📚 A-Level CAIE Statistics: Case Study Practical Exercises | A-Level CAIE 统计学：案例分析实战演练
In A-Level CAIE Statistics (9709), success comes not just from memorising formulas but from applying them to real-world data. This article walks you through a series of integrated case studies that mirror the style of exam questions, helping you develop a structured approach to statistical problem-solving. Each case focuses on a core topic and illustrates how descriptive tools, probability models and inferential methods are used in practice.
在 A-Level CAIE 统计学（9709）中，成功不仅靠记忆公式，更在于将之应用于真实数据。本文通过一系列模拟考试风格的综合性案例，帮助你构建系统化的统计解题思路。每个案例聚焦一个核心主题，展示描述工具、概率模型和推断方法如何在实践中发挥作用。
1. Descriptive Statistics: Analysing Exam Scores | 描述性统计：分析考试成绩
Ten students sat a mathematics test. Their marks, out of 100, were: 45, 52, 60, 60, 68, 72, 75, 80, 85, 90. We begin with a quick numerical summary to understand the centre and spread of these scores.
十名学生参加了一场数学测验，满分 100 分，成绩如下：45, 52, 60, 60, 68, 72, 75, 80, 85, 90。我们先做一个快速的数字摘要，以了解这些分数的集中趋势和离散程度。
The mean is the total divided by the count. The sum is 687, so the mean x&# between the 5th and 6th values: (68 + 72)/2 = 70. The mode is 60, since it appears twice.
均值是总和除以个数。总和为 687，因此均值 x̄ = 687 / 10 = 68.7。将数据排序后，中位数位于第 5 和第 6 个值之间：(68 + 72)/2 = 70。众数是 60，因为它出现了两次。
riance is approximately 213.12, giving a standard deviation s = √213.12 ≈ 14.6. The interquartile range (IQR) is Q3 − Q1. With Q1 ≈ 60 and Q3 ≈ 80, the IQR is roughly 20.
为了测量离散程度，我们计算标准差。首先求出各值与均值之差，平方，求和，再除以 n - 1 = 9。样本方差约为 213.12，因此标准差 s = √213.12 ≈ 14.6。四分位距 IQR = Q3 − Q1，其中 Q1 ≈ 60、Q3 ≈ 80，故 IQR 约为 20。
A box-and-whisker plot would show the median at 70, a box from 60 to 80, and whiskers extending to 45 and 90. This reveals a slight left skew, as the lower whisker is longer than the upper one.
箱线图会显示中位数在 70，箱体从 60 延伸到 80，触须延伸至 45 和 90。由于下触须比上触须长，这表明分布略呈左偏。
| Statistic | Value |
| Mean | 68.7 |
| Median | 70 |
| Standard Deviation | 14.6 |
| IQR | 20 |
2. Probability: Risk Assessment in Card Games | 概率：纸牌游戏中的风险评估
A card is drawn at random from a standard deck of 52 playing cards. We want to find the probability that it is a heart or an ace, and then the conditional probability that it is an ace given that it is a heart.
从一副标准 52 张扑克牌中随机抽取一张牌。我们希望求出它是红心或是 A 的概率，以及在已知它是红心的条件下它是 A 的条件概率。
Let H be the event 'heart' and A be the event 'ace'. There are 13 hearts and 4 aces, with one card — the ace of hearts — tion rule: P(H ∪ A) = P(H) + P(A) - P(H ∩ A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13.
设 H 为“红心”事件，A 为“A”事件。红心有 13 张，A 有 4 张，其中红心 A 被重复计算。使用加法法则：P(H ∪ A) = P(H) + P(A) - P(H ∩ A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13。
P(H ∪ A) = 4/13
The conditional probability P(A | H) asks: of all hearts, what fraction are aces? Since only one heart is an ace, P(A | H) = 1/13. This matches the formula P(A ∩ H)/P(H) = (1/52) / (13/52) = 1/13.
条件概率 P(A | H) 询问：在所有红心中，A 占多大比例？因为只有一张红心是 A，所以 P(A | H) = 1/13。这与公式 P(A ∩ H)/P(H) = (1/52) / (13/52) = 1/13 一致。
In real-world risk assessment, such conditional probabilities help update beliefs when new information arrives — for instance, reassessing the likelihood of a defect after a positive test result.
在现实的风险评估中，这类条件概率有助于在新信息出现时更新信念——例如，在得到阳性测试结果后重新评估缺陷的可能性。
3. Binomial Distribution: Quality Control on a Production Line | 二项分布：生产线质量控制
A factory produces USB cables, and the long-term defect rate is 2%. A quality inspector randomly selects 20 cables. We can model the number of defective cables, X, with a binomial distribution: X ~ B(20, 0.02).
一家工厂生产 USB 数据线，长期缺陷率为 2%。质检员随机抽取 20 根数据线。我们可以用二项分布来模拟缺陷数量 X：X ~ B(20, 0.02)。
X ~ B(20, 0.02)
The probability of exactly two defectives is P(X = 2) = ²⁰C₂ × (0.02)² × (0.98)¹⁸. Using a calculator, ²⁰C₂ = 190, so P(X = 2) ≈ 190 × 0.0004 × 0.6676 ≈ 0.0508.
恰好有两件次品的概率为 P(X = 2) = ²⁰C₂ × (0.02)² × (0.98)¹⁸。使用计算器，²⁰C₂ = 190，因此 P(X = 2) ≈ 190 × 0.0004 × 0.6676 ≈ 0.0508。
P(X = 2) ≈ 0.0508
obability of at most one defective, we sum P(X = 0) and P(X = 1). P(X = 0) = (0.98)²⁰ ≈ 0.6676, and P(X = 1) = 20 × 0.02 × (0.98)¹⁹ ≈ 0.2725. Adding gives about 0.9401. Thus there is a 94% r defectives, suggesting that if two or more appear, the process might need investigation.
要求最多有一件次品的概率，我们将 P(X = 0) 与 P(X = 1) 相加。P(X = 0) = (0.98)²⁰ ≈ 0.6676，P(X = 1) = 20 × 0.02 × (0.98)¹⁹ ≈ 0.2725。两者之和约为 0.9401。因此有 94% 的可能性出现一件或更少次品，这意味着如果出现两件或以上，可能需要对生产过程进行调查。
4. Normal Distribution: Lifespan of Light Bulbs | 正态分布：灯泡寿命
d standard deviation σ = 40 hours. What proportion of bulbs last more than 850 hours? And what lifetime guarantee can the manufacturer give so that only 5% of bulbs fail before the guaranteed time?
某品牌 LED 灯泡的寿命服从正态分布，均值 μ = 800 小时，标准差 σ = 40 小时。有多少比例的灯泡寿命超过 850 小时？制造商若想保证仅有 5% 的灯泡在质保期前失效，应设定多长的质保时间？
normal tables, P(Z > 1.25) = 1 - Φ(1.25) ≈ 1 - 0.8944 = 0.1056. So about 10.6% of bulbs exceed 850 hours.
我们将 850 标准化为 Z 值：Z = (850 - 800) / 40 = 1.25。查阅标准正态分布表，P(Z > 1.25) = 1 - Φ(1.25) ≈ 1 - 0.8944 = 0.1056。因此约有 10.6% 的灯泡寿命超过 850 小时。
P(X > 850) ≈ 0.1056
ma; = 800 + (−1.645)(40) = 800 − 65.8 = 734.2 hours. A guarantee of roughly 734 hours would ensure only 5% of bulbs fail earlier.
对于质保时间，我们需要求 t 使得 P(X < t) = 0.05。对应的 Z 值约为 −1.645。去标准化：t = μ + zσ = 800 + (−1.645)(40) = 800 − 65.8 = 734.2 小时。大约 734 小时的质保期可确保仅有 5% 的灯泡提前失效。
Guarantee time ≈ 734 hours
5. Sampling and Estimation: Estimating Average Income | 抽样与估计：估计平均收入
A researcher surveys 100 randomly selected households in a city and finds the sample mean monthly income to be $5200 with a sample standard deviation of $800. Construct a 95% confidence interval for the true population mean income.
研究人员随机调查了某城市 100 户家庭，样本月均收入为 5200 美元，样本标准差为 800 美元。请构建总体均值收入的 95% 置信区间。
rmal approximation. The standard error of the mean is s / √n = 800 / 10 = 80. The 95% confidence interval is x&# 96. This gives 5200 ± 1.96 × 80 = 5200 ± 156.8, i.e. (5043.2, 5356.8).
由于样本量较大（n = 100），我们使用正态近似。均值的标准误为 s / √n = 800 / 10 = 80。95% 置信区间为 x̄ ± z × SE，其中 z = 1.96。计算得 5200 ± 1.96 × 80 = 5200 ± 156.8，即 (5043.2, 5356.8)。
95% CI for μ: (5043.2, 5356.8)
We are 95% confident that the true average monthly income lies within this interval. If the researcher wanted a narrower interval, increasing the sample size would reduce the standard error.
我们有 95% 的置信度认为真实的月均收入落在此区间内。如果研究人员希望区间更窄，增加样本量可减小标准误。
6. Hypothesis Testing: Testing a New Drug | 假设检验：测试新药
A pharmaceutical company claims its new painkiller relieves headaches in under 30 minutes on average. A consumer group doubts this and tests the drug on 50 patients, recording a sample mean relief time of 28.5 minutes with a standard deviation of 5 minutes. Test the company's claim at the 5% significance level.
一家制药公司声称其新型止痛药平均可在 30 分钟内缓解头痛。某消费者组织对此表示怀疑，对 50 名患者进行测试，得到样本平均缓解时间为 28.5 分钟，标准差为 5 分钟。在 5% 显著性水平下检验该公司的声明。
Let μ represent the true mean relief time. The hypotheses are H₀: μ = 30 and H&# ) = −1.5 / 0.7071 ≈ −2.12. The critical value for a one-tailed test at 5% is −1.645. Since −2.12 < −1.645, we reject H₀.
令 μ 表示真实的平均缓解时间。假设为 H₀: μ = 30，H₁: μ < 30（单尾检验）。检验统计量 Z = (28.5 - 30) / (5/√50) = −1.5 / 0.7071 ≈ −2.12。在 5% 显著性水平下，单尾检验的临界值为 −1.645。由于 −2.12 < −1.645，我们拒绝 H₀。
Z = −2.12, Reject H₀
There is sufficient evidence at the 5% level to support the consumer group's belief that the true mean relief time is less than 30 minutes. The result is statistically significant, and the company's claim may be contested.
在 5% 显著性水平下有充分证据支持消费者组织的观点，即真实平均缓解时间少于 30 分钟。该结果具有统计显著性，公司的声明可能受到质疑。
7. Correlation and Regression: Advertising Spend vs. Sales | 相关与回归：广告支出与
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