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

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

Welcome to this detailed walkthrough of an IGCSE Cambridge Statistics unit test mock paper. In this article, we will break down each question, providing step-by-step solutions and key concepts to help you master the exam content and build confidence for the final examination.

欢迎阅读这份 IGCSE Cambridge 统计单元测试模拟卷的详细解析。在本文中，我们将逐步拆解每道题目，提供分步解答和关键概念，助你掌握考试内容，为最终考试建立信心。

1. Mock Paper Overview | 模拟卷概览

This mock paper is designed to cover the core topics of the Cambridge IGCSE Statistics syllabus (0470). The paper includes questions on data collection, representation, measures of central tendency and dispersion, probability, binomial distribution, and correlation & regression. It carries a total of 60 marks, and the suggested completion time is 1 hour 15 minutes.

本模拟卷覆盖剑桥 IGCSE 统计课程（0470）的核心主题。试题涉及数据收集、数据表示、集中趋势和离散程度的度量、概率、二项分布以及相关与回归。卷面总分60分，建议完成时间为75分钟。

2. Data Collection and Sampling | 数据收集与抽样

The first question asks students to identify the sampling method used in a survey: “A researcher selects every 10th student from an alphabetical register.” This is an example of systematic sampling, because the sample is chosen at regular intervals from an ordered list.

第一题要求识别调查中使用的抽样方法：“研究人员按字母顺序名单每10名学生抽取一人。”这属于系统抽样，因为样本是按照固定间隔从一个有序名单中选取的。

Another part requires identifying a suitable source of data: “To find the average daily temperature in London for July 2023, you would use the Meteorological Office records.” This is secondary data, as it was originally collected by another organisation.

另一部分要求指出合适的数据来源：“要找到伦敦2023年7月的日均气温，应使用气象局的记录。”这是二手数据，因为它最初是由其他机构收集的。

3. Organising Data and Charts | 组织数据与图表

A frequency table shows the number of books read by 30 students in a month. Values for x (books) are 0, 1, 2, 3, 4 with corresponding frequencies 4, 7, 9, 6, 4.

一张频数表列出了30名学生在一个月内阅读的书籍数量。变量 x（书数）取值 0,1,2,3,4，对应的频数分别为 4,7,9,6,4。

The total frequency is 30, so no data is missing. To represent this discrete data, a bar chart is appropriate: draw bars of equal width, spaced equally, with heights corresponding to frequency. The horizontal axis is labelled “Number of books” and the vertical axis “Frequency”.

总频数为 30，没有缺失数据。为了表示这类离散数据，条形图是合适的：绘制等宽的条形，间距相等，条形高度对应频数。横轴标为“书籍数量”，纵轴标为“频数”。

Interpreting a pie chart: if the sector for “3 books” has a central angle of 72°, the proportion is 72/360 = 1/5. Since the total is 30, the number of students who read exactly 3 books is 30 × 1/5 = 6, which matches the table.

解读饼图：若“3本书”类别的圆心角为72°，其比例为72/360 = 1/5。总人数为30，因此恰读3本书的学生人数为30 × 1/5 = 6，与表格相符。

4. Measures of Central Tendency | 集中趋势度量

Using the same frequency table, we calculate the mean number of books read. Mean = Σfx / Σf. Σfx = (0×4) + (1×7) + (2×9) + (3×6) + (4×4) = 0 + 7 + 18 + 18 + 16 = 59. Therefore the mean is 59/30 ≈ 1.97 books.

沿用同一频数表，计算平均阅读书籍数。均值 = Σfx / Σf。Σfx = (0×4)+(1×7)+(2×9)+(3×6)+(4×4)=59，故均值 = 59/30 ≈ 1.97 本。

To find the median, the position is at (30+1)/2 = 15.5ᵗʰ value. Build cumulative frequencies: 4 (x=0), 11 (x=1), 20 (x=2)·s·so the 15.5ᵗʰ value lies in the x = 2 category. Hence the median is 2 books.

求中位数时，位置在(30+1)/2=15.5个数据。累计频数：4（x=0），11（x=1），20（x=2）——第15.5个数据落在 x=2 组内，因此中位数为 2 本书。

The mode is the value with the highest frequency, which is 2 books (frequency 9).

众数是频数最高的值，即 2 本书（频数9）。

5. Measures of Dispersion | 离散程度度量

The range = maximum – minimum = 4 – 0 = 4 books. To obtain the interquartile range (IQR), we first find Q₁ and Q₃. For 30 values, Q₁ is the 8ᵗʰ value (30×0.25 = 7.5 → 8ᵗʰ) and Q₃ is the 23ʳᵈ value (30×0.75 = 22.5 → 23ʳᵈ). From the cumulative frequency, the 8ᵗʰ value is 1, and the 23ʳᵈ value is 3. Therefore IQR = 3 – 1 = 2 books.

极差 = 最大值 – 最小值 = 4 – 0 = 4 本书。求四分位距(IQR)需先找到 Q₁ 和 Q₃。对30个数据，Q₁ 是第8个值(30×0.25=7.5→第8个)，Q₃ 是第23个值(30×0.75=22.5→第23个)。由累计频数知第8个值为1，第23个值为3，因此 IQR = 3 – 1 = 2 本书。

Although variance and standard deviation are not always required on a small discrete set, they can be computed. Σfx² = 0×0+1×7+4×9+9×6+16×4 = 161. Using the formula s² = [Σfx² – (Σfx)²/n] / (n–1), we get s² = [161 – 59²/30] / 29 ≈ [161 – 115.633] / 29 ≈ 1.564. Hence s ≈ √1.564 ≈ 1.25 books.

虽然小型离散数据集不一定要求计算方差与标准差，但仍可算出。Σfx² = 161。使用公式 s² = [Σfx² – (Σfx)²/n] / (n–1)，得 s² ≈ 1.564，故标准差 s ≈ 1.25 本。

6. Basic Probability | 概率基础

A bag contains 5 red, 3 blue and

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