📚 Year 12 Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥AS统计单元测试模拟卷解析
Welcome to this detailed breakdown of a mock unit test for Cambridge AS Level Statistics (S1). The paper covers data handling, probability, counting techniques, discrete random variables, and the binomial and normal distributions. We will examine each question step by step, highlight common errors, and consolidate the techniques needed for exam success.
欢迎阅读这份剑桥AS统计（S1）单元测试模拟卷的详细解析。试卷涵盖数据处理、概率、计数技巧、离散随机变量以及二项分布和正态分布。我们将逐步分析每一道题，指出常见错误，并巩固考试成功所需的解题技巧。
1. Mock Paper Overview | 模拟卷概览
This mock unit test contains seven structured questions with a total of 50 marks, designed to be completed in 60 minutes. The questions are typical of Paper 5 (Probability & Statistics 1) and require a mix of calculator use, formula application, and clear written reasoning.
本模拟单元测试包含七道结构题，满分50分，要求在60分钟内完成。题目风格贴近试卷5（概率与统计1），需要综合运用计算器、公式推导和清晰的书面推理。
You should show all steps of working, as marks are allocated for method as well as final answers. When using a calculator for statistical measures, always write down the numbers you input and the values obtained.
你需要展示全部计算步骤，因为方法分和最终答案分都会给出。使用计算器求统计量时，务必写下输入的数字和数据，以便考官追踪。
2. Question 1: Descriptive Statistics | 第1题：描述性统计
The frequency table below shows the marks of 40 students in a test. We are asked to estimate the mean and standard deviation, and to draw a cumulative frequency graph to find the median and interquartile range.
以下频率表给出了40名学生的考试成绩。题目要求估计均值与标准差，并绘制累积频率图以求出中位数和四分位距。
We first add columns for the class midpoint (x), fx and fx². Using midpoints ensures a good approximation for grouped data.
我们首先添加组中值(x)、fx 和 fx² 列。对于分组数据，使用组中值能够获得良好的近似。
| Marks | Frequency (f) | Midpoint (x) | fx | fx² |
|---|---|---|---|---|
| 0–19 | 3 | 9.5 | 28.5 | 270.75 |
| 20–39 | 7 | 29.5 | 206.5 | 6091.75 |
| 40–59 | 12 | 49.5 | 594.0 | 29403.00 |
| 60–79 | 10 | 69.5 | 695.0 | 48302.50 |
| 80–99 | 8 | 89.5 | 716.0 | 64082.00 |
| Total | 40 | 2240.0 | 148150.00 |
The estimated mean is x̄ fx / Σf = 2240 ÷ 40 = 56. For the standard deviation we use the formula s = √[ Σfx²/n − (x̄)² ].
估计均值 x̄ = Σfx / Σf = 2240 ÷ 40 = 56。标准差使用公式 s = √[ Σfx²/n − (x̄)² ]。
s = √148150/40 − 56² = √3703.75 − 3136 = √567.75 ≈ 23.83
n, we construct the cumulative frequency: 3, 10, 22, 32, 40. The 20th value lies in the 40– Using linear interpolation: median = 39.5 + ((20−10)/12)×20 = 39.5 + 16.7 = 56.2 (approx).
为确定中位数，我们构建累积频率列：3, 10, 22, 32, 40。第20个值落在40–59区间。使用线性插值：中位数 = 39.5 + ((20−10)/12)×20 = 39.5 + 16.7 ≈ 56.2。
Quartiles are found similarly: Q₁ is the 10th value (in 20–39) and Q₃ is the 30th value (in 60– nterquartile range IQR = Q₃ − Q₁ gives a measure of spread not affected by extreme values.
四分位数同理可得：Q₁ 为第10个值（20–39区间），Q₃ 为第30个值（60–79区间）。四分位距 IQR = Q₃ − Q₁，提供了一种不受极端值影响的离散度度量。
3. Question 2: Probability Rules | 第2题：概率法则
Events A and B satisfy P(A) = 0.6, P(B
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