing and Checking Your Answers | 可视化与检验答案
art (b), shading the central area around the mean reminds you to subtract two cumulative values. For part (c), marking the cut‑off on the right tail and labelling 5% ensures you look for the correct tail. Symmetry checks like verifying that Φ(–z) + Φ(z) = 1 can also catch arithmetic mistakes.
快速画一条标注均值的钟形曲线有助于避免符号错误。对于(a)部分，在最左端涂上阴影则直观看到概率很小。对于(b)部分，涂出均值周围的中央区域，提醒你需将两个累积值相减。对于(c)部分，在右尾标出截断点并注明5%，确保你查找的是正确尾部。对称性检查（如验证Φ(–z) + Φ(z) = 1）也能发现计算错误。
9. Common Errors and How to Avoid Them | 常见错误与避免方法
- Forgetting to standardise: No probability table works with raw X values. Always compute z first.
忘记标准化：概率表无法直接处理原始X值，务必先计算z。 - Wrong table direction: Reading Φ(z) as the right‑tail area. Remember that standard normal tables usually give the area to the left.
查表方向错误：将Φ(z)当作右尾面积。标准正态表通常给出左侧面积。 - Misusing the inverse: When finding a right‑tail cut‑off, convert to a left‑tail probability (1 − tail) before looking up z.
逆向使用不当：在求右尾截断值时，先转换为左尾概率（1 − 尾面积）再查z。 - Rounding incorrectly: Keep intermediate z‑scores to at least three decimal places; round the final answer according to context.
舍入不当：中间z分数至少保留三位小数，最终答案则根据情境舍入。
10. Extending the Problem: Exam-Style Variations | 拓展：考试中的变体
Examiners often adapt Exercise 14B.4 to include two‑tailed intervals, hypothesis testing or sampling distributions. For instance, you might be asked: 'Find the probability that the mean mass of a sample of 10 apples exceeds 125 g.' That introduces the standard error σ/√n and tests your understanding of the Central Limit Theorem. The same standardisation logic applies, but with μ unchanged and σ replaced by σ/√n.
出题人常常将习题14B.4改编为双尾区间、假设检验或抽样分布题。例如可能会问：“求10个苹果样本的平均质量超过125克的概率。”这就引入了标准误σ/√n，并考察你对中心极限定理的理解。标准化的逻辑不变，只是μ保持不变，σ则替换为σ/√n。
11. Summary of Key Techniques | 关键技巧总结
n exercise efficiently requires a clear sequence: (1) Identify μ and σ; (2) Convert each x‑value to a z‑score; (3) Use Φ(z) to find left‑ tract Φ(smaller z) from Φ(larger z); (5) For inverse problems, determine the required left‑ is routine with a variety of contexts and always sketch the curve to guide your reasoning.
高效解决正态分布习题需要清晰的步骤：（1）确定μ和σ；（2）将每个x值转化为z分数；（3）用Φ(z)求左尾概率；（4）对于区间，用Φ(较大z)减去Φ(较小z)；（5）对于逆问题，确定所需左尾概率，查出对应的z，再用x = μ + zσ逆标准化。在不同情境中反复练习这一流程，并始终画出曲线以辅助推理。
12. Final Thoughts | 结语
Exercise 14B.4 may look simple, but it encapsulates the core skills of normal distribution problem solving that appear throughout the IB course. By mastering standardisation, table reading and inverse logic, you build a foundation for confidence in statistics and beyond. Revisit this structure whenever you face a similar question, and remember that careful step‑by‑step reasoning always outperforms rushed guessing.
习题14B.4看似简单，却浓缩了贯穿IB课程的正态分布核心解题技能。通过掌握标准化、查表和逆向推理，你将为统计及其他领域打下自信的基础。每当你遇到类似问题时，回顾这一解题框架，记住：步步为营的推理永远胜过仓促的猜测。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图