Common Mistakes in A-Level Mathematics (9709) Paper 5 – June 2022 Exam Report | A-Level数学（9709）卷五2022年6月考试报告易错点总结

📚 Common Mistakes in A-Level Mathematics (9709) Paper 5 – June 2022 Exam Report | A-Level数学（9709）卷五2022年6月考试报告易错点总结

The June 2022 examiner report for Cambridge International A-Level Mathematics (9709) Paper 5 (Probability & Statistics 2) highlighted a number of recurring errors that prevented candidates from achieving top marks. Many of these mistakes were not due to a lack of knowledge, but rather a failure to apply concepts precisely, interpret questions correctly, or meet the specific conditions required by statistical methods. This article summarises the most critical pitfalls, using examples from the report, to help future candidates avoid similar errors and improve their exam technique.

剑桥国际A-Level数学（9709）卷五（概率与统计2）2022年6月的考官报告指出了一系列反复出现的错误，这些错误阻碍了考生取得高分。其中许多失误并非因为知识欠缺，而是未能精确应用概念、正确理解题意或满足统计方法所需的特定条件。本文汇总了报告中最关键的易错点，并辅以示例，以帮助未来的考生避免类似错误，提升应试技巧。

1. Misusing Poisson Approximation Conditions | 误用泊松近似条件

Many candidates attempted to use the Poisson distribution to approximate a binomial probability without first checking that n is large and p is small. The standard rule of thumb is n > 50 and np < 5, or alternatively n > 20 and p < 0.1. In the June 2022 report, examiners noted that even when candidates did mention the conditions, they sometimes failed to verify them with numerical values from the question, leading to mark penalties. It is essential to write explicitly: ‘n = … is large and p = … is small, so the Poisson approximation is suitable.’

许多考生在未检查 n 很大且 p 很小的前提下，就直接用泊松分布近似二项概率。标准的经验法则是 n > 50 且 np < 5，或者 n > 20 且 p < 0.1。2022年6月的报告中，考官指出，即使考生提到了这些条件，有时也忘记用题目中的具体数值进行验证，导致扣分。务必明确写出：’n = … 很大，p = … 很小，因此适合使用泊松近似。’

X ~ B(n, p) → X ~ Po(λ) where λ = np

X ~ B(n, p) → X ~ Po(λ)，其中 λ = np

2. Forgetting Continuity Corrections in Normal Approximations | 正态近似中遗忘连续性校正

When using a normal distribution to approximate a discrete distribution such as the binomial or Poisson, a continuity correction must be applied. Examiners repeatedly observed candidates writing P(X ≤ 10) as P(Y < 10) for the approximating normal variable Y, or omitting the ±0.5 adjustment altogether. For a binomial X ~ B(n, p), the correct approximation for P(X ≤ a) is P(Y < a + 0.5) where Y ~ N(np, npq). This adjustment often made the difference between a nearly correct answer and a completely accurate one.

使用正态分布近似离散分布（如二项分布或泊松分布）时，必须应用连续性校正。考官一再发现，考生在近似正态变量 Y 下，将 P(X ≤ 10) 写成 P(Y < 10)，或完全忽略 ±0.5 的调整。对于二项分布 X ~ B(n, p)，P(X ≤ a) 的正确近似为 P(Y < a + 0.5)，其中 Y ~ N(np, npq)。这个调整往往决定了答案接近正确还是完全准确。

P(X ≤ 45) ≈ P(Y < 45.5) and P(X ≥ 60) ≈ P(Y > 59.5)

P(X ≤ 45) ≈ P(Y < 45.5)，P(X ≥ 60) ≈ P(Y > 59.5)

3. Stating Distributions Incompletely | 分布写得不完整

A frequent mistake was failing to state the full distribution in hypothesis testing or probability calculations. Candidates often wrote ‘X ~ Po’ without specifying the mean, or ‘X ~ B’ without giving n and p. Even if the parameters are stated elsewhere, the distribution must be defined clearly at the point of use. The report emphasised that marks are reserved for correct notation: X ~ B(20, 0.15) or X ~ Po(3.6), for example. Incomplete statements were treated as errors.

一个常见错误是在假设检验或概率计算中没有完整地写出分布。考生经常写 ‘X ~ Po’ 却不指明均值，或写 ‘X ~ B’ 却不给出 n 和 p。即便参数已在别处说明，在使用点也必须明确写出分布。报告强调，分数的获得有赖于正确符号：例如 X ~ B(20, 0.15) 或 X ~ Po(3.6)。不完整的写法即被视为错误。

4. Hypothesis Test Setup Errors | 假设检验的设定错误

Setting up null and alternative hypotheses incorrectly was a major source of lost marks. Candidates often confused the direction of the inequality in one‑tailed tests, writing H₁: p > 0.3 when the context required H₁: p < 0.3. Others used a two‑tailed test when a one‑tailed test was indicated by the wording of the question. It is vital to read phrases like ‘test whether the proportion has decreased’ as a signal for a lower‑tailed test. The hypotheses must be stated using parameters, not sample statistics: H₀: μ = 50, H₁: μ < 50.

错误设立原假设和备择假设是丢分的一大原因。考生经常在单尾检验中弄错不等号方向，在题目要求 H₁: p < 0.3 时却写成 H₁: p > 0.3。还有人根据题目的措辞应当用单尾检验，却采用了双尾检验。仔细阅读类似 ‘检验比例是否下降’ 的表述，这是下尾检验的提示。假设必须用参数而非样本统计量来表述：H₀: μ = 50，H₁: μ < 50。

5. Misinterpreting Confidence Intervals | 置信区间的错误解读

Examiners noted that many candidates wrote probabilistic statements about the parameter being inside the interval, such as ‘there is a 95% chance that the true mean lies in the interval’. This is incorrect. A confidence interval is a statement about the method: if we repeated the sampling many times, 95% of the intervals constructed would contain the true parameter. Candidates should express the interpretation as: ‘We are 95% confident that the interval (a, b) captures the population mean.’

考官注意到，许多考生对参数落在区间内写下了概率性的陈述，例如 ‘真实均值有95%的可能性落在此区间内’。这是错误的。置信区间陈述的是方法本身：如果多次重复抽样，那么所构造的区间中有95%会包含真实的参数。考生应这样表达解释：’我们有95%的信心认为区间 (a, b) 包含了总体均值。’

6. Misreading Statistical Tables | 读错统计表格

The Poisson and normal distribution tables were often misread, especially for cumulative probabilities. When asked for P(X = 4) using a cumulative Poisson table, some candidates simply read off the value for λ = 2.5 at x = 4, forgetting that the table gives cumulative probabilities P(X ≤ x). Thus they needed to compute P(X ≤ 4) − P(X ≤ 3). Similarly, in the normal distribution table, using the wrong tail or misreading Φ(z) for negative z led to avoidable errors. Practice with table formats is essential.

泊松分布和正态分布表格经常被读错，尤其是涉及累积概率时。在要求用泊松累积表求 P(X = 4) 时，有些考生直接读取 λ = 2.5 时 x = 4 的表格值，却忘记了表格提供的是累积概率 P(X ≤ x)。因此，他们需要计算 P(X ≤ 4) − P(X ≤ 3)。类似地，正态分布表中用错尾部，或对负 z 值读错 Φ(z)，也会导致本可避免的错误。熟悉表格格式的练习至关重要。

7. Ignoring the Domain of a Probability Density Function | 忽视概率密度函数的定义域

For continuous random variables, a pdf f(x) is defined only over a specified interval. The June 2022 report highlighted that many candidates forgot to restrict their calculations to this domain, resulting in probabilities greater than 1 or negative values. When finding the median or quartiles, it is necessary to check that the solution falls within the support of the distribution. Any answer outside the domain must be rejected, and working must clearly state the valid interval.

对于连续随机变量，概率密度函数 f(x) 仅定义在特定区间上。2022年6月的报告强调，很多考生忘记了将计算限制在此定义域内，导致出现大于1的概率或负值。当求中位数或四分位数时，有必要检查解是否落在分布的支持集内。任何位于域外的答案都必须排除，计算过程应明确写出有效区间。

8. Confusing One‑Tailed and Two‑Tailed Tests | 混淆单尾与双尾检验

Another common fault was using a two‑tailed test when the question clearly asked for a one‑tailed test, or vice versa. In some cases, candidates calculated the p‑value for a two‑tailed test but then compared it to the significance level for a one‑tailed test, leading to an incorrect conclusion. The wording ‘test whether the mean has increased’ requires a one‑tailed test with H₁: μ > μ₀. Examiners advised paying close attention to the alternative hypothesis implied by the context.

另一个常见错误是，题目明确要求单尾检验时却用了双尾检验，或反过来。部分考生按双尾检验计算了 p 值，却将其与单尾检验的显著性水平进行比较，从而得出错误结论。措辞 ‘检验均值是否增加’ 要求进行单尾检验，H₁: μ > μ₀。考官建议要密切关注上下文所暗示的备择假设。

9. Confusing p‑Value with the Significance Level | 混淆 p 值与显著性水平

Many candidates failed to distinguish between the p‑value and the significance level α. They would correctly compute a p‑value but then state that because the p‑value is less than α, H₀ is accepted. The correct conclusion is to reject H₀. Furthermore, some wrote that the p‑value is the probability that H₀ is true, which is a serious conceptual mistake. The p‑value is the probability of observing a test statistic at least as extreme as the one obtained, assuming H₀ is true.

许多考生未能区分 p 值与显著性水平 α。他们正确计算出了 p 值，却接着说因为 p 值小于 α，所以接受 H₀。正确的结论应该是拒绝 H₀。此外，还有人写道 p 值是 H₀ 为真的概率，这是严重的概念性错误。p 值是在 H₀ 成立的条件下，观察到至少与实际结果同样极端的检验统计量的概率。

10. Misapplying the Central Limit Theorem | 中心极限定理的应用不当

When dealing with the sample mean of a non‑normal population, the Central Limit Theorem allows the use of a normal distribution provided the sample size is sufficiently large (usually n ≥ 30). The examiner report noted that some candidates either applied the normal approximation without quoting the CLT, or used it for small samples from a highly skewed population, which is invalid. Additionally, when finding probabilities involving the sample mean X̅, the standard deviation must be σ/√n, not σ. Forgetting this resulted in grossly inaccurate answers.

在处理非正态总体的样本均值时，只要样本量足够大（通常 n ≥ 30），中心极限定理允许使用正态分布。考官报告指出，有些考生要么不引用 CLT 就直接用正态近似，要么对于高度偏态总体的抽样，样本量很小时还依然使用，这是不合理的。此外，在求涉及样本均值 X̅ 的概率时，标准差必须是 σ/√n 而非 σ。忘记这一点会导致答案严重失准。

X̅ ~ N(μ, σ²/n) approximately for large n

当 n 很大时，近似有 X̅ ~ N(μ, σ²/n)

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