GCSE CAIE Statistics: Unit Test Mock Paper Analysis | GCSE CAIE 统计：单元测试模拟卷解析

📚 GCSE CAIE Statistics: Unit Test Mock Paper Analysis | GCSE CAIE 统计：单元测试模拟卷解析

This article walks you through a complete GCSE CAIE Statistics unit test mock paper, providing detailed solutions and commentary for each question. Use this analysis to sharpen your exam technique and deepen your understanding of key statistical concepts.

本文将带你完整解析一份 GCSE CAIE 统计单元测试模拟卷，逐题给出详细解答与点评。通过这份解析，你可以打磨考试技巧，加深对核心统计概念的理解。

1. Mock Paper Overview | 模拟卷概览

The mock paper contains nine structured questions, each targeting essential topics from the CAIE IGCSE Statistics syllabus. The table below shows the distribution of questions, marks, and the bilingual topic coverage.

本模拟卷包含九道结构化试题，每道题都针对 CAIE IGCSE 统计学大纲中的核心主题。下表展示了题目、分值以及中英双语主题的分布。

You are advised to spend about 1 hour 15 minutes on this mock test. Each question is worth the indicated marks, and full working must be shown. In the following sections, every question is broken down with step‑by‑step reasoning and common pitfalls to avoid.

建议你花大约 1 小时 15 分钟完成这份模拟卷。每题分值如表中所示，并且必须展示完整的解题过程。在后续小节中，我们将逐一拆解每道题，给出逐步推理以及需要避免的常见错误。

2. Sampling Methods | 抽样方法

Question 1: A school has 800 students, stratified by year group: 300 in Year 10 and 500 in Year 11. You need a sample of 40 students to survey about study habits. Determine how many students should be selected from each year group, and describe a suitable selection procedure. State one advantage of this sampling method.

问题1：某学校共有 800 名学生，按年级分层：10 年级 300 人，11 年级 500 人。你需要抽取 40 名学生就学习习惯进行调查。确定每个年级应抽取的人数，并描述一种合适的选取流程。请说明该抽样方法的一个优点。

To obtain a proportional stratified sample, calculate the fraction for each stratum: Year 10 sample size = (300/800) × 40 = 15; Year 11 sample size = (500/800) × 40 = 25. This ensures the sample reflects the population structure.

为获得比例分层样本，计算每层的比例：10 年级样本容量 = (300/800) × 40 = 15；11 年级样本容量 = (500/800) × 40 = 25。这保证了样本能反映总体结构。

Within each year group, assign a unique number to every student. Use a random number generator or random number table to select the required 15 students from Year 10 and 25 from Year 11. This combination of stratification and simple random sampling reduces bias and improves representativeness.

在每个年级内，给每位学生分配一个唯一编号。使用随机数生成器或随机数表从 10 年级选出所需的 15 人，从 11 年级选出 25 人。这种分层与简单随机抽样相结合的方法可以减少偏差，提高代表性。

A key advantage is that the sample will accurately mirror the proportion of each year group in the school, so any conclusions about study habits are less likely to be distorted by an under‑ or over‑represented group.

一个关键优点是样本能准确反映学校中各年级的比例，因此有关学习习惯的任何结论都不太可能因某个群体代表不足或过多而失真。

3. Graphical Representation of Data | 数据的图表表示

Question 2: The table below shows the distribution of test scores out of 50 for 80 students. Draw a histogram to display the data, clearly calculating frequency density for each class interval.

问题2：下表显示了 80 名学生在满分 50 分测验中的成绩分布。绘制直方图展示数据，并清晰计算每个组距的频率密度。

For histograms with unequal class widths, the vertical axis must show frequency density = frequency ÷ class width. The class widths here are 10, 10, 10, 10, and 11 for the interval 40–50.

对于组距不相等的直方图，垂直轴必须显示频率密度 = 频数 ÷ 组距。此处组距分别为 10、10、10、10，而 40–50 组的组距为 11。

Calculate each frequency density: 0–9: 4/10 = 0.4; 10–19: 12/10 = 1.2; 20–29: 28/10 = 2.8; 30–39: 24/10 = 2.4; 40–50: 12/11 ≈ 1.09. Draw bars with these heights over the given class boundaries. Label axes clearly, and remember the area of each bar is proportional to the frequency.

计算各频率密度：0–9 组：4/10 = 0.4；10–19 组：12/10 = 1.2；20–29 组：28/10 = 2.8；30–39 组：24/10 = 2.4；40–50 组：12/11 ≈ 1.09。以这些高度在给定的组界上绘制条形。清楚地标记坐标轴，记住每个条形的面积与频数成正比。

A common mistake is to plot frequency instead of frequency density when class widths differ. Always check the interval widths before drawing; using frequency directly would visually distort the distribution.

常见错误是当组距不同时，绘制了频数而非频率密度。在绘图前一定要检查组距宽度；直接使用频数会在视觉上扭曲分布形态。

4. Measures of Central Tendency | 集中趋势的度量

Question 3(a): For the dataset 5, 8, 9, 7, 6, 10, 7, 12, 9, 7, calculate the mean, median, and mode. Interpret which measure best represents the data.

问题3(a)：对于数据集 5, 8, 9, 7, 6, 10, 7, 12, 9, 7，计算均值、中位数和众数。解释哪种度量最能代表该数据。

First, sort the data: 5, 6, 7, 7, 7, 8, 9, 9, 10, 12. Mean: (5+6+7+7+7+8+9+9+10+12) / 10 = 80/10 = 8. Median: (7+8)/2 = 7.5. Mode: 7 (appears three times). The mean is slightly pulled up by the value 12, so the median or mode might be more typical of the majority of scores. Since the data is nearly symmetric with one mild outlier, the mean still works well, but the median is more robust.

首先将数据排序：5, 6, 7, 7, 7, 8, 9, 9, 10, 12

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