📚 A-Level CCEA Statistics: Unit Test Mock Exam Analysis | A-Level CCEA 统计：单元测试模拟卷解析
Preparing for the CCEA A-Level Statistics unit test can be challenging without regular exposure to exam-style questions. This article presents a mock exam analysis structured around typical unit test topics, offering step-by-step solutions, common pitfalls, and revision strategies. By working through these model answers, students can strengthen their understanding and boost their confidence for the actual assessment.
如果没有定期接触考试风格的题目，备考 CCEA A-Level 统计单元测试可能会很有挑战性。本文围绕典型的单元测试主题，提供一份模拟卷解析，包括分步解答、常见错误和复习策略。通过演练这些范例答案，学生可以加深理解并增强应对真实考试的信心。
1. Overview of the CCEA Statistics Unit Test | CCEA 统计单元测试概览
The CCEA Statistics unit test typically covers data presentation, probability, discrete and continuous distributions, hypothesis testing, correlation and regression, and possibly chi-squared tests. The mock exam in this analysis contains two sections: Section A with short-answer questions worth 30 marks, and Section B with three longer structured questions worth 30 marks. Time allowed is 1 hour 30 minutes. You will need a calculator and access to statistical tables.
CCEA 统计单元测试通常涵盖数据呈现、概率、离散和连续分布、假设检验、相关与回归，可能还包括卡方检验。本分析中的模拟卷包含两部分：A 部分为简答题，共 30 分；B 部分为三道结构化的长题，共 30 分。考试时间为 1 小时 30 分钟。你需要计算器及统计表格。
2. Data Handling and Summary Statistics | 数据处理与汇总统计
Example Question: The following data show the number of hours 10 students spent revising: 5, 7, 8, 6, 10, 12, 9, 11, 7, 5. Find the mean, median, interquartile range, and sample standard deviation.
例题：以下数据显示了 10 名学生用于复习的小时数：5, 7, 8, 6, 10, 12, 9, 11, 7, 5。求均值、中位数、四分位距和样本标准差。
To calculate the mean, sum the values: 5+7+8+6+10+12+9+11+7+5 = 80, so mean = 80/10 = 8 hours. The ordered data is 5,5,6,7,7,8,9,10,11,12. Median is the average of the 5th and 6th values: (7+8)/2 = 7.5. Q₁ is the median of the lower half (5,5,6,7,7) → 6, Q₃ is the median of the upper half (8,9,10,11,12) → 10. Thus IQR = Q₃ – Q₁ = 10 – 6 = 4. For sample standard deviation, use s = √[Σ(x – x̄)²/(n–1)]. Compute squared deviations: (5–8)²=9, (7–8)²=1, (8–8)²=0, (6–8)²=4, (10–8)²=4, (12–8)²=16, (9–8)²=1, (11–8)²=9, (7–8)²=1, (5–8)²=9. Sum = 54. s = √(54/9) = √6 ≈ 2.449 hours.
计算均值：总和为5+7+8+6+10+12+9+11+7+5=80，均值=80/10=8小时。排序后为5,5,6,7,7,8,9,10,11,12。中位数为第5和第6个数值的平均：(7+8)/2=7.5。下四分位数Q₁为低半部(5,5,6,7,7)的中位数=6，上四分位数Q₃为高半部(8,9,10,11,12)的中位数=10，因此IQR=10–6=4。样本标准差使用 s = √[Σ(x – x̄)²/(n–1)]。计算离差平方和：54，s = √(54/9) = √6 ≈ 2.449小时。
A common mistake is to use n instead of n–1 when calculating sample standard deviation. The CCEA specification requires the sample standard deviation formula unless stated otherwise. Also, ensure you do not confuse the median with the mean, especially when data contain outliers.
一个常见错误是在计算样本标准差时使用 n 而不是 n–1。除非另有说明，CCEA 大纲要求使用样本标准差公式。同时，要确保不混淆中位数和均值，尤其当数据存在异常值时。
3. Probability and Venn Diagrams | 概率与维恩图
Example: For events A and B, P(A)=0.4, P(B)=0.3 and P(A∩B)=0.1. Find P(A∪B), P(A|B), and determine whether A and B are independent.
例题：对于事件A和B，已知P(A)=0.4，P(B)=0.3，P(A∩B)=0.1。求P(A∪B)、P(A|B)，并判断A与B是否独立。
P(A∪B) = P(A)+P(B)–P(A∩B) = 0.4+0.3–0.1 = 0.6. This can also be checked using a Venn diagram. P(A|B) = P(A∩B)/P(B) = 0.1/0.3 = 1/3. Since P(A|B) = 1/3 ≈ 0.333 ≠ 0.4 = P(A), the events are not independent. Alternatively, check P(A∩B) ≠ P(A)×P(B), because 0.1 ≠ 0.12.
P(A∪B) = P(A)+P(B)–P(A∩B) = 0.4+0.3–0.1 = 0.6。可以通过维恩图交叉验证。P(A|B) = P(A∩B)/P(B) = 0.1/0.3 = 1/3。由于P(A|B)=1/3≈0.333 ≠ P(A)=0.4，因此事件不独立。另一种判断方法：P(A∩B)=0.1 ≠ P(A)P(B)=0.12，所以不独立。
Pay close attention to conditional probability wording, e.g., ‘given that’ indicates a reduced sample space. Always check whether probabilities sum to 1 across a partition.
注意条件概率的表述，例如“已知…”表示缩减的样本空间。务必检查各分支概率之和是否等于1。
4. Discrete Probability Distributions | 离散概率分布
Example: A fair tetrahedral die with faces labelled 1, 2, 3 and 4 is rolled twice. The random variable X is the larger of the two scores. Tabulate the probability distribution of X and find E(X) and Var(X).
例题：一个公平的四面体骰子，面标记1,2,3,4，投掷两次。随机变量X为两次得分中较大的一个。列出X的概率分布表，并求E(X)和Var(X)。
There are 4×4=16 equally likely outcomes. Count the pairs: (1,1) → X=1; (1,2),(2,1),(2,2) → X=2; (1,3),(2,3),(3,1),(3,2),(3,3) → X=3; all other outcomes give X=4. Counting: f(1)=1, f(2)=3, f(3)=5, f(4)=7. So P(X=x): 1/16, 3/16, 5/16, 7/16. Then E(X)=Σx·P(X=x)=1×(1/16)+2×(3/16)+3×(5/16)+4×(7/16) = (1+6+15+28)/16 = 50/16 = 3.125. E(X²) = 1²×(1/16)+4×(3/16)+9×(5/16)+16×(7/16) = (1+12+45+112)/16 = 170/16 = 10.625. Var(X)=E(X²)–[E(X)]² = 10.625 – 9.765625 = 0.859375.
共有 4×4=16 种等可能结果。统计出现次数：(1,1)→X=1；(1,2),(2,1),(2,2)→X=2；(1,3),(2,3),(3,1),(3,2),(3,3)→X=3；其余结果X=4。计数得：f(1)=1, f(2)=3, f(3)=5, f(4)=7。因此P(X=x)：1/16, 3/16, 5/16, 7/16。E(X)=Σx·P(X=x)=1×(1/16)+2×(3/16)+3×(5/16)+4×(7/16)=50/16=3.125。E(X²)=1²×(1/16)+4×(3/16)+9×(5/16)+16×(7/16)=170/16=10.625。Var(X)=E(X²)–[E(X)]²=10.625–9.765625=0.859375。
E(X) = Σx·P(X=x), Var(X) = E(X²) − [E(X)]²
5. Binomial and Poisson Distributions | 二项分布与泊松分布
Example: A factory produces components and 5% are defective. A random sample of 20 components is inspected. Find the probability that exactly 2 are defective, and the probability that at most 2 are defective. Use a Poisson approximation and comment on its accuracy.
例题：一家工厂生产的元件有5%存在缺陷。随机抽取20个元件进行检查。求恰好有2个缺陷品的概率，以及至多2个缺陷品的概率。使用泊松近似并评价其准确性。
Let X ~ B(20, 0.05). Using the binomial formula: P(X=k) = ²⁰Cₖ (0.05)ᵏ (0.95)²⁰⁻ᵏ. P(X=2) = ²⁰C₂ × 0.05² × 0.95¹⁸ = 190 × 0.0025 × 0.397 (approx.) ≈ 0.1887. P(X=0) = 1 × 1 × 0.3585 = 0.3585, P(X=1) = 20 × 0.05 × 0.3774 = 0.3774. Therefore P(X ≤ 2) = 0.3585+0.3774+0.1887 = 0.9246. For Poisson approximation, λ = np = 1. P(Y=k) = e⁻¹(1)ᵏ/k!. P(Y=0)=0.3679, P(Y=1)=0.3679, P(Y=2)=0.1839, so P(Y≤2)=0.9197. The approximation is reasonable because n is large and p is small, but exact binomial gives slightly higher probability.
设 X ~ B(20, 0.05)。二项公式：P(X=k) = ²⁰Cₖ (0.05)ᵏ (0.95)²⁰⁻ᵏ。P(X=2)=²⁰C₂×0.05²×0.95¹⁸=190×0.0025×0.397≈0.1887。P(X=0)=0.3585，P(X=1)=0.3774，所以P(X≤2)=0.3585+0.3774+0.1887=0.9246。泊松近似：λ=np=1，P(Y=k)=e⁻¹×1ᵏ/k!。P(Y=0)=0.3679，P(Y=1)=0.3679，P(Y=2)=0.1839，故P(Y≤2)=0.9197。由于n较大、p较小，近似效果尚可，但精确二项概率略高。
P(X = k) = ⁿCₖ pᵏ (1−p)ⁿ⁻ᵏ, Poisson: P(Y = k) = (e⁻λ λᵏ)/k!
6. Normal Distribution and Standardisation | 正态分布与标准化
Example: The mass of a bag of flour is normally distributed with mean 500 g and standard deviation 15 g. Find the proportion of bags weighing less than 485 g. Also, determine the weight above which the heaviest 5% of bags lie.
例题：一袋面粉的质量服从正态分布，均值为500克，标准差为15克。求质量低于485克的袋子比例。同时，确定最重的5%袋子对应的质量下限。
First, standardise: z = (485 − 500)/15 = −1.00. From tables, Φ(−1.00) = 0.1587. Thus, about 15.87% of bags weigh less than 485 g. For the top 5%, we need the 95th percentile. Using the z-table, z = 1.6449 (or 1.645). Set (x − 500)/15 = 1.645, so x = 500 + 1.645×15 = 524.675 g. Bags heavier than 524.7 g are in the heaviest 5%.
首先标准化：z = (485−500)/15 = −1.00。查表得 Φ(−1.00)=0.1587，因此约15.87%的袋子低于485克。对于最重的5%，需要第95百分位数。查z值表，z=1.6449（或1.645）。令(x−500)/15=1.645，得x=500+1.645×15=524.675克。质量高于524.7克的袋子属于最重的5%。
z = (x − μ)/σ, then P(X < x) = Φ(z)
Always sketch the bell curve to confirm the tail area you are working with. Remember that the total area is 1, and symmetry helps in reverse lookups.
务必画出正态曲线草图，确认所处理的是哪一部分尾部面积。牢记总面积等于1，利用对称性有助于反向查表。
7. Sampling and Confidence Intervals | 抽样与置信区间
Example: The lifetime of a light bulb has known standard deviation σ = 40 hours. A random sample of 50 bulbs gives a mean lifetime of 800 hours. Construct a 95% confidence interval for the population mean. What if σ is unknown and the sample standard deviation is 42 hours instead?
例题：已知灯泡寿命总体标准差σ=40小时。随机抽取50个灯泡，样本平均寿命为800小时。构建总体均值的95%置信区间。如果σ未知且样本标准差为42小时，结果会如何？
When σ is known, the 95% CI is: x̄ ± z₀.₀₂₅ × (σ/√n). z₀.₀₂₅ = 1.96. Standard error = 40/√50 ≈
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