Year 13 CIE Statistics: Unit Test Mock Paper Walkthrough | CIE A2 统计单元测试模拟卷解析

📚 Year 13 CIE Statistics: Unit Test Mock Paper Walkthrough | CIE A2 统计单元测试模拟卷解析

This walkthrough breaks down a complete S2 unit test mock paper for the CIE Year 13 Statistics syllabus. The paper covers Poisson distributions, linear combinations of normal variables, continuous random variables, confidence intervals, hypothesis tests for proportions, and tests for the difference of two means. Each question is solved step by step with paired English and Chinese explanations to deepen your understanding and exam technique. Use this as a revision tool to check your speed, accuracy, and clarity of statistical reasoning.

本解析完整讲解一份针对 CIE Year 13 统计（S2）课程设计的单元测试模拟卷。试卷涵盖泊松分布、正态变量的线性组合、连续随机变量、置信区间、比例假设检验以及两均值差检验。每道题逐步求解并配以中英对照的讲解，帮助你加深理解，提升应试技巧。请把这份解析当作复习工具，检验你的解题速度、准确性和统计思路的清晰度。

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1. Introduction & Paper Structure | 模拟卷结构与考点概览

This mock paper is designed to reflect the style of CIE A Level Statistics 2 (9709/62) and covers the main assessment objectives. The total mark is 50 and the suggested time is 75 minutes. Questions 1–3 test probability and distribution modelling; Questions 4–6 focus on statistical inference – estimation and hypothesis testing. All required tables (normal, Poisson) are assumed to be supplied, and you should use critical values from the standard normal distribution in hypothesis tests unless stated otherwise.

本模拟卷对标 CIE A Level 统计2（9709/62）的命题风格，覆盖主要评估目标。卷面总分50分，建议答题时间75分钟。第1–3题考查概率与分布建模；第4–6题侧重统计推断——估计和假设检验。假定考试中会提供必要的正态分布表和泊松分布表；除非另有说明，在假设检验中应使用标准正态分布的临界值。

Each question is accompanied by a detailed walkthrough. We define variables, state assumptions, show key formulae with centred equations, and discuss common pitfalls. The bilingual layout follows the principle: every English explanation point is immediately followed by its Chinese counterpart, ensuring both language learners and dual-language students can follow without switching between paragraphs.

每道题都配有详细讲解。我们会定义变量、陈述假设条件、以居中公式展示关键等式并讨论常见易错点。中英双语编排遵循“每个英文解释点后立即接中文对应段落”的原则，确保英语学习者与双语学生都能顺畅跟上，无需在段落间反复切换。

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2. Q1: Poisson Distribution & Approximations | 题目1：泊松分布与近似计算

Question recap: Number of cars per 5 min ~ Po(3.5). (a) P(X ≤ 2). (b) For a 10-minute interval, mean becomes 7, find P(X = 5). (c) Mean rises to 6 per 5 min; use a suitable approximation for P(X > 10).

题目回顾：每5分钟汽车数服从 Po(3.5)。(a) 求 P(X ≤ 2)。(b) 10分钟内均值为7，求 P(X = 5)。(c) 均值升至6每5分钟，使用适当近似求 P(X > 10)。

(a) X ~ Po(3.5). P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2). Apply the Poisson formula.

(a) X ~ Po(3.5)。P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)。使用泊松公式。

P(X = k) = e⁻³·⁵ (3.5ᵏ) / k!

P(X=0) = e⁻³·⁵ ≈ 0.0302, P(X=1) = 3.5 × e⁻³·⁵ ≈ 0.1057, P(X=2) = (3.5² / 2) × e⁻³·⁵ ≈ 0.1850. Sum = 0.3209. The probability that at most 2 cars arrive in a 5‑minute interval is 0.321 (3 s.f.).

P(X=0) = e⁻³·⁵ ≈ 0.0302, P(X=1) = 3.5 × e⁻³·⁵ ≈ 0.1057, P(X=2) = (3.5² / 2) × e⁻³·⁵ ≈ 0.1850。总和 = 0.3209。即5分钟内最多到达2辆车的概率为0.321（保留三位有效数字）。

(b) For a 10‑minute interval the mean λ = 2 × 3.5 = 7. Let Y ~ Po(7). P(Y = 5) = e⁻⁷ × 7⁵ / 5! = e⁻⁷ × 16807 / 120 ≈ 0.1277. So the probability of exactly 5 cars is 0.128 (3 s.f.).

(b) 10分钟区间的均值 λ = 2 × 3.5 = 7。设 Y ~ Po(7)。P(Y = 5) = e⁻⁷ × 7⁵ / 5! = e⁻⁷ × 16807 / 120 ≈ 0.1277。因此恰好到达5辆车的概率为0.128（3 s.f.）。

(c) Now mean λ = 6 for 5 minutes; we can use a normal approximation because λ is reasonably large. X ~ Po(6) ≈ N(μ = 6, σ² = 6). Using continuity correction, P(X > 10) = P(X ≥ 11) ≈ P(X > 10.5). Standardise: z = (10.5 – 6) / √6 ≈ 4.5 / 2.449 = 1.837. From tables, P(Z > 1.837) = 1 – 0.9669 = 0.0331. Therefore the approximate probability is 0.0331.

(c) 此时5分钟均值 λ = 6；因 λ 足够大，可用正态近似。X ~ Po(6) ≈ N(μ = 6, σ² = 6)。使用连续性校正，P(X > 10) = P(X ≥ 11) ≈ P(X > 10.5)。标准化：z = (10.5 – 6) / √6 ≈ 4.5 / 2.449 = 1.837。查表得 P(Z > 1.837) = 1 – 0.9669 = 0.0331。因此近似概率为0.0331。

Common mistake: Forgetting continuity correction when approximating a discrete distribution by a continuous one. Always adjust by 0.5 before standardising.

常见错误：用连续分布近似离散分布时忘记连续性校正。务必在标准化前用0.5修正。

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3. Q2: Linear Combinations of Normal Variables | 题目2：正态变量的线性组合

Question recap: Apple masses: variety A ~ N(150, 10²), B ~ N(180, 12²). (a) Total of 5 A apples, P(total > 780). (b) P(B mass – A mass ≥ 20). (c) Bag with 3 A and 2 B; P(total < 850).

题目回顾：苹果质量：品种A ~ N(150, 10²)，品种B ~ N(180, 12²)。(a) 5个A的总重，P(总重 > 780)。(b) P(B质量 – A质量 ≥ 20)。(c) 袋中有3个A和2个B；P(总重 < 850)。

(a) Let T = A₁ + A₂ + … + A₅. Since apples are independent, T ~ N(5×150, 5×10²) = N(750, 500). Standardise: z = (780 – 750) / √500 = 30 / 22.3607 ≈ 1.342. P(T > 780) = P(Z > 1.342) = 1 – 0.9100 = 0.0900.

(a) 设 T = A₁ + A₂ + … + A₅。各苹果独立，故 T ~ N(5×150, 5×10²) = N(750, 500)。标准化：z = (780 – 750) / √500 = 30 / 22.3607 ≈ 1.342。P(T > 780) = P(Z > 1.342) = 1 – 0.9100 = 0.0900。

(b) Define D = B – A. Both normal and independent, D ~ N(180 – 150, 12² + 10²) = N(30, 244). For P(D ≥ 20): z = (20 – 30) / √244 = –10 / 15.6205 ≈ –0.640. P(Z > –0.640) = 1 – P(Z < –0.640) = 1 – 0.2611 = 0.7389. Thus the probability that a B apple is at least 20 g heavier is 0.739.

(b) 定义 D = B – A。两者独立且正态，D ~ N(180 – 150, 12² + 10²) = N(30, 244)。求 P(D ≥ 20)：z = (20 – 30) / √244 = –10 / 15.6205 ≈ –0.640。P(Z > –0.640) = 1 – 0.2611 = 0.7389。即品种B苹果比品种A至少重20g的概率为0.739。

(c) Total mass S = A₁+A₂+A₃ + B₁+B₂. Mean = 3×150 + 2×180 = 450 + 360 = 810. Variance = 3×10² + 2×12² = 300 + 288 = 588. S ~ N(810, 588). P(S < 850): z = (850 – 810) / √588 = 40 / 24.2487 ≈ 1.649. P(Z < 1.649) = 0.9504. The bag weighs less than 850 g with probability 0.950.

(c) 总重 S = A₁+A₂+A₃ + B₁+B₂。均值 = 3×150 + 2×180 = 810。方差 = 3×10² + 2×12² = 588。S ~ N(810, 588)。P(S < 850)：z = (850 – 810) / √588 = 40 / 24.2487 ≈ 1.649。P(Z < 1.649) = 0.9504。这袋苹果总重小于850g的概率为0.950。

Key insight: When combining normal variables, variances add regardless of signs. Always check independence – a condition clearly given in the question.

关键点：组合正态变量时，方差无论在加减运算中均相加。务必检查独立性——题目中明确给出相关条件。

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4. Q3: Continuous Random Variable (PDF, Mean, Median) | 题目3：连续随机变量（密度函数、均值、中位数）

Question recap: f(x) = kx(4 – x) for 0 ≤ x ≤ 4, 0 otherwise. (a) Show k = 3/32. (b) Find E(X) and Var(X). (c) Find the median.

题目回顾：f(x) = kx(4 – x)，0 ≤ x ≤ 4，其余为0。(a) 证明 k = 3/32。(b) 求 E(X) 和 Var(X)。(c) 求中位数。

(a) Total area must equal 1: ∫₀⁴ kx(4–x) dx = 1. Integrate: ∫₀⁴ (4kx – kx²) dx = [2kx² – (k/3)x³]₀⁴ = 2k(16) – (k/3)(64) = 32k – 64k/3 = (96k – 64k)/3 = 32k/3. Set 32k/3 = 1 ⇒ k = 3/32. Shown.

(a) 总面积为1：∫₀⁴ kx(4–x) dx = 1。积分：∫₀⁴ (4kx – kx²) dx = [2kx² – (k/3)x³]₀⁴ = 2k(16) – (k/3)(64) = 32k – 64k/3 = (96k – 64k)/3 = 32k/3。令其等于1得 k = 3/32。证毕。

(b) E(X) = ∫ x f(x) dx = (3/32) ∫₀⁴ x²(4–x) dx = (3/32) ∫₀⁴ (4x² – x³) dx = (3/32) [ (4/3)x³ – (1/4)x⁴ ]₀⁴ = (3/32) [ (4/3)×64 – (1/4)×256 ] = (3/32) [ 85.333… – 64 ] = (3/32) × 21.333… = (3/32) × (64/3) = 2. So E(X) = 2.

(b) E(X) = ∫ x f(x) dx = (3/32) ∫₀⁴ x²(4–x) dx = (3/32) ∫₀⁴ (4x² – x³) dx = (3/32) [ (4/3)x³ – (1/4)x⁴ ]₀⁴。代入得 (3/32) [ (4/3)×64 – (1/4)×256 ] = (3/32) [ 85.333… – 64 ] = (3/32) × 21.333… = (3/32) × (64/3) = 2。故 E(X) = 2。

E(X²) = ∫ x² f(x) dx = (3/32) ∫₀⁴ x³(4–x) dx = (3/32) ∫₀⁴ (4x³ – x⁴) dx = (3/32) [ x⁴ – (1/5)x⁵ ]₀⁴ = (3/32) [ 256 – (1024/5) ] = (3/32) [ (1280 – 1024)/5 ] = (3/32) × (256/5) = 768 / 160 = 4.8. Var(X) = E(X²) – [E(X)]² = 4.8 – 4 = 0.8.

E(X²) = ∫ x² f(x) dx = (3/32) ∫₀⁴ x³(4–x) dx = (3/32) ∫₀⁴ (4x³ – x⁴) dx = (3/32) [ x⁴ – (1/5)x⁵ ]₀⁴ = (3/32) [ 256 – (1024/5) ] = (3/32) × (256/5) = 4.8。Var(X) = 4.8 – 2² = 0.8。

(c) Median m satisfies ∫₀ᵐ f(x) dx = 0.5. ∫₀ᵐ (3/32)(4x – x²) dx = (3/32)[2x² – x³/3]₀ᵐ = (3/32)(2m² – m³/3) = 0.5. Multiply by 32: 3(2m² – m³/3) = 16 → 6m² – m³ = 16 → m³ – 6m² + 16 = 0. Try m = 2: 8 – 24 + 16 = 0, so m = 2 is a root. Factor: (m – 2)(m² – 4m – 8) = 0. The quadratic gives m = 2 ± 2√3 ≈ –1.46 or 5.46. Only m in [0,4] is 2. Hence median = 2.

(c) 中位数 m 满足 ∫₀ᵐ f(x) dx = 0.5。积分得 (3/32)[2x² – x³/3]₀ᵐ = (3/32)(2m² – m³/3) = 0.5。乘以32：3(2m² – m³/3) = 16 → 6m² – m³ = 16 → m³ – 6m² + 16 = 0。试 m=2：8–24+16=0，故 m=2 是一个根。分解因式：(m–2)(m²–4m–8)=0，二次部分给出 m=2±2√3≈-1.46或5.46，仅 m=2 在 [0,4] 内。因此中位数为2。

Note: Because the pdf is symmetric about x=2 (quadratic in x with roots 0 and 4), both mean and median equal 2, which serves as a quick check.

注意：此概率密度函数关于 x=2 对称（x 的二次函数，根为0和4），因此均值和中位数均等于2，可用作快速验证。

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5. Q4: Confidence Interval for the Mean | 题目4：均值的置信区间

Question recap: Heights ~ N(μ, 6.5²). n=64, sample mean = 176.2 cm. (a) 95% CI for μ. (b) Minimum n for CI width < 4 cm. (c) Assumption.

题目回顾：身高 ~ N(μ, 6.5²)。n=64，样本均值 176.2 cm。(a) 求 μ 的95%置信区间。(b) 使区间宽度小于4 cm所需的样本量。(c) 所需假设。

(a) Since population standard deviation is known and sample size is large, use z‑interval. 95% confidence ⇒ z = 1.96. The formula is x̄ ± z × σ/√n. 176.2 ± 1.96 × 6.5 / √64 = 176.2 ± 1.96 × 6.5 / 8 = 176.2 ± 1.96 × 0.8125 = 176.2 ± 1.5925. So the CI is (174.6 cm, 177.8 cm) to 3 s.f.

(a) 总体标准差已知且样本量足够大，使用 z 区间。95% 置信度 ⇒ z = 1.96。公式为 x̄ ± z × σ/√n。176.2 ± 1.96 × 6.5/8 = 176.2 ± 1.5925。置信区间为 (174.6 cm, 177.8 cm)（保留三位有效数字）。

(b) Width = 2 × z × σ/√n. We require 2 × 1.96 × 6.5 / √n < 4 → 25.48 / √n < 4 → √n > 6.37 → n > 40.5769. So the smallest integer sample size is 41.

(b) 区间宽度 = 2 × z × σ/√n。要求 2 × 1.96 × 6.5 / √n < 4 → 25.48 / √n < 4 → √n > 6.37 → n > 40.5769。故最小整数样本量为41。

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