📚 IB Math Normal Distribution Practice: Exercise 24D.2 | IB数学正态分布练习：24D.2
Exercise 24D.2 is a typical problem in the IB Mathematics Analysis and Approaches (AA) or Applications and Interpretation (AI) courses, designed to test your understanding of sampling distributions and probability calculations involving the sample mean. This type of question often appears in the Statistics and Probability topic and requires a systematic approach to transform a real‑world scenario into a standard normal probability problem. In this article we will break down every step needed to tackle Exercise 24D.2 with confidence, explain the underlying theory, and highlight common mistakes so you can score full marks in similar exam questions.
练习 24D.2 是 IB 数学分析与方法（AA）或应用与解释（AI）课程中的一道典型题目，旨在检验你对抽样分布以及涉及样本均值的概率计算的理解。这类问题经常出现在统计与概率专题中，需要采用系统的方法将实际情景转化为标准正态分布的概率问题。本文将逐步拆解攻克练习 24D.2 所需的每一个步骤，解释背后的理论知识，并点出常见错误，帮助你在类似的考题中拿满分数。
1. The Problem Statement of Exercise 24D.2 | 练习24D.2的题目陈述
For the purpose of this detailed walkthrough, we consider the following real‑style IB problem: A factory produces light bulbs. The lifetimes of the bulbs are normally distributed with a mean of 800 hours and a standard deviation of 40 hours. A quality control inspector randomly selects 25 bulbs. Find the probability that the mean lifetime of these 25 bulbs lies between 790 hours and 810 hours. This is a classic exercise on the sampling distribution of the sample mean.
为便于详细讲解，我们采用一道贴近 IB 真题风格的题目：某工厂生产灯泡，灯泡寿命服从正态分布，平均寿命为 800 小时，标准差为 40 小时。质检员随机抽取 25 个灯泡，求这 25 个灯泡的平均寿命介于 790 小时到 810 小时之间的概率。这是一道关于样本均值抽样分布的经典练习。
2. Identifying Parameters and Variables | 识别参数与变量
given information using proper notation. Let X be the lifetime of a single bulb. We have X ~ andard deviation σ = 40. The sample size is n = 25. We are interested in the sample mean X̄ (read as “X‑bar”). The goal is to compute P(790 < X̄ < 810).
首先用规范的符号写下所有已知信息。设 X 为单个灯泡的寿命，则 X ~ N(μ = 800, σ² = 40²)。因此总体均值 μ = 800，总体标准差 σ = 40。样本容量 n = 25。我们关注的是样本均值 X̄（读作“X bar”）。目标是计算 P(790 < X̄ < 810)。
3. The Sampling Distribution of the Sample Mean | 样本均值的抽样分布
Because the population is normally distributed, the sample mean X̄ also follows a normal distribution, regardless of the sample size. The mean of X̄ is exactly the population mean μ. The standard deviation of X̄, often called the standard error, is σ/√n. Therefore, X̄ ~ N(μ, σ²/n), which in this case becomes X̄ ~ 0, 8²). This exact normality is a key property that makes the calculation straightforward.
因为总体服从正态分布，无论样本量多大，样本均值 X̄ 都精确服从正态分布。X̄ 的均值恰好等于总体均值 μ，其标准差通常称为标准误，大小为 σ/√n。因此，X̄ ~ N(μ, σ²/n)，代入数值后得到 X̄ ~ N(800, 40²/25) = N(800, 8²)。这一精确正态性是使计算变得简洁的关键性质。
4. The Standard Error and Its Interpretation | 标准误及其解读
The standard error is σ/√n = 40/√25 the sample mean is expected to vary from sample to sample. Compared with the original population standard deviation of 40, the standard error of 8 is much smaller, reflecting the reduced variability when we average 25 observations. Visualising the sampling distribution as a narrower normal curve helps intuition.
标准误为 σ/√n = 40/√25 = 40/5 = 8。这个数值反映了样本均值在不同样本之间的波动大小。与原来的总体标准差 40 相比，标准误 8 要小得多，这说明对 25 个观测值取平均后变异性显著降低。把抽样分布想象成一条更窄的正态曲线有助于建立直观感觉。
5. Standardising the Sample Mean | 将样本均值标准化
To find P(790 < X̄ < 810), we convert the sample mean limits to Z‑scores using the formula Z = (X̄ - μ) / (σ/√n). For X̄ = 790, Z₁ = (790 - 800) / 8 = -1.25. For X̄ (810 - 800) / 8 = 1.25. The probability problem becomes P(-1.25 < Z < 1.25), where Z ~ N(0, 1).
为求 P(790 < X̄ < 810)，我们用公式 Z = (X̄ − μ) / (σ/√n) 将样本均值边界转化为 Z 分数。当 X̄ = 790 时，Z₁ = (790 − 800) / 8 = −1.25；当 X̄ = 810 时，Z₂ = (810 − 800) / 8 = 1.25。概率问题就转化为 P(−1.25 < Z < 1.25)，其中 Z ~ N(0, 1)。
Z = (X̄ − μ) / (σ/√n)
6. Using the Standard Normal Distribution Table | 使用标准正态分布表
056. Therefore, P(-1.25 < Z < 1.25) = Φ(1.25) - Φ(-1.25) = 0.8944 - 0.1056 = 0.7888.
我们需要用到标准正态累积概率 Φ(z) = P(Z < z)，分别计算 z = 1.25 和 z = −1.25 时的值。查标准正态分布表或用计算器可得 Φ(1.25) ≈ 0.8944。由对称性知，Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056。因此，P(−1.25 < Z < 1.25) = Φ(1.25) − Φ(−1.25) = 0.8944 − 0.1056 = 0.7888。
7. Calculating the Probability Step by Step | 逐步计算概率
he sample mean life of 25 bulbs lies between 790 and 810 hours is approximately 0.7888, or 78.9%. This step‑by‑step working is essential for earning method marks.
整合各个步骤：标准误 = 8，Z₁ = (790 − 800)/8 = −1.25，Z₂ = (810 − 800)/8 = 1.25。查表得 Z = 1.25 左侧面积为 0.8944，Z = −1.25 左侧面积为 0.1056。所求概率为 0.8944 − 0.1056 = 0.7888。因此，25 个灯泡的样本平均寿命落在 790 小时到 810 小时之间的概率约为 0.7888，即 78.9%。这样逐步展示解题过程对于拿到方法分至关重要。
8. Interpreting the Result in Context | 在实际语境中解释结果
A probability of 0.7888 means that if we repeatedly take random samples of 25 bulbs, about 79% of those samples would have a mean lifetime between 790 and 810 hours. This high probability suggests that a sample mean in this range is quite common given the production parameters. If an inspector obtained a sample mean outside this interval, it might indicate a shift in the process, although formal hypothesis testing would be required for a definitive conclusion.
概率 0.7888 意味着如果我们反复抽取容量为 25 的随机样本，大约有 79% 的样本的平均寿命会落在 790 到 810 小时之间。这一高概率表明，在给定的生产参数下，出现在该区间的样本均值十分常见。若质检员获得的样本均值落在区间之外，则可能暗示生产过程发生了偏移，不过要得出确切结论还需借助正式的假设检验。
9. Common Pitfalls and How to Avoid Them | 常见错误及如何避免
One frequent mistake is using the population standard deviation σ instead of the standard error σ/√n when calculating the Z‑score. Another is misreading the inequality direction or forgetting to subtract the lower tail area. Students also sometimes incorrectly assume the sample mean has the same variability as individual observations. Always draw a shaded normal curve and label the standard error clearly. Double‑check that you have used n, not just n, in the denominator of the standard error.
一个常见错误是在计算 Z 分数时使用总体标准差 σ 而不是标准误 σ/√n。另一个错误是读错不等号方向，或者忘记减去下尾面积。还有学生有时错误地认为样本均值的变异程度与单个观测值相同。务必画出正态曲线并标出阴影区域，清晰标注标准误。再次确认分母中使用的是 √n 而不仅仅是 n。
Always write the standard error as σ/√n explicitly.
始终将标准误明确写为 σ/√n。
Remember that the variance of X̄ is σ²/n, standard deviation is σ/√n.
记住 X̄ 的方差是 σ²/n，标准差是 σ/√n。
10. Extension: What If the Population Standard Deviation Is Unknown? | 延伸：如果总体标准差未知？
In many IB questions, σ may be unknown and estimated by the sample standard deviation s. The sampling distribution of X̄ then follows a t‑distribution with n-1 degrees of freedom. The standard error would be s/√n. The process is otherwise similar, but you would use the t‑table instead of the standard normal table. Recognising when to switch to the t‑distribution is a crucial skill examined at the Higher Level.
在很多 IB 考题中，σ 可能是未知的，需要用样本标准差 s 来估计。此时 X̄ 的抽样分布服从自由度为 n−1 的 t 分布，标准误变为 s/√n。除此以外解题过程类似，但要改用 t 分布表而不是标准正态分布表。判断何时需改用 t 分布是 IB 高水平考试中要求的一项关键技能。
11. Exam Tips for Similar Exercises | 类似练习的考试技巧
When faced with a sampling distribution problem, write down the distribution of the population and then the distribution of X̄. Always state the standard error. Show the standardisation formula clearly. If a calculator is allowed, you may compute the probability directly using normalcdf, but still exhibit the Z‑score working for the method mark. Time management is critical; practising Exercise 24D.2‑style problems will speed up your recognition of patterns.
面对抽样分布问题时，先写下总体的分布，再写出 X̄ 的分布。务必标明标准误，清晰展示标准化公式。若允许使用计算器，可以直接用 normalcdf 计算概率，但仍需写出 Z 分数计算过程以获取方法分。时间管理很关键；多练像练习 24D.2 这类题目能加快你对解题模式的识别。
12. Summary and Key Takeaways | 总结与关键要点
To master IB Math Exercise 24D.2, remember the three‑step framework: identify the population parameters, determine the sampling distribution of the mean with standard error σ/√n, and standardise to Z‑scores. The problem centred on P(790 < X̄ < 810) for a normal population, yielding a probability of 0.7888. The same logic extends to confidence intervals and hypothesis tests. Keep a formula sheet handy and always link your numerical answer back to the context.
要攻克 IB 数学练习 24D.2，请记住三步框架：识别总体参数，确定样本均值的抽样分布及标准误 σ/√n，然后标准化为 Z 分数。本题围绕正态总体下的 P(790 < X̄ < 810) 展开，得出概率 0.7888。同样的逻辑也适用于置信区间和假设检验。手边备好公式表，并始终将数值答案与实际背景联系起来。
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