A-Level CAIE Statistics: Worked Case Study Practice | A-Level CAIE 统计：案例分析实战演练

📚 A-Level CAIE Statistics: Worked Case Study Practice | A-Level CAIE 统计：案例分析实战演练

In A-Level CAIE Statistics (Papers 5 and 6), case study questions often combine several statistical techniques within one real-world context. Practising worked examples helps you recognise which method to use and how to set out your solution clearly. This article walks through eight representative case studies, from data summary to hypothesis testing, with full calculations and exam tips.

在 A-Level CAIE 统计学（Paper 5 和 Paper 6）中，案例分析题常常在一个真实情境中综合多个统计方法。练习完整例题有助于你识别该使用哪种方法，并学会清晰书写解题过程。本文通过八个典型案例，从数据汇总到假设检验，给出完整计算与考试技巧。

1. Understanding the Case Study Approach | 理解案例分析方法

A statistics case study usually gives a short scenario followed by several parts testing different skills. The key is to read the scenario carefully, identify the variables, and decide whether the question is about data description, probability, distributions, or inference.

统计学案例通常先给出一个简短情境，然后分几个小问考查不同技能。关键是仔细阅读情境，识别变量，并判断题目属于数据描述、概率、分布还是统计推断。

Always show your method step by step. In CAIE Statistics, method marks are awarded for correct formulas, substitution, and interpretation, not only for the final answer. Write down the model you use, for example X ~ B(12, 0.1) or X ~ N(50, 4²), before calculating.

解题时必须逐步展示方法。在 CAIE 统计学中，即使最终答案错误，只要公式正确、代入正确、解释合理，也能获得方法分。计算前先写下所用模型，例如 X ~ B(12, 0.1) 或 X ~ N(50, 4²)。

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2. Case 1: Summarising Data from a Frequency Table | 案例一：频数表数据汇总

A teacher records the marks of 50 students in a test using the grouped frequency table below. The midpoints are used because exact raw marks are not available.

一名教师用下面的分组频数表记录 50 名学生的测验成绩。由于没有原始分数，使用组中值进行计算。

Marks x	0 ≤ x < 10	10 ≤ x < 20	20 ≤ x < 30	30 ≤ x < 40	40 ≤ x < 50
Frequency f	4	8	15	13	10

We first compute Σfx and Σfx². Using the midpoints 5, 15, 25, 35, 45, we get:

首先计算 Σfx 和 Σfx²。使用组中值 5, 15, 25, 35, 45，得到：

Σfx = 4×5 + 8×15 + 15×25 + 13×35 + 10×45 = 1420

is 1420 ÷ 50 = 28.4. For variance, Σfx² = 4×5² + 8×15² + 15×25² + 13×35² + 10×45² = 47450.

因此平均数为 1420 ÷ 50 = 28.4。对于方差，Σfx² = 4×5² + 8×15² + 15×25² + 13×35² + 10×45² = 47450。

Variance = 47450 ÷ 50 − 28.4² = 142.44

1.93. Always use the grouped data formula with midpoints, and state your answers to an appropriate degree of accuracy.

标准差为 √142.44 ≈ 11.93。务必使用带组中值的分组数据公式，并将答案保留到合适的精度。

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3. Case 2: Probability with Tree Diagrams | 案例二：树状图概率

A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. Draw a tree diagram to find the probability that at least one red ball is drawn, and the probability that the second ball is blue.

一个袋子里有 5 个红球和 3 个蓝球。随机不放回地抽取两个球。画出树状图，求至少抽到一个红球的概率，以及第二个球是蓝球的概率。

3/8. After drawing one ball, the totals change because sampling is without replacement. The four joint probabilities are:

第一次抽取 P(R) = 5/8，P(B) = 3/8。由于是不放回抽样，抽走一个球后总数会改变。四种联合概率为：

P(RR) = 5/8 × 4/7 = 20/56, P(RB) = 5/8 × 3/7 = 15/56

P(BR) = 3/8 × 5/7 = 15/56, P(BB) = 3/8 × 2/7 = 6/56

he second ball is blue occurs in RB or BB, so P(second blue) = 15/56 + 6/56 = 21/56 = 3/8.

至少一个红球意味着不是两个都是蓝球，所以 P(至少一个红球) = 1 − P(BB) = 1 − 6/56 = 50/56 = 25/28。第二个球是蓝球出现在 RB 或 BB 中，因此 P(第二个蓝球) = 15/56 + 6/56 = 21/56 = 3/8。

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4. Case 3: Discrete Random Variables and Expectation | 案例三：离散随机变量与期望

) = 0.2, P(X = 1) = 0.3, P(X = 2) = 0.4, P(X = 3) = 0.1. Find E(X), Var(X), and E(4X − 2).

离散随机变量 X 具有以下概率分布：P(X = 0) = 0.2，P(X = 1) = 0.3，P(X = 2) = 0.4，P(X = 3) = 0.1。求 E(X)、Var(X) 以及 E(4X − 2)。

First compute E(X) as ΣxP(X = x).

首先按 ΣxP(X = x) 计算 E(X)。

E(X) = 0×0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 1.4

For variance, compute E(X²) first.

对于方差，先计算 E(X²)。

E(X²) = 0²×0.2 + 1²×0.3 + 2²×0.4 + 3²×0.1 = 2.8

Var(X) = E(X²) − [E(X)]² = 2.8 − 1.4² = 0.84

− 2) = 4E(X) − 2 = 4 × 1.4 − 2 = 3.6. This property is very common in CAIE exam questions.

利用期望的线性性质，E(4X − 2) = 4E(X) − 2 = 4 × 1.4 − 2 = 3.6。这一性质在 CAIE 考试中非常常见。

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5. Case 4: Binomial Distribution in Context | 案例四：二项分布情境应用

A machine produces components with a 10% defective rate. A random sample of 12 components is taken. Let X be the number of defective components, so X ~ B(12, 0.1). Find the probability that exactly 2 components are defective, at most 1 is defective, and at least 1 is defective.

一台机器生产的零件次品率为 10%。随机抽取 12 个零件。设 X 为次品数量，则 X ~ B(12, 0.1)。求恰好有 2 个次品、至多有 1 个次品以及至少有 1 个次品的概率。

For a binomial distribution, P(X = r) = C(n, r) pʳ qⁿ⁻ʳ, where q = 1 − p. Here n = 12, p = 0.1, q = 0.9.

对于二项分布，P(X =

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