📚 Year 11 CAIE Statistics Mock Unit Test Walkthrough | Year 11 CAIE 统计单元测试模拟卷解析
This walkthrough is designed to help Year 11 students prepare for a CAIE Statistics unit test. The mock paper below covers the core topics typically assessed in the first half of the IGCSE or AS Statistics course: data representation, measures of central tendency and spread, basic probability, discrete random variables, and the binomial distribution. Each question is followed by a full solution with examiner-style commentary. Use this resource to identify common pitfalls, refine your calculator skills, and build confidence in writing clear, step-by-step reasoning.
本解析旨在帮助 Year 11 学生备考 CAIE 统计单元测试。下面的模拟卷涵盖了通常在上半学期评估的核心主题：数据表示、集中趋势与离差度量、基础概率、离散随机变量以及二项分布。每道题后附有完整解答和考官风格的评注。请利用这一资源发现常见易错点、提升计算器使用技巧，并建立清晰、分步骤推理的应试信心。
1. Mock Paper Structure | 模拟卷结构
This mock test consists of five compulsory questions totalling 50 marks, to be completed in 60 minutes under exam conditions. A scientific calculator is required. The questions progress from descriptive statistics through probability theory to discrete distributions, mirroring the gradient of difficulty in a typical CAIE paper.
本模拟卷包含五道必答题，满分 50 分，需在 60 分钟考试条件下完成。需要使用科学计算器。题目从描述性统计逐步过渡到概率论和离散分布，反映了典型 CAIE 试卷的难度梯度。
| Question | Topic | Marks |
|---|---|---|
| 1 | Stem-and-leaf diagrams, median, quartiles, outliers | 10 |
| 2 | Histogram, frequency density, mean estimation | 10 |
| 3 | Venn diagrams, conditional probability, independence | 10 |
| 4 | Discrete random variables, E(X), Var(X), linear coding | 10 |
| 5 | Binomial distribution, hypothesis testing concept (prep) | 10 |
2. Question 1: Stem-and-Leaf and Outlier Detection | 茎叶图与异常值检测
The following data represent the reaction times (in milliseconds) of 20 participants in a psychology experiment: 245, 268, 272, 275, 278, 281, 283, 285, 288, 291, 293, 295, 298, 302, 305, 310, 318, 325, 340, 390. (a) Draw an ordered stem-and-leaf diagram. (3 marks) (b) Find the median, lower quartile Q₁, and upper quartile Q₃. (3 marks) (c) Determine whether 390 is an outlier, using the IQR rule (1.5 × IQR). (4 marks)
以下数据代表 20 名参与者在心理学实验中的反应时间（单位毫秒）：245, 268, 272, 275, 278, 281, 283, 285, 288, 291, 293, 295, 298, 302, 305, 310, 318, 325, 340, 390。(a) 绘制排序后的茎叶图。(3 分) (b) 求中位数、下四分位数 Q₁ 和上四分位数 Q₃。(3 分) (c) 使用 IQR 规则（1.5 × IQR）判断 390 是否为异常值。(4 分)
(a) Solution: Use the hundreds and tens digits as the stem (24, 25, 26, …, 39) and the units digit as the leaf. Since all three-digit numbers share the same stem logic, we group by 24|5 means 245. The ordered diagram must have leaves sorted ascending within each stem. The correct stem-and-leaf is: 24 | 5 ; 26 | 8 ; 27 | 2 5 8 ; 28 | 1 3 5 8 ; 29 | 1 3 5 8 ; 30 | 2 5 ; 31 | 0 8 ; 32 | 5 ; 34 | 0 ; 39 | 0. Include a key: 24 | 5 = 245 ms.
(a) 解答：使用百位和十位数字作为茎（24, 25, 26, …, 39），个位数字作为叶。由于所有三位数共享相同的茎逻辑，我们按 24|5 表示 245 的方式分组。排序图必须每根茎内的叶按升序排列。正确的茎叶图是：24 | 5 ; 26 | 8 ; 27 | 2 5 8 ; 28 | 1 3 5 8 ; 29 | 1 3 5 8 ; 30 | 2 5 ; 31 | 0 8 ; 32 | 5 ; 34 | 0 ; 39 | 0。附上图例：24 | 5 = 245 ms。
(b) Solution: n = 20 (even). Median = average of 10th and 11th values = (291 + 293) / 2 = 292. Q₁ is the median of the lower half (first 10 values): the 5th and 6th values are 278 and 281, so Q₁ = (278 + 281) / 2 = 279.5. Q₃ is the median of the upper half: the 15th and 16th values are 305 and 310, so Q₃ = (305 + 310) / 2 = 307.5.
(b) 解答：n = 20（偶数）。中位数 = 第 10 和第 11 个值的平均值 = (291 + 293) / 2 = 292。Q₁ 是下半部分（前 10 个值）的中位数：第 5 和第 6 个值为 278 和 281，所以 Q₁ = (278 + 281) / 2 = 279.5。Q₃ 是上半部分的中位数：第 15 和第 16 个值为 305 和 310，所以 Q₃ = (305 + 310) / 2 = 307.5。
(c) Solution: IQR = Q₃ − Q₁ = 307.5 − 279.5 = 28. Lower fence = Q₁ − 1.5 × IQR = 279.5 − 42 = 237.5. Upper fence = Q₃ + 1.5 × IQR = 307.5 + 42 = 349.5. Since 390 > 349.5, the value 390 is an outlier. Common mistake: forgetting to multiply 1.5 by IQR before adding to Q₃, or misidentifying quartiles for an even dataset.
(c) 解答：IQR = Q₃ − Q₁ = 307.5 − 279.5 = 28。下限 = Q₁ − 1.5 × IQR = 279.5 − 42 = 237.5。上限 = Q₃ + 1.5 × IQR = 307.5 + 42 = 349.5。因为 390 > 349.5，所以数值 390 是一个异常值。常见错误：忘记在加上 Q₃ 之前将 1.5 与 IQR 相乘，或对于偶数数据集错误识别四分位数。
3. Question 2: Histogram and Frequency Density | 直方图与频率密度
The table summarises the distances (km) travelled by 60 delivery vans in one day. (a) Complete the table by calculating frequency densities. (3 marks) (b) Draw a histogram to represent the data. (4 marks) (c) Estimate the mean distance travelled. (3 marks)
下表汇总了 60 辆送货车一天的行驶距离（公里）。(a) 通过计算频率密度完成表格。(3 分) (b) 画直方图表示数据。(4 分) (c) 估算平均行驶距离。(3 分)
| Distance (km) | Frequency | Class width | Frequency density |
|---|---|---|---|
| 0 ≤ d < 20 | 8 | 20 | ? |
| 20 ≤ d < 40 | 14 | 20 | ? |
| 40 ≤ d < 60 | 18 | 20 | ? |
| 60 ≤ d < 100 | 16 | 40 | ? |
| 100 ≤ d < 140 | 4 | 40 | ? |
(a) Solution: Frequency density = Frequency ÷ Class width. Row 1: 8 ÷ 20 = 0.4. Row 2: 14 ÷ 20 = 0.7. Row 3: 18 ÷ 20 = 0.9. Row 4: 16 ÷ 40 = 0.4. Row 5: 4 ÷ 40 = 0.1. Always check that total area (sum of frequency × class width checks out: 8+14+18+16+4=60).
(a) 解答：频率密度 = 频率 ÷ 组距。第 1 行：8 ÷ 20 = 0.4。第 2 行：14 ÷ 20 = 0.7。第 3 行：18 ÷ 20 = 0.9。第 4 行：16 ÷ 40 = 0.4。第 5 行：4 ÷ 40 = 0.1。始终检查总面积（频率总和 × 组距验证：8+14+18+16+4=60）。
(b) Histogram: The horizontal axis must use a continuous linear scale. Draw bars with no gaps: 0–20 (height 0.4), 20–40 (height 0.7), 40–60 (height 0.9), 60–100 (height 0.4), 100–140 (height 0.1). The bar for 60–100 is wider but lower in frequency density because the class width is doubled. Label axes: Frequency density (vertical) and Distance (km, horizontal).
(b) 直方图：横轴必须使用连续的线性刻度。绘制无间隙的条形：0–20（高度 0.4）、20–40（高度 0.7）、40–60（高度 0.9）、60–100（高度 0.4）、100–140（高度 0.1）。60–100 的条形更宽但频率密度更低，因为组距翻倍。标注坐标轴：纵轴为频率密度，横轴为距离 (km)。
(c) Mean estimation: Use midpoints: 10, 30, 50, 80, 120. Σfx = (10×8) + (30×14) + (50×18) + (80×16) + (120×4) = 80 + 420 + 900 + 1280 + 480 = 3160. Estimated mean = Σfx / Σf = 3160 ÷ 60 = 52.67 km (or 52.7 km to 3 s.f.).
(c) 均值估算：使用组中值：10、30、50、80、120。Σfx = (10×8) + (30×14) + (50×18) + (80×16) + (120×4) = 80 + 420 + 900 + 1280 + 480 = 3160。估算平均值 = Σfx / Σf = 3160 ÷ 60 = 52.67 km（或 52.7 km 保留三位有效数字）。
4. Question 3: Venn Diagrams and Conditional Probability | 维恩图与条件概率
In a sixth-form college of 120 students, 65 study Mathematics (M), 48 study Physics (P), and 22 study both. (a) Draw a Venn diagram showing these frequencies. (3 marks) (b) Find the probability that a randomly chosen student studies neither subject. (2 marks) (c) Given that a student studies Mathematics, find the probability they also study Physics. (3 marks) (d) Determine, with a reason, whether events M and P are independent. (2 marks)
在一所有 120 名学生的高中，65 人学习数学 (M)，48 人学习物理 (P)，22 人同时学习两门。(a) 画维恩图显示这些频数。(3 分) (b) 求随机选择一名学生既不学数学也不学物理的概率。(2 分) (c) 已知一名学生学习数学，求他也学习物理的概率。(3 分) (d) 判断事件 M 和 P 是否独立，并说明理由。(2 分)
(a) Solution: Intersection = 22. Mathematics only = 65 − 22 = 43. Physics only = 48 − 22 = 26. Neither = 120 − (43 + 22 + 26) = 120 − 91 = 29. Venn diagram: two overlapping circles inside rectangle, labelled M and P, with 43 in M-only, 22 in overlap, 26 in P-only, and 29 outside.
(a) 解答：交集 = 22。仅数学 = 65 − 22 = 43。仅物理 = 48 − 22 = 26。两门都不学 = 120 − (43 + 22 + 26) = 120 − 91 = 29。维恩图：矩形内有两个重叠的圆，标注 M 和 P，分别在 M 独有区域填 43、重叠区域填 22、P 独有区域填 26，外部填 29。
(b) P(neither): 29 / 120 = 0.2417 (or 29/120 simplified).
(b) 两门都不学的概率：29 / 120 = 0.2417（或 29/120 约分形式）。
(c) P(P|M): P(P|M) = P(P ∩ M) / P(M) = (22/120) / (65/120) = 22/65 = 0.3385 (about 0.338). The given-condition restricts the sample space to the 65 Mathematics students. Of those, 22 also do Physics.
(c) P(P|M)：P(P|M) = P(P ∩ M) / P(M) = (22/120) / (65/120) = 22/65 = 0.3385（约 0.338）。给定条件将样本空间限制在 65 名数学学生中。其中 22 人也学习物理。
(d) Independence check: For independence, P(P|M) must equal P(P). P(P) = 48/120 = 0.4. P(P|M) = 22/65 ≈ 0.338. Since 0.338 ≠ 0.4, the events are not independent. Alternatively, check P(M ∩ P) = P(M)×P(P): 22/120 ≠ (65/120)×(48/120). Thus, they are dependent.
(d) 独立性检验：若独立，P(P|M) 必须等于 P(P)。P(P) = 48/120 = 0.4。P(P|M) = 22/65 ≈ 0.338。因为 0.338 ≠ 0.4，所以事件不独立。或者检查 P(M ∩ P) = P(M)×P(P)：22/120 ≠ (65/120)×(48/120)。因此，它们是相关的。
5. Question 4: Discrete Random Variables and Linear Coding | 离散随机变量与线性编码
The discrete random variable X has the following probability distribution. (a) Find the value of k. (2 marks) (b) Calculate E(X) and Var(X). (4 marks) (c) The random variable Y is defined as Y = 3X − 5. Find E(Y) and Var(Y). (4 marks)
离散随机变量 X 具有如下概率分布。(a) 求 k 的值。(2 分) (b) 计算 E(X) 和 Var(X)。(4 分) (c) 随机变量 Y 定义为 Y = 3X − 5。求 E(Y) 和 Var(Y)。(4 分)
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X=x) | 0.2 | k | 0.3 | 0.1 |
(a) Solution: Sum of probabilities = 1 → 0.2 + k + 0.3 + 0.1 = 1 → k + 0.6 = 1 → k = 0.4.
(a) 解答：概率之和 = 1 → 0.2 + k + 0.3 + 0.1 = 1 → k + 0.6 = 1 → k = 0.4。
(b) E(X): E(X) = Σ x·P(X=x) = (1×0.2) + (2×0.4) + (3×0.3) + (4×0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3. E(X²) = (1²×0.2) + (4×0.4) + (9×0.3) + (16×0.1) = 0.2 + 1.6 + 2.7 + 1.6 = 6.1. Var(X) = E(X²) − [E(X)]² = 6.1 − (2.3)² = 6.1 − 5.29 = 0.81.
(b) E(X)：E(X) = Σ x·P(X=x) = (1×0.2) + (2×0.4) + (3×0.3) + (4×0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3。E(X²) = (1²×0.2) + (4×0.4) + (9×0.3) + (16×0.1) = 0.2 + 1.6 + 2.7 + 1.6 = 6.1。Var(X) = E(X²) − [E(X)]² = 6.1 − (2.3)² = 6.1 − 5.29 = 0.81。
(c) Linear coding: E(Y) = E(3X − 5) = 3E(X) − 5 = 3(2.3) − 5 = 6.9 − 5 = 1.9. Var(Y) = Var(3X − 5) = 3² Var(X) = 9 × 0.81 = 7.29. Remember: adding/subtracting a constant does not affect variance, but multiplying scales variance by the square of the multiplier.
(c) 线性编码：E(Y) = E(3X − 5) = 3E(X) − 5 = 3(2.3) − 5 = 6.9 − 5 = 1.9。Var(Y) = Var(3X − 5) = 3² Var(X) = 9 × 0.81 = 7.29。记住：加减常数不影响方差，但乘以一个系数会使方差按该系数的平方缩放。
6. Question 5: Binomial Distribution Application | 二项分布应用
A fair die is rolled 8 times. Success is defined as rolling a 6. (a) State the parameters n and p of the binomial distribution. (2 marks) (b) Find the probability of exactly two successes. (3 marks) (c) Find the probability of at least one success. (3 marks) (d) Find the expected number of successes and the standard deviation. (2 marks)
一个公平的骰子被掷 8 次。将掷出 6 点定义为成功。(a) 陈述二项分布的参数 n 和 p。(2 分) (b) 求恰好两次成功的概率。(3 分) (c) 求至少一次成功的概率。(3 分) (d) 求成功次数的期望值和标准差。(2 分)
(a) Parameters: n = 8 (fixed number of trials). p = 1/6 (probability of rolling a 6 on a fair die). Each roll is independent, so B(8, 1/6) applies.
(a) 参数：n = 8（固定试验次数）。p = 1/6（公平骰子掷出 6 的概率）。每次掷骰子独立，因此适用 B(8, 1/6)。
(b) P(X = 2): Using the binomial formula: P(X = r) = ⁿCᵣ × pʳ × (1−p)ⁿ⁻ʳ. P(X=2) = ⁸C₂ × (1/6)² × (5/6)⁶ = 28 × (1/36) × (15625/46656) = 28 × 15625 / (36 × 46656). Compute stepwise: 15625 / 46656 ≈ 0.334898; multiply by 28 ≈ 9.377; divide by 36 ≈ 0.2605. Exact fraction: 28×15625 / 1679616 = 437500 / 1679616 ≈ 0.2604. Round to 0.260 (3 s.f.).
(b) P(X = 2)：使用二项式公式：P(X = r) = ⁿCᵣ × pʳ × (1−p)ⁿ⁻ʳ。P(X=2) = ⁸C₂ × (1/6)² × (5/6)⁶ = 28 × (1/36) × (15625/46656) = 28 × 15625 / (36 × 46656)。分步计算：15625 / 46656 ≈ 0.334898；乘以 28 ≈ 9.377；除以 36 ≈ 0.2605。精确分数：28×15625 / 1679616 = 437500 / 1679616 ≈ 0.2604。四舍五入为 0.260（3 位有效数字）。
(c) P(X ≥ 1): P(at least one success) = 1 − P(X = 0). P(X=0) = ⁸C₀ × (1/6)⁰ × (5/6)⁸ = 1 × 1 × (5/6)⁸ = (390625 / 1679616) ≈ 0.2326. Therefore, P(X ≥ 1) = 1 − 0.2326 = 0.7674 (or 0.767 to 3 s.f.).
(c) P(X ≥ 1)：P(至少一次成功) = 1 − P(X = 0)。P(X=0) = ⁸C₀ × (1/6)⁰ × (5/6)⁸ = 1 × 1 × (5/6)⁸ = (390625 / 1679616) ≈ 0.2326。因此，P(X ≥ 1) = 1 − 0.2326 = 0.7674（或 0.767 至三位有效数字）。
(d) Expectation and SD: E(X) = n × p = 8 × (1/6) = 4/3 ≈ 1.333 successes. Var(X) = n × p × (1−p) = 8 × (1/6) × (5/6) = 40/36 = 10/9 ≈ 1.1111. Standard deviation = √(Var(X)) = √(10/9) = √10 / 3 ≈ 3.1623 / 3 ≈ 1.054 (3 s.f.).
(d) 期望值与标准差：E(X) = n × p = 8 × (1/6) = 4/3 ≈ 1.333 次成功。Var(X) = n × p × (1−p) = 8 × (1/6) × (5/6) = 40/36 = 10/9 ≈ 1.1111。标准差 = √(Var(X)) = √(10/9) = √10 / 3 ≈ 3.1623 / 3 ≈ 1.054（3 位有效数字）。
7. Common Mistakes and Examiner Tips | 常见错误与考官提示
In descriptive statistics, students often miscount data points when determining median position for even n, or forget that quartiles for discrete ungrouped data are found without interpolation. In histograms, the classic error is plotting frequency instead of frequency density on the vertical axis – always check class widths. In probability, failing to reduce the sample space in conditional probability is a frequent source of lost marks. For discrete random variables, ensure ΣP(X=x) = 1 is verified before computing E(X). In binomial questions, clearly stating n and p earns easy marks, and the ‘at least one’ shortcut (1 − P(X=0)) should be automatic.
在描述性统计中，学生常在确定偶数 n 的中位数位置时计错数据个数，或忘记离散未分组数据的四分位数不需使用插值法。在直方图中，典型错误是在纵轴上绘制频率而非频率密度——务必检查组距。在概率中，条件概率未能缩小样本空间是常见失分点。对于离散随机变量，计算 E(X) 前务必验证 ΣP(X=x) = 1。在二项分布题目中，清晰地陈述 n 和 p 可获得容易的分数，而“至少一次”的快捷方法（1 − P(X=0)）应成为本能反应。
When using a calculator for binomial coefficients, confirm you are using the nCr function correctly (8C2 = 28, not 56). Always write the full formula before substituting numbers: this shows method and can earn partial marks even if arithmetic slips. Keep probabilities as fractions where possible, then convert to decimals for final answers as specified in the question. Rounding: use 3 significant figures unless the question states otherwise.
使用计算器求二项式系数时，确认你正确使用了 nCr 功能（8C2 = 28，而非 56）。在代入数字前务必写出完整公式：这展示了方法，即使计算有误也能获得部分分数。尽可能将概率保留为分数，然后根据题目要求将最终答案转换为小数。舍入规则：除非题目另有说明，使用三位有效数字。
8. Mark Scheme Alignment and Self-Assessment | 评分标准对标与自我评估
Use the following simplified mark scheme to self-assess your mock test attempt. Award marks for method (M), accuracy (A), and final answer (B) as indicated. In the real CAIE exam, communication marks are also awarded for clear notation and correct use of statistical terminology.
使用以下简化评分标准对你的模拟卷作答进行自我评估。按指示为方法 (M)、准确性 (A) 和最终答案 (B) 给分。在实际 CAIE 考试中，清晰的符号书写和正确使用统计术语也能获得表达分。
| Q | Part | Criteria |
|---|---|---|
| 1 | a–c | Stem-and-leaf correct with key (B3); median and quartiles (B3); IQR + fences + outlier conclusion (M2 A2) |
| 2 | a–c | Frequency densities correct (B3); histogram bars correct width/height + labelled axes (B4); mean method + answer (M2 A1) |
| 3 | a–d | Venn diagram complete (B3); P(neither) (B2); conditional probability formula + answer (M2 A1); independence reasoning (B2) |
| 4 |
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