KS3 CAIE Statistics: Unit Test Mock Paper Walkthrough | KS3 CAIE 统计：单元测试模拟卷解析

📚 KS3 CAIE Statistics: Unit Test Mock Paper Walkthrough | KS3 CAIE 统计：单元测试模拟卷解析

This walkthrough covers a full KS3 CAIE Statistics unit test mock paper, providing step-by-step solutions and explanations for every question. It is designed to help students revise key concepts such as data representation, averages, probability, graph interpretation and critical evaluation of charts.

本解析涵盖一份完整的 KS3 CAIE 统计单元测试模拟卷，为每道题目提供逐步解答与详细讲解。旨在帮助学生复习数据表示、平均数、概率、图表解读以及批判性评估图表等核心概念。

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1. Interpreting Pictograms | 解读象形图

A pictogram shows the number of books read by four students in one month. Each complete book icon represents 2 books. Ali has 3 full icons, Ben has 2 full icons and 1 half icon, Chloe has 4 full icons and Dina has 1 full icon.

一个象形图记录四名学生在一个月内阅读的书籍数量。每个完整的书本图标代表 2 本书。Ali 有 3 个完整图标，Ben 有 2 个完整图标和 1 个半个图标，Chloe 有 4 个完整图标，Dina 有 1 个完整图标。

Convert each student’s icons into a number of books: Ali = 3 × 2 = 6 books, Ben = 2.5 × 2 = 5 books, Chloe = 4 × 2 = 8 books, Dina = 1 × 2 = 2 books.

将每名学生的图标转换为书籍数量：Ali = 3 × 2 = 6 本，Ben = 2.5 × 2 = 5 本，Chloe = 4 × 2 = 8 本，Dina = 1 × 2 = 2 本。

Chloe read the most books. The total for the group = 6 + 5 + 8 + 2 = 21 books. Chloe read 8 – 2 = 6 more books than Dina.

Chloe 读书最多。四人合计 = 6 + 5 + 8 + 2 = 21 本书。Chloe 比 Dina 多读 8 – 2 = 6 本。

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2. Reading Bar Charts | 阅读条形图

A bar chart displays the number of students in each of four Year 5 classes: Class 5A = 25, Class 5B = 30, Class 5C = 20, Class 5D = 28.

一个条形图显示了五年级四个班的学生人数：5A 班 25 人，5B 班 30 人，5C 班 20 人，5D 班 28 人。

Total number of students = 25 + 30 + 20 + 28 = 103. The largest class is 5B with 30 students, and the smallest is 5C with 20.

学生总人数 = 25 + 30 + 20 + 28 = 103。人数最多的班级是 5B（30 人），最少的是 5C（20 人）。

Mean number of students per class = total ÷ number of classes = 103 ÷ 4 = 25.75. The range = maximum – minimum = 30 – 20 = 10.

每班平均学生人数 = 总人数 ÷ 班级数 = 103 ÷ 4 = 25.75。极差 = 最大值 – 最小值 = 30 – 20 = 10。

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3. Calculating Mean, Median, Mode and Range | 计算平均数、中位数、众数和极差

The data set shows the daily hours of screen time for seven students: 8, 12, 6, 10, 14, 8, 10.

数据集显示七名学生每日屏幕时间（小时）：8, 12, 6, 10, 14, 8, 10。

Arrange in order: 6, 8, 8, 10, 10, 12, 14. There are 7 values.

按顺序排列：6, 8, 8, 10, 10, 12, 14。共有 7 个数据。

Mean = (6 + 8 + 8 + 10 + 10 + 12 + 14) ÷ 7 = 68 ÷ 7 ≈ 9.71 (to 2 d.p.)

平均数 = (6 + 8 + 8 + 10 + 10 + 12 + 14) ÷ 7 = 68 ÷ 7 ≈ 9.71（保留两位小数）

The median is the 4th value: 10. The modes are 8 and 10 (bimodal). The range = 14 – 6 = 8.

中位数是第 4 个值：10。众数是 8 和 10（双众数）。极差 = 14 – 6 = 8。

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4. Pie Chart Angles and Proportions | 饼图角度与比例

A pie chart shows the favourite fruits of 360 primary students: apples 150°, bananas 90°, oranges 80°, grapes 40°.

某饼图展示 360 名小学生最喜爱的水果：苹果 150°，香蕉 90°，橙子 80°，葡萄 40°。

Since the total angle is 360°, each degree represents one student. So the number of students liking each fruit equals its angle.

因为总角度为 360°，每度代表一名学生。因此喜爱每种水果的人数等于其角度。

Apples: 150 students (150/360 × 100 ≈ 41.7%), Bananas: 90 (25%), Oranges: 80 (≈22.2%), Grapes: 40 (≈11.1%). Apples are the most popular.

苹果：150 人（150/360 × 100 ≈ 41.7%），香蕉：90 人（25%），橙子：80 人（约 22.2%），葡萄：40 人（约 11.1%）。苹果最受欢迎。

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5. Estimating the Mean from a Frequency Table | 根据频数表估算平均数

A frequency table groups test scores: 1-10 marks, frequency 4; 11-20, frequency 6; 21-30, frequency 7; 31-40, frequency 3. There are 20 students in total.

一张频数表将测验分数分组：1-10 分，频数 4；11-20，频数 6；21-30，频数 7；31-40，频数 3。共有 20 名学生。

First find the midpoint of each class interval: (1+10)÷2 = 5.5; (11+20)÷2 = 15.5; (21+30)÷2 = 25.5; (31+40)÷2 = 35.5.

先求每个区间的中点：(1+10)÷2 = 5.5；(11+20)÷2 = 15.5；(21+30)÷2 = 25.5；(31+40)÷2 = 35.5。

Estimated total = (5.5 × 4) + (15.5 × 6) + (25.5 × 7) + (35.5 × 3) = 22 + 93 + 178.5 + 106.5 = 400

估算总分 = (5.5 × 4) + (15.5 × 6) + (25.5 × 7) + (35.5 × 3) = 22 + 93 + 178.5 + 106.5 = 400

Estimated mean = 400 ÷ 20 = 20 marks. This method assumes that the values in each group are evenly distributed around the midpoint.

估算平均数 = 400 ÷ 20 = 20 分。该方法假设各组中的数值围绕中点均匀分布。

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6. Probability Scale and Simple Probability | 概率尺度与简单概率

A bag contains 5 red balls, 3 blue balls and 2 green balls. One ball is chosen at random. Total outcomes = 10.

一个袋子装有 5 个红球、3 个蓝球和 2 个绿球。随机抽取一个球。可能结果总数 = 10。

P(red) = 5/10 = 1/2. P(not blue) = P(red or green) = (5+2)/10 = 7/10, which is also 1 – 3/10.

P（红）= 5/10 = 1/2。P（不是蓝）= P（红或绿）= (5+2)/10 = 7/10，也等于 1 – 3/10。

On a probability scale from 0 to 1, an impossible event is marked at 0, a certain event at 1, and P(blue) = 3/10 = 0.3 would be placed about one-third of the way from 0 to 1.

在从 0 到 1 的概率尺度上，不可能事件标记为 0，必然事件为 1，P（蓝）= 3/10 = 0.3 应置于从 0 到 1 约三分之一处。

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7. Two-Way Tables | 双向表

A two-way table records whether students bring a packed lunch or have a school dinner: Boys: packed 20, dinner 15; Girls: packed 25, dinner 10.

某双向表记录了学生自带午餐还是吃学校餐：男生：自带 20，校餐 15；女生：自带 25，校餐 10。

Total students = 20 + 15 + 25 + 10 = 70. The probability a randomly chosen student is a girl = (25+10)/70 = 35/70 = 1/2.

学生总数 = 20 + 15 + 25 + 10 = 70。随机选一名学生为女生的概率 = (25+10)/70 = 35/70 = 1/2。

P(student brings packed lunch) = (20+25)/70 = 45/70 = 9/14. Given a student is a boy, P(he has school dinner) = 15/(20+15) = 15/35 = 3/7.

P（学生自带午餐）= (20+25)/70 = 45/70 = 9/14。若已知该生为男生，他吃校餐的概率 = 15/(20+15) = 15/35 = 3/7。

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8. Line Graphs and Trends | 折线图与趋势

A line graph plots the noon temperature over 10 days: Day1 22°C, Day2 24°C, Day3 23°C, Day4 25°C, Day5 27°C, Day6 26°C, Day7 28°C, Day8 29°C, Day9 30°C, Day10 28°

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