CIE AS Statistics Unit Test Mock Paper Analysis | CIE AS 统计单元测试模拟卷解析

📚 CIE AS Statistics Unit Test Mock Paper Analysis | CIE AS 统计单元测试模拟卷解析

Welcome to this detailed walkthrough of a mock unit test for CIE AS Level Statistics (9709/51). This paper covers core topics including data representation, probability, distributions, and sampling. By working through each question and understanding the common pitfalls, you can reinforce your knowledge and boost your exam confidence.

欢迎阅读这篇CIE AS统计（9709/51）单元模拟卷的详细解析。本试卷涵盖了数据表示、概率、分布与抽样等核心主题。通过解析每一题并吃透常见易错点，你可以巩固知识，提升应考信心。

1. Histograms and Frequency Density | 直方图与频率密度

Question 1 asked you to construct a histogram for the time taken by 200 students to complete a puzzle. The data were grouped into intervals: 0-10, 10-15, 15-20, 20-30, 30-60 minutes, with frequencies 20, 35, 50, 60, 35 respectively.

第1题要求你为200名学生完成拼图的时间绘制直方图。数据分组为：0-10、10-15、15-20、20-30、30-60分钟，对应频数为20、35、50、60、35。

Since the class widths are not all equal (10, 5, 5, 10, 30), you must work with frequency density. The formula is: frequency density = frequency ÷ class width. The calculated frequency densities are 2, 7, 10, 6, and 1.167 (approx).

由于组距不相等（10、5、5、10、30），必须使用频率密度。公式为：频率密度 = 频数 ÷ 组距。计算得到的频率密度分别为 2、7、10、6 和约 1.167。

A common mistake is to plot frequency on the vertical axis instead of frequency density, which distorts the distribution. Always label the y‑axis clearly as ‘Frequency density’ when class widths differ.

常见错误是在纵轴绘制频数而非频率密度，这会扭曲分布。当组距不同时，务必将y轴明确标注为“频率密度”。

The next part of the question required an estimate of the median. Cumulative frequencies are 20, 55, 105, 165, 200. The median position is the 100th value, which lies in the interval 15–20.

题目下一部分要求估算中位数。累积频数为20、55、105、165、200。中位数位置为第100个值，落在区间15–20。

The linear interpolation formula is:

Median = L + ((n/2 − F)/f) × w

线性插值公式为：

中

Published by TutorHao | Year 12 统计 Revision Series | aleveler.com

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