📚 A-Level Maths: Example Responses for MA04 Unit S2 | A-Level 数学：MA04 单元 S2 题型解析与范例
MA04 Unit S2 (Statistics 2) tests your ability to handle probability distributions, approximations, and hypothesis tests. This article walks through key question types with model answers, showing you exactly what examiners look for. Each section pairs English commentary with Chinese explanations to strengthen both language and mathematical understanding.
MA04 单元 S2（统计 2）考查处理概率分布、近似计算和假设检验的能力。本文通过经典题型和范例答案，展示评卷者关注的重点。每个小节均采用英中双语对照讲解，帮助巩固语言与数学思维。
1. Overview of S2 and the MA04 Assessment | S2 与 MA04 考试概览
S2 builds on the descriptive statistics of S1, focusing on discrete and continuous probability models. You will encounter binomial, Poisson, and normal distributions, continuity corrections, and formal hypothesis tests. Mark schemes reward method marks for clear statement of distributions, correct use of formulas, and final answers rounded sensibly.
S2 在 S1 描述性统计的基础上深化，重点为离散与连续概率模型。你将处理二项、泊松、正态分布，以及连续性校正和规范的假设检验。评分标准奖赏清楚写明分布、正确使用公式和合理取整的最终答案。
The MA04 paper typically contains about 7 questions mixing calculation and interpretation. Example responses must show full workings; simply writing a final probability is never enough. Always define the random variable first, e.g., X ~ B(n, p) or X ~ Po(λ).
MA04 试卷通常包含约 7 题，融合计算与解释。范例答案必须展示完整步骤；仅仅写下最终概率远远不够。务必先定义随机变量，如 X ~ B(n, p) 或 X ~ Po(λ)。
2. Binomial Distribution: Calculation and Interpretation | 二项分布：计算与解释
Example: A factory produces bulbs. The probability a bulb is defective is 0.05. A sample of 20 bulbs is taken. Find the probability exactly 2 are defective.
例题：某工厂生产灯泡，次品率为 0.05。随机抽取 20 只灯泡，求恰好有 2 只次品的概率。
Model response: Let X be the number of defective bulbs, X ~ B(20, 0.05). We require P(X = 2). Using the formula: P(X = 2) = C(20, 2) × (0.05)2 × (0.95)18. Calculate C(20,2)=190, then evaluate to about 0.1887. So the probability is approximately 0.189 (3 s.f.).
范例作答：设 X 为次品数量，X ~ B(20, 0.05)。需求 P(X = 2)。使用公式：P(X = 2) = C(20, 2) × (0.05)2 × (0.95)18。计算组合数 190，得出约 0.1887。概率约为 0.189（三位有效数字）。
Often you need cumulative probabilities P(X ≤ a). The exam expects you to use your calculator’s binomial CD or statistical tables efficiently. Always note whether the question asks “less than”, “at most”, “more than” or “at least” and adjust accordingly.
常需计算累积概率 P(X ≤ a)。考试期望你高效使用计算器的二项积累功能或统计表。注意区分“小于”、“至多”、“大于”、“至少”并相应调整。
3. Poisson Distribution: Modelling Rare Events | 泊松分布：稀有事件建模
Example: Calls arrive at a switchboard at an average rate of 4 per minute. Find the probability of receiving exactly 3 calls in a randomly chosen minute.
例题：某总机接听电话的平均速率为每分钟 4 个。求随机一分钟内恰好接到 3 个电话的概率。
Model response: Let Y be the number of calls per minute, Y ~ Po(4). Then P(Y = 3) = (e−4 × 43) / 3! = 0.1954 (4 d.p.). State the distribution first to secure the method mark; then substitute correctly.
范例作答：设 Y 为每分钟电话数，Y ~ Po(4)。则 P(Y = 3) = (e−4 × 43) / 3! = 0.1954（四位小数）。先写分布以获得方法分，再正确代入公式。
For changed time intervals, scale λ proportionally. If the period is extended to 2 minutes, λ becomes 8. This is a common pitfall; many students forget to scale the parameter correctly.
若时间区间改变，需按比例缩放 λ。若扩展至 2 分钟，λ 变为 8。这是常见易错点，许多同学忘记正确调整参数。
4. Poisson Approximation to Binomial | 泊松近似二项分布
When n is large and p is small, the binomial B(n, p) can be approximated by Po(np). The standard condition is n > 50 and p < 0.1. In your response, you must state the approximation and justify it.
当 n 大且 p 小，二项分布 B(n, p) 可用泊松 Po(np) 近似。常用条件为 n > 50 且 p < 0.1。作答时必须写出近似并说明理由。
Example: A rare disease affects 0.2% of a population. A random sample of 1000 people is tested. Find the approximate probability that more than 3 test positive.
例题：某罕见病患病率为 0.2%。随机检测 1000 人，求超过 3 人呈阳性的近似概率。
Response: Let X ~ B(1000, 0.002). Here n is large, p small, np=2. So X ≈ Po(2). Then P(X > 3) = 1 − P(X ≤ 3). Using tables or calculator, P(X ≤ 3) = 0.8571, giving 0.1429. Thus the approximate probability is 0.143.
作答：设 X ~ B(1000, 0.002)。n 大、p 小，np=2，因此 X ≈ Po(2)。则 P(X > 3) = 1 − P(X ≤ 3)。查表或计算器得 P(X ≤ 3) = 0.8571，故近似概率为 0.143。
5. Continuous Random Variables: PDF and CDF | 连续型随机变量：概率密度函数与累积分布函数
Example: A continuous random variable X has probability density function f(x) = 0.2 − k x for 0 < x < 4, and 0 otherwise. Find k, the cumulative distribution function F(x), and P(1 < X < 2).
例题：连续随机变量 X 的概率密度函数为 f(x) = 0.2 − k x（0 < x < 4），其他区间为 0。求 k、累积分布函数 F(x) 及 P(1 < X < 2)。
Model answer: First use ∫04 f(x) dx = 1. That gives [0.2x − 0.5k x2]04 = 0.8 − 8k = 1, so 8k = −0.2, k = −0.025. Thus f(x) = 0.2 + 0.025x. For F(x): F(x) = ∫0x (0.2+0.025t) dt = 0.2x + 0.0125x2 (0 < x < 4). Then P(1
范例解答：利用 ∫04 f(x) dx = 1，得 [0.2x − 0.5k x2]04 = 0.8 − 8k = 1，所以 k = −0.025。于是 f(x) = 0.2 + 0.025x。求 F(x)：F(x) = ∫0x (0.2+0.025t) dt = 0.2x + 0.0125x2 (0 < x < 4)。最后 P(1
Always check that F(x) increases from 0 to 1 over the domain. The median and quartiles can be found by solving F(m) = 0.5. Showing integration steps clearly is essential for method marks.
务必验证 F(x) 在定义域内从 0 递增到 1。中位数和四分位数可通过解 F(m) = 0.5 求得。清晰地展示积分步骤是获得方法分的关键。
6. Normal Approximation to Binomial and Poisson | 正态近似二项与泊松分布
For large n, a binomial distribution can be approximated by a normal distribution if np > 5 and nq > 5. You must apply a continuity correction because a discrete distribution is being approximated by a continuous one. The same logic extends to Poisson when λ > 10.
对大样本，若 np > 5 且 nq > 5，二项分布可用正态分布近似。由于用连续分布近似离散分布，必须进行连续性校正。当 λ > 10 时，同样用正态近似泊松。
Example: X ~ B(200, 0.42). Use a normal approximation to find P(X ≤ 75).
例题：X ~ B(200, 0.42)。用正态近似求 P(X ≤ 75)。
Response: Mean μ = np = 84, variance σ2 = npq = 200×0.42×0.58 = 48.72. So σ = √48.72 ≈ 6.98. For continuity correction, P(X ≤ 75) becomes P(X < 75.5). Standardising: z = (75.5 − 84) / 6.98 = −1.218. Using normal tables, Φ(z) = 0.1115 (approx). So probability ≈ 0.1115.
作答：均值 μ = 84，方差 σ2 = 48.72，σ ≈ 6.98。进行连续性校正，P(X ≤ 75) 转化为 P(X < 75.5)。标准化：z = (75.5 − 84) / 6.98 = −1.218。查表得 Φ(z) ≈ 0.1115。概率约为 0.1115。
Forgetting the continuity correction is the most common error. If the question states “use a suitable approximation”, always check conditions and then state the correction explicitly.
忘记连续性校正是最常见的错误。若题目要求“使用合适近似”，务必检验条件，并明确写出校正步骤。
7. Hypothesis Testing: One-tailed Tests | 假设检验：单尾检验
Example: A manufacturer claims that at most 6% of their items are faulty. A random sample of 100 items finds 11 faulty. Test at the 5% significance level whether the claim is supported.
例题：某厂家声称次品率不超过 6%。随机抽取 100 件产品发现 11 件次品。在 5% 显著性水平下检验该声称是否成立。
Model answer: Let p be the population proportion of faulty items. H0: p = 0.06; H1: p > 0.06 (one-tailed test). Under H0, if X is number faulty, X ~ B(100, 0.06). Significance level α = 0.05. Find the smallest r such that P(X ≥ r | p=0.06) ≤ 0.05. Using cumulative tables, P(X ≤ 11) = 0.9696, so P(X ≥ 12) = 0.0304 < 0.05. P(X ≥ 11) = 0.0702 > 0.05. Critical region is X ≥ 12. Observed value is 11, which does not lie in the critical region. Therefore we do not reject H0. There is insufficient evidence to say the proportion exceeds 6%.
范例解答：设 p 为总体次品率。H0: p = 0.06；H1: p > 0.06（单尾）。在 H0 下，设 X 为次品数量，X ~ B(100, 0.06)。显著性水平 α = 0.05。寻找最小 r 使 P(X ≥ r | p=0.06) ≤ 0.05。查表得 P(X ≤ 11) = 0.9696，故 P(X ≥ 12) = 0.0304 < 0.05；而 P(X ≥ 11) = 0.0702 > 0.05。临界域为 X ≥ 12。实际观测值为 11，不在临界域内，因此不拒绝 H0。没有足够证据表明次品率超过 6%。
The conclusion must be stated in the context of the problem, not just “reject/do not reject H0”. Also show the critical region or p-value clearly.
结论必须置于问题背景中表述，而不能只写“拒绝/不拒绝 H0”。同时要清晰展示临界域或 p 值。
8. Hypothesis Testing: Two-tailed Tests | 假设检验：双尾检验
Example: A coin is tossed 50 times, obtaining 22 heads. Test at the 10% significance level whether the coin is fair.
例题：抛一枚硬币 50 次，得到 22 次正面。在 10% 显著性水平下检验硬币是否公平。
Response: Let p = probability of heads. H0: p = 0.5; H1: p ≠ 0.5 (two-tailed). Under H0, X ~ B(50, 0.5). For a two-tailed 10% test, each tail carries 5%. We need critical values such that P(X ≤ c1) ≤ 0.05 and P(X ≥ c2) ≤ 0.05. From binomial tables: P(X ≤ 18) = 0.0325, P(X ≤ 19) = 0.0595; so lower critical value is 18. By symmetry, upper critical value is 50 − 18 = 32. The observed 22 lies outside the critical region (18 to 32). Hence we do not reject H0. There is insufficient evidence to suggest the coin is biased.
作答：设 p 为正面概率。H0: p = 0.5；H1: p ≠ 0.5（双尾）。H0 下，X ~ B(50, 0.5)。10% 双尾检验每尾占 5%。找临界值使 P(X ≤ c1) ≤ 0.05 且 P(X ≥ c2) ≤ 0.05。查表：P(X ≤ 18) = 0.0325，P(X ≤ 19) = 0.0595，故下临界值为 18。由对称得上临界值为 32。观测值 22 不在临界域 (18 ~ 32) 内，因此不拒绝 H0。没有足够证据表明硬币存在偏差。
When using binomial symmetry, ensure that p = 0.5 before relying on symmetry. For other p, both tails must be calculated independently. Alternatively, you may calculate a p-value and compare with α/2.
利用二项分布对称性时，务必先确认 p = 0.5。对于其他 p 值，需独立计算两尾。也可计算 p 值并与 α/2 比较。
9. Common Mistakes and Examiner Tips | 常见错误与阅卷提示
Mistake 1: Not defining the random variable clearly. Always start with “Let X be … X ~ …”.
错误 1：未清晰定义随机变量。务必以“设 X 为…，X ~ …”开头。
Mistake 2: Misinterpreting “more than” and “at least”. “More than 5” means X > 5, i.e. P(X ≥ 6). Write the inequality explicitly.
错误 2：混淆“多于”与“至少”。“多于 5”表示 X > 5，即 P(X ≥ 6)。应明确写出不等式。
Mistake 3: Forgetting continuity correction in normal approximations. Mark schemes often penalise heavily.
错误 3：在正态近似中忘记连续性校正。评分方案常对此严厉扣分。
Mistake 4: Using approximate distribution without checking conditions. Write a brief justification: “n is large and p is small, so Poisson approximation is appropriate.”
错误 4：未检验条件直接使用近似分布。应简要说明理由：“n 大且 p 小，故泊松近似适用。”
Mistake 5: Conclusions detached from context. Always say “there is evidence at the 5% level that …” or “we cannot reject the claim that …”.
错误 5：结论脱离背景。总应表述为“在 5% 水平下有证据表明…”或“无法拒绝…的声称”。
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