📚 CAIE Year 12 Statistics: Case Study Drill | CAIE Year 12 统计：案例分析实战演练
Welcome to a hands-on case study designed to reinforce your CAIE Year 12 Statistics knowledge. We will work through a realistic scenario involving the lifetimes of batteries produced by a company. By applying data representation, probability, discrete random variables, binomial and normal distributions, and more, you will see how AS-level topics come together in a practical investigation.
欢迎参加这次专为巩固 CAIE Year 12 Statistics 知识而设计的实战案例分析。我们将围绕一家公司生产的电池寿命数据展开真实场景演练，通过应用数据展示、概率、离散随机变量、二项分布、正态分布等多个模块，你将看到 AS 阶段各知识点如何在实际调查中协同运作。
1. Case Background | 案例背景
BatteryLife Ltd manufactures AA batteries and claims that their mean lifetime is 30 hours with a standard deviation of 2 hours. The lifetimes are assumed to be normally distributed. To monitor quality, a random sample of 20 batteries is taken from a day's production and tested. The recorded lifetimes (in hours) are as follows:
BatteryLife 公司生产 AA 电池，声称其平均寿命为 30 小时，标准差为 2 小时，且寿命服从正态分布。为监控质量，他们从一天的产量中随机抽取 20 节电池进行测试，记录下的寿命（小时）数据如下：
30.1, 29.5, 28.9, 31.2, 30.5, 27.8, 32.0, 29.9, 30.3, 28.5, 31.8, 29.0, 30.7, 28.1, 32.2, 30.0, 29.3, 31.5, 28.8, 30.9
We will use this dataset to explore descriptive and inferential statistical methods as specified in the CAIE AS syllabus.
我们将利用这一数据集，逐步探索 CAIE AS 教学大纲中所要求的描述性统计和推断性统计方法。
2. Data Representation: Stem-and-Leaf and Histogram | 数据展示：茎叶图和直方图
First, we organise the data using a stem-and-leaf plot. The integer part forms the stem and the decimal part the leaf. Sorting the values gives:
首先，我们用茎叶图整理数据：整数部分作茎，小数部分作叶。排序后的数据如下：
27 | 8
28 | 1 5 8 9
29 | 0 3 5 9
30 | 0 1 3 5 7 9
31 | 2 5 8
32 | 0 2
A frequency distribution for a histogram can be created using equal class widths of 1 hour starting from 27.0. The frequency densities are shown in the table below:
可建立组距为 1 小时（从 27.0 起）的频数分布，用于绘制直方图。各组的频数和频数密度见下表：
| Class interval (hours) | Frequency | Frequency density |
| 27.0 ≤ x < 28.0 | 1 | 1 |
| 28.0 ≤ x < 29.0 | 4 | 4 |
| 29.0 ≤ x < 30.0 | 4 | 4 |
| 30.0 ≤ x < 31.0 | 6 | 6 |
| 31.0 ≤ x < 32.0 | 3 | 3 |
| 32.0 ≤ x < 33.0 | 2 | 2 |
These representations give a clear picture of the distribution's shape, showing a slight skew but generally centred around 30 hours.
这些展示清晰地呈现了分布的形状，可见稍有偏斜但整体集中在 30 小时附近。
3. Measures of Centre and Spread: Mean and Variance | 集中趋势和离散度量：均值和方差
the average of the 10th and 11th values: (30.0 + 30.1)/2 = 30.05 hours. The sample mean x̄ is also 30.05 hours (sum = 601, n = 20).
对已排序的数据，中位数是第 10 和第 11 个数值的平均：(30.0 + 30.1)/2 = 30.05 小时。样本均值 x̄ 也是 30.05 小时（总和 = 601, n = 20）。
The sum of squared deviations from the mean is 32.47, so the sample variance is:
离均差的平方和为 32.47，因此样本方差为：
s² = Σ(xᵢ − x̄)² / (n − 1) = 32.47 / 19 ≈ 1.7089
e claimed population mean but shows slightly less variability than the stated σ = 2 hours.
样本标准差 s ≈ 1.307 小时。这些估计量表明样本接近声称的总体均值，但变异性略小于标称的 σ = 2 小时。
4. Basic Probability: Calculating Event Probabilities | 概率基础：计算事件概率
A battery is classified as defective if its lifetime is below 28 hours. From the sample, three batteries (27.8, 28.1, 28.5) fall below 28 h, giving an empirical probability:
若电池寿命低于 28 小时即判为次品。样本中有三节（27.8, 28.1, 28.5）低于 28 小时，经验概率为：
P(defective) = 3 / 20 = 0.15
Under the manufacturer' < 28) = P(Z < −1) = 0.1587. The sample estimate is reasonably consistent with this value.
根据制造商的正态模型 N(30, 2²)，理论概率为 P(X < 28) = P(Z < −1) = 0.1587。样本估计值与此较为吻合。
5. Conditional Probability and Independence | 条件概率与独立事件
If two batteries are drawn at random without replacement from the 20, we can find the probability both are defective:
若从 20 节电池中随机不放回抽取两节，可求二者均次品的概率：
P(both defective) = (3/20) × (2/19) = 6/380 ≈ 0.0158
itional probability that the second is defective given the first is defective is P(2nd def | differs from the unconditional probability 0.15, the events are not independent, which is expected when sampling without replacement.
在已知第一个为次品的条件下，第二个为次品的概率为 P(第二个次品 | 第一个次品) = 2/19 ≈ 0.1053。由于该条件概率不同于无条件概率 0.15，因此这两个事件并非独立，符合不放回抽样的预期。
6. Discrete Random Variables: Probability Distribution and Expectation | 离散随机变量：概率分布和期望
defective, 0 otherwise. Using p = 0.15 (from the sample), the probability distribution is:
对单个电池定义离散随机变量 D：若为次品则 D = 1，否则 D = 0。利用样本所得 p = 0.15，其概率分布为：
P(D = 1) = 0.15, P(D = 0) = 0.85
Expectation: E(D) = 1×0.15 + 0×0.85 = 0.15.
期望值：E(D) = 1×0.15 + 0×0.85 = 0.15。
Variance: Var(D) = 1²×0.15 + 0²×0.85 − (0.15)² = 0.15 − 0.0225 = 0.1275.
方差：Var(D) = 1²×0.15 + 0²×0.85 − (0.15)² = 0.15 − 0.0225 = 0.1275。
The standard deviation is √0.1275 ≈ 0.357. These measures will be useful in binomial modelling.
标准差为 √0.1275 ≈ 0.357。这些指标在二项分布建模中会用到。
7. Binomial Distribution: Modelling Defectives | 二项分布：次品率建模
Suppose a quality inspector randomly selects 10 batteries from the day's production. The number of defectives X can be modelled as B(10, 0.15), assuming independence and a constant defect probability.
假设质检员从当日产品中随机抽取 10 节电池。在独立且次品概率不变的条件下，次品数 X 可建模为 B(10, 0.15)。
For example, the probability of exactly 2 defectives is:
例如，恰好出现 2 个次品的概率为：
P(X = 2) = ¹⁰C₂ (0.15)² (0.85)⁸ = 45 × 0.0225 × 0.2725 ≈ 0.2759
ve is P(X ≤ 1) = P(X=0) + P(X=1) = (0.85)¹⁰ + 10×(0.15)(0.85)⁹ ≈ 0.1969 + 0.3474 = 0.5443.
至多 1 个次品的概率为 P(X ≤ 1) = P(X=0) + P(X=1) = (0.85)¹⁰ + 10×(0.15)(0.85)⁹ ≈ 0.1969 + 0.3474 = 0.5443。
= 10 × 0.15 × 0.85 = 1.275. These values help the company set acceptable tolerance limits.
E(X) = 10 × 0.15 = 1.5, Var(X) = 10 × 0.15 × 0.85 = 1.275。这些值有助于企业设定可接受的容差限。
8. Geometric Distribution: First Defective | 几何分布：首次出现次品
If batteries are tested one by one until the first defective is found, the number of
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