A-Level CIE Statistics: Unit Test Mock Paper Walkthrough | A-Level CIE 统计：单元测试模拟卷解析

📚 A-Level CIE Statistics: Unit Test Mock Paper Walkthrough | A-Level CIE 统计：单元测试模拟卷解析

This article provides a detailed step-by-step walkthrough of a mock unit test designed for the CIE A-Level Statistics syllabus. The paper covers key topics including probability calculations, discrete random variables, the binomial distribution, normal distribution and the normal approximation to the binomial. Each question is broken down with clear working and commentary, followed by a parallel explanation in Chinese to support bilingual learners. Common pitfalls and examination tips are highlighted so that you can strengthen your understanding and boost your confidence for the real exam.

本文详细解析了一份针对 CIE A-Level 统计课程设计的单元测试模拟卷。试卷涵盖概率计算、离散随机变量、二项分布、正态分布以及二项分布的正态近似等核心主题。每道题目都配有清晰的解题步骤和点评，并提供对应的中文解释，帮助双语学习者深入理解。文中还特别指出常见错误和应试技巧，助你巩固知识、提升实考信心。

1. Overview of the Mock Paper | 模拟试卷概览

The mock test consists of six compulsory questions, carrying a total of 50 marks. It is designed to be completed in approximately 60 minutes, mirroring the style and difficulty of a typical CIE Statistics 1 unit assessment. The questions are as follows:

这套模拟卷包含六道必答题，总分 50 分，设计时长为 60 分钟，反映了 CIE 统计学 S1 单元测试的常见风格与难度。题目分布如下：

Question \| 题号	Topic \| 主题	Marks \| 分值
1	Probability without replacement / 不放回概率	7
2	Discrete random variable and expectation / 离散随机变量与期望	8
3	Binomial distribution – exact probabilities / 二项分布精确概率	8
4	Binomial distribution – mean, variance and interpretation / 二项分布均值、方差与应用	8
5	Normal distribution: standardisation and probabilities / 正态分布标准化与概率	9
6	Normal approximation to the binomial with continuity correction / 二项分布正态近似与连续性校正	10

Throughout this walkthrough we will use standard statistical tables from the CIE formula booklet, so you can practise as you would in the real examination. Let us now work through each question.

在本次解析中，我们将全程使用 CIE 公式手册中的标准统计表，让你像在真实考试中一样进行练习。现在让我们逐题突破。

2. Question 1: Probability without Replacement | 问题 1：不放回概率

A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. Find the probability that (a) both are red, (b) exactly one is red, (c) at least one is red.

一个袋子里有 5 个红球和 3 个蓝球。从中随机不放回地抽取两个球。求下列事件的概率：(a) 两个都是红球，(b) 恰好一个红球，(c) 至少一个红球。

Step 1: Draw a tree diagram or use combinations. For (a), the probability the first ball is red is 5/8. Given the first is red, the probability the second is red becomes 4/7. Hence, P(RR) = (5/8) × (4/7) = 20/56 = 5/14.

步骤 1：画出树状图或使用组合数。对于 (a)，第一个球是红球的概率为 5/8。在第一个球是红球的条件下，第二个球也是红球的概率变为 4/7。因此，P(两红) = (5/8) × (4/7) = 20/56 = 5/14。

Step 2: For (b) exactly one red, the successful outcomes are RB or BR. P(RB) = (5/8) × (3/7) = 15/56. P(BR) = (3/8) × (5/7) = 15/56. So P(exactly one red) = 15/56 + 15/56 = 30/56 = 15/28.

步骤 2：对于 (b) 恰好一个红球，有利结果有 RB 或 BR 两种顺序。P(RB) = (5/8) × (3/7) = 15/56。P(BR) = (3/8) × (5/7) = 15/56。因此 P(恰好一个红球) = 30/56 = 15/28。

Step 3: For (c) at least one red, use complement: P(at least one red) = 1 – P(both blue). P(BB) = (3/8) × (2/7) = 6/56 = 3/28. So the answer is 1 – 3/28 = 25/28.

步骤 3：对于 (c) 至少一个红球，用补集：P(至少一个红球) = 1 – P(两个都是蓝球)。P(两蓝) = (3/8) × (2/7) = 6/56 = 3/28。答案为 1 – 3/28 = 25/28。

Always check that the sum of all mutually exclusive probabilities (RR, RB, BR, BB) equals 1: 5/14 + 15/56 + 15/56 + 3/28 = 1. This confirms your calculation.

务必检查所有互斥事件的概率之和是否等于 1：5/14 + 15/56 + 15/56 + 3/28 = 1。这能验证你的计算是否正确。

3. Question 2: Discrete Random Variable and Expectation | 问题 2：离散随机变量与期望

A fair six-sided die is rolled. The random variable X is defined as the square of the score obtained, i.e. X = score². Construct the probability distribution of X, and find E(X) and Var(X).

投掷一枚公平的六面骰子。随机变量 X 定义为所得点数的平方，即 X = 点数²。构造 X 的概率分布，并求 E(X) 和 Var(X)。

Step 1: List possible scores and corresponding X values. Score: 1,2,3,4,5,6. X = 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36. Each score has probability 1/6, so P(X = x) = 1/6 for each of these six values.

步骤 1：列出可能的点数及对应的 X 值。点数 1,2,3,4,5,6。X 分别等于 1,4,9,16,25,36。每个点数出现的概率都是 1/6，因此 P(X = x) = 1/6 对所有这六个值成立。

Step 2: Calculate E(X) = Σ x·P(X=x) = (1+4+9+16+25+36) × (1/6) = 91/6 ≈ 15.1667.

步骤 2：计算数学期望 E(X) = Σ x·P(X=x) = (1+4+9+16+25+36) × (1/6) = 91/6 ≈ 15.1667。

Step 3: Calculate E(X²) = Σ x²·P(X=x). x² values: 1²=1, 4²=16, 9²=81, 16²=256, 25²=625, 36²=1296. Sum = 1+16+81+256+625+1296 = 2275. So E(X²) = 2275/6 ≈ 379.1667.

步骤 3：计算 E(X²) = Σ x²·P(X=x)。x² 的值分别为 1,16,81,256,625,1296，总和为 2275。于是 E(X²) = 2275/6 ≈ 379.1667。

Step 4: Var(X) = E(X²) – [E(X)]² = (2275/6) – (91/6)² = (2275/6) – (8281/36) = (13650/36) – (8281/36) = 5369/36 ≈ 149.1389.

步骤 4：方差 Var(X) = E(X²) – [E(X)]² = (2275/6) – (91/6)² = (2275/6) – (8281/36) = 5369/36 ≈ 149.1389。

Alternatively, you could use the formula Var(X) = (1/6)Σ(x – μ)², but the computational formula is faster here. Remember to present the distribution clearly in a table for full marks.

你也可以使用公式 Var(X) = (1/6)Σ(x – μ)²，但在此使用计算式更快捷。为拿满分，记得用表格清晰地展示分布。

4. Question 3: Binomial Distribution – Exact Probabilities | 问题 3：二项分布精确概率

It is known that 15% of light bulbs produced by a factory are defective. A quality inspector randomly selects 12 bulbs. Let X be the number of defective bulbs. (a) Write down the distribution of X, stating its parameters. (b) Find P(X = 2). (c) Find P(X ≤ 1). (d) Find P(X ≥ 3).

已知某工厂生产的灯泡中有 15% 是次品。质检员随机抽取 12 只灯泡。设 X 为次品数量。(a) 写出 X 的分布并指明参数。(b) 求 P(X = 2)。(c) 求 P(X ≤ 1)。(d) 求 P(X ≥ 3)。

Step 1: X follows a binomial distribution with n = 12 and p = 0.15. We write X ~ B(12, 0.15). The probability mass function is P(X = x) = ¹²Cₓ × (0.15)ˣ × (0.85)¹²⁻ˣ.

步骤 1：X 服从二项分布，参数为 n = 12, p = 0.15。记作 X ~ B(12, 0.15)。概率质量函数为 P(X = x) = ¹²Cₓ × (0.15)ˣ × (0.85)¹²⁻ˣ。

Step 2: For P(X = 2), we compute ¹²C₂ × (0.15)² × (0.85)¹⁰. ¹²C₂ = 66. (0.15)² = 0.0225. (0.85)¹⁰ = 0.19687 (using calculator). Hence P(X = 2) = 66 × 0.0225 × 0.19687 ≈ 0.2924.

步骤 2：对于 P(X = 2)，计算 ¹²C₂ × (0.15)² × (0.85)¹⁰。¹²C₂ = 66，(0.15)² = 0.0225，(0.85)¹⁰ ≈ 0.19687。因此 P(X = 2) ≈ 66 × 0.0225 × 0.19687 = 0.2924。

Step 3: P(X ≤ 1) = P(X = 0) + P(X = 1). P(X = 0) = (0.85)¹² ≈ 0.1422. P(X = 1) = 12 × 0.15 × (0.85)¹¹ ≈ 12 × 0.15 × 0.1673 = 0.3012. So P(X ≤ 1) ≈ 0.1422 + 0.3012 = 0.4434.

步骤 3：P(X ≤ 1) = P(X = 0) + P(X = 1)。P(X = 0) = (0.85)¹² ≈ 0.1422。P(X = 1) = 12 × 0.15 × (0.85)¹¹ ≈ 12 × 0.15 × 0.1673 = 0.3012。故 P(X ≤ 1) ≈ 0.4434。

Step 4: P(X ≥ 3) = 1 – P(X ≤ 2). P(X ≤ 2) = P(X=0)+P(X=1)+P(X=2) = 0.1422 + 0.3012 + 0.2924 = 0.7358. So P(X ≥ 3) = 1 – 0.7358 = 0.2642.

步骤 4：P(X ≥ 3) = 1 – P(X ≤ 2)。P(X ≤ 2) = 0.1422 + 0.3012 + 0.2924 = 0.7358。因此 P(X ≥ 3) = 1 – 0.7358 = 0.2642。

Always use the cumulative binomial tables if they are provided. In the CIE exam, you are expected to use tables where available to reduce calculator errors.

如果有累积二项分布表，务必加以利用。在 CIE 考试中，当表格可用时应优先使用，以降低计算器错误。

5. Question 4: Binomial – Mean, Variance and Interpretation | 问题 4：二项分布的均值、方差与解读

For the same binomial random variable X ~ B(12, 0.15), (a) write down E(X) and Var(X). (b) In a production run of 80 such batches (each of size 12), the inspector recorded the number of defective bulbs per batch. A summary of results gave a total of 148 defective bulbs. Determine whether this observation suggests the defect rate has changed from 15%.

对于上述二项随机变量 X ~ B(12, 0.15)，(a) 写出 E(X) 和 Var(X)。(b) 在一次生产运行中，质检员记录了 80 批（每批 12 只）的次品数量，总计发现 148 个次品。判断此观察结果是否暗示次品率已不是 15%。

Step 1: For X ~ B(12, 0.15), the mean is E(X) = np = 12 × 0.15 = 1.8. The variance is Var(X) = np(1–p) = 12 × 0.15 × 0.85 = 1.53.

步骤 1：X ~ B(12, 0.15) 的均值 E(X) = np = 12 × 0.15 = 1.8。方差 Var(X) = np(1–p) = 12 × 0.15 × 0.85 = 1.53。

Step 2: For 80 batches, the total number of defective bulbs, T, is the sum of 80 independent B(12, 0.15) variables. Thus E(T) = 80 × 1.8 = 144, and Var(T) = 80 × 1.53 = 122.4. Standard deviation σₜ = √122.4 ≈ 11.06.

步骤 2：80 批的总次品数 T 是 80 个独立 B(12, 0.15) 随机变量之和。故 E(T) = 80 × 1.8 = 144，Var(T) = 80 × 1.53 = 122.4，标准差 σₜ = √122.4 ≈ 11.06。

Step 3: The observed total is 148. Under the assumed p = 0.15, T is approximately normal with mean 144 and sd 11.06. Compare the z-score: z = (148 – 144) / 11.06 ≈ 0.36. This is not an extreme value (within ±2). Therefore, the observation does not provide strong evidence that the defect rate has changed from 15%.

步骤 3：实际观察总数为 148。在 p = 0.15 的假设下，T 近似服从均值为 144、标准差为 11.06 的正态分布。计算 z 值：z = (148 – 144) / 11.06 ≈ 0.36，并非异常值（位于 ±2 之内）。因此，该观察结果并未提供有力证据表明次品率已偏离 15%。

This is a gentle introduction to the idea of hypothesis testing using distribution parameters. In later topics, you will formalise this with p-values and significance levels.

这是利用分布参数进行假设检验思想的温和引入，后续章节会正式进入 p 值与显著性水平的学习。

6. Question 5: Normal Distribution – Standardisation and Probabilities | 问题 5：正态分布标准化与概率

The weights of apples from an orchard are normally distributed with mean μ = 150 g and standard deviation σ = 20 g. Calculate the probability that a randomly chosen apple (a) weighs more than 180 g, (b) weighs between 130 g and 165 g, (c) is in the heaviest 10%. (Give the weight boundary.)

某果园苹果的重量服从正态分布，均值 μ = 150 g，标准差 σ = 20 g。计算随机抽取一个苹果：(a) 重量超过 180 g 的概率；(b) 重量介于 130 g 与 165 g 之间的概率；(c) 属于最重 10% 的重量分界值。

Step 1: Standardise using Z = (X – μ) / σ. For (a), P(X > 180) = P(Z > (180–150)/20) = P(Z > 1.5). From tables, Φ(1.5) = P(Z < 1.5) = 0.9332, so P(Z > 1.5) = 1 – 0.9332 = 0.0668.

步骤 1：用 Z = (X – μ)/σ 标准化。(a) P(X > 180) = P(Z > 1.5)。查表得 Φ(1.5) = 0.9332，故 P(Z > 1.5) = 1 – 0.9332 = 0.0668。

Step 2: For (b), convert both bounds: z₁ = (130–150)/20 = –1.0, z₂ = (165–150)/20 = 0.75. P(130 < X < 165) = P(–1.0 < Z < 0.75) = Φ(0.75) – Φ(–1.0). Φ(0.75) ≈ 0.7734, Φ(–1.0) = 1 – Φ(1.0) = 0.1587. So probability = 0.7734 – 0.1587 = 0.6147.

步骤 2：(b) 转换界值：z₁ = –1.0，z₂ = 0.75。P(130 < X < 165) = P(–1.0 < Z < 0.75) = Φ(0.75) – Φ(–1.0) = 0.7734 – 0.1587 = 0.6147。

Step 3: For (c) heaviest 10%, we need the value x such that P(X > x) = 0.10. This implies P(Z > z) = 0.10, so Φ(z) = 0.90. From inverse normal tables, z₀.₉₀ ≈ 1.2816. Then x = μ + zσ = 150 + 1.2816 × 20 = 150 + 25.63 = 175.63 g. Round to 176 g (or 175.6 g depending on context).

步骤 3：(c) 最重的 10%，求 x 使 P(X > x) = 0.10。这就要求 z 满足 Φ(z) = 0.90。查逆正态表得 z₀.₉₀ ≈ 1.2816。则 x = μ + zσ = 150 + 25.63 = 175.63 g，可约至 176 g。

Always draw a sketch of the normal curve and shade the relevant region. This reduces sign errors and makes it easier to visualise the required table look-up.

始终画出正态曲线简图并标出阴影区域，这样能减少符号错误，并使查表过程更直观。

7. Question 6: Normal Approximation to the Binomial – Continuity Correction | 问题 6：二项分布的正态近似及连续性校正

A fair coin is tossed 200 times. Let Y be the number of heads obtained. (a) State the exact distribution of Y. (b) Show that Y can be approximated by a normal distribution and give the parameters. (c) Use the approximation with a continuity correction to estimate P(90 ≤ Y ≤ 110).

一枚公平硬币抛掷 200 次，记 Y 为出现正面的次数。(a) 写出 Y 的精确分布。(b) 说明 Y 可用正态分布近似，并给出参数。(c) 运用带连续性校正的近似方法估计 P(90 ≤ Y ≤ 110)。

Step 1: Y ~ B(200, 0.5) exactly. The conditions for normal approximation are satisfied because np = 100 > 5 and n(1–p) = 100 > 5.

步骤 1：Y 精确服从 B(200, 0.5)。正态近似

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A-Level CIE Statistics: Unit Test Mock Paper Walkthrough | A-Level CIE 统计：单元测试模拟卷解析