📚 Year 7 CAIE Statistics: Unit Test Mock Paper Walkthrough | Year 7 CAIE 统计：单元测试模拟卷解析
This article provides a complete walkthrough of a Year 7 CAIE Statistics mock unit test. Each question is presented with a clear solution and key learning points. Work through all ten questions to revise data collection, representation, averages, and range — all aligned with the Cambridge Lower Secondary curriculum.
本文提供一套 Year 7 CAIE 统计单元模拟测试题的完整解析。每道题目均配有清晰的解题过程和关键知识点讲解。请逐一完成这十道题目，以复习数据收集、数据展示、平均数和极差等内容——所有题目均贴合剑桥初中课程要求。
1. Finding the Mode | 找出众数
The shoe sizes of ten students are recorded: 4, 5, 5, 6, 4, 5, 7, 6, 5, 8. Identify the mode of this data set.
十名学生的鞋码记录如下：4, 5, 5, 6, 4, 5, 7, 6, 5, 8。请找出这组数据的众数。
To find the mode, count how many times each number appears. The number 4 appears twice, 5 appears four times, 6 appears twice, 7 appears once, and 8 appears once. The mode is the value that occurs most often, which is 5.
找众数时，需要数出每个数值出现的次数。数字 4 出现了两次，5 出现了四次，6 出现了两次，7 出现了一次，8 出现了一次。众数是出现次数最多的数值，因此众数为 5。
2. Finding the Median | 找出中位数
The test scores of seven pupils are: 11, 14, 9, 18, 13, 10, 16. Work out the median score.
七名学生的测验分数为：11, 14, 9, 18, 13, 10, 16。请计算中位数分数。
First, write the numbers in ascending order: 9, 10, 11, 13, 14, 16, 18. With seven values, the median is the middle number, which is the 4th value — 13. So the median score is 13.
首先将数字按升序排列：9, 10, 11, 13, 14, 16, 18。一共有七个数值，中位数是位于正中间的数，即第 4 个数——13。因此中位数分数为 13。
3. Calculating the Mean | 计算平均数
A survey asks eight friends how many pets they own. The responses are: 2, 0, 1, 3, 2, 4, 1, 3. Calculate the mean number of pets.
一项调查询问了八位朋友各自养了多少只宠物，回答如下：2, 0, 1, 3, 2, 4, 1, 3。请计算平均宠物数量。
Add all the values together: 2 + 0 + 1 + 3 + 2 + 4 + 1 + 3 = 16. There are 8 responses. Divide the total by the number of values: 16 ÷ 8 = 2. The mean number of pets is 2.
将所有数值相加：2 + 0 + 1 + 3 + 2 + 4 + 1 + 3 = 16。一共有 8 个回答。用总和除以数值的个数：16 ÷ 8 = 2。平均宠物数量为 2 只。
4. Finding the Range | 找出极差
The maximum daily temperatures (in °C) for one week are: 12, 15, 11, 18, 14, 13, 17. What is the range of these temperatures?
一周内每日最高气温（单位：°C）为：12, 15, 11, 18, 14, 13, 17。这些气温的极差是多少？
The range is the difference between the largest and smallest values. The highest temperature is 18 °C and the lowest is 11 °C. Subtract: 18 – 11 = 7. The range is 7 °C.
极差是最大值与最小值之间的差值。最高气温为 18 °C，最低气温为 11 °C。相减得：18 – 11 = 7。极差为 7 °C。
5. Tally Charts and Frequency Tables | 划记表与频数表
Twenty students were asked their favourite colour. The raw data is: Red, Blue, Blue, Green, Red, Red, Yellow, Blue, Green, Red, Blue, Yellow, Red, Green, Blue, Red, Blue, Green, Red, Blue. Complete a frequency table and state which colour is the mode.
向二十名学生询问了他们最喜欢的颜色，原始数据为：红、蓝、蓝、绿、红、红、黄、蓝、绿、红、蓝、黄、红、绿、蓝、红、蓝、绿、红、蓝。请完成频数表，并说出众数是哪种颜色。
- Red: 7 students
- Blue: 8 students
- Green: 3 students
- Yellow: 2 students
The colour with the highest frequency is Blue (8 students), so Blue is the mode.
- 红色：7 名学生
- 蓝色：8 名学生
- 绿色：3 名学生
- 黄色：2 名学生
频数最高的颜色是蓝色（8 名学生），因此众数为蓝色。
6. Interpreting a Bar Chart | 解读条形图
A bar chart shows the number of books read by five students in a month: Anna 5, Ben 3, Chloe 7, Daniel 4, Emily 6. Use the chart to answer: Who read the most books? How many more books did Chloe read than Ben?
一幅条形图显示了五名学生在一个月内阅读书籍的数量：Anna 5 本，Ben 3 本，Chloe 7 本，Daniel 4 本，Emily 6 本。请根据图表回答：谁读的书最多？Chloe 比 Ben 多读了多少本书？
From the bar chart, the tallest bar is for Chloe, with a frequency of 7. So Chloe read the most books. Ben read 3 books. The difference is 7 – 3 = 4. Chloe read 4 more books than Ben.
从条形图中可以看出，最高的条形对应 Chloe，频数为 7。因此 Chloe 读的书最多。Ben 读了 3 本书。两人的差值为 7 – 3 = 4。Chloe 比 Ben 多读了 4 本书。
7. Pie Chart Angles and Interpretation | 饼图角度与解读
In a survey, 30 students choose their favourite fruit: Apple 10, Banana 12, Orange 8. Calculate the angle for each sector if the data is shown in a pie chart. Then state the fraction of students who chose Banana.
在一项调查中，30 名学生选择了自己最喜欢的水果：苹果 10 人，香蕉 12 人，橙子 8 人。如果要用饼图表示，请计算每个扇形的角度，并写出选择香蕉的学生所占的比例。
Total number of students = 30. Each student represents 360° ÷ 30 = 12°. Apple: 10 × 12° = 120°. Banana: 12 × 12° = 144°. Orange: 8 × 12° = 96°. The fraction for Banana is 12/30, which simplifies to 2/5.
学生总人数为 30。每名学生对应的角度为 360° ÷ 30 = 12°。苹果：10 × 12° = 120°。香蕉：12 × 12° = 144°。橙子：8 × 12° = 96°。选择香蕉的人数为 12/30，化简为 2/5。
8. Comparing Data Sets Using Mean and Range | 利用平均数与极差比较数据
Two groups of students took the same test. Group A scored: 10, 12, 14, 16, 18. Group B scored: 13, 13, 14, 15, 15. Calculate the mean and range for each group. Which group performed more consistently?
两组学生参加了同一场测验。A 组分数为：10, 12, 14, 16, 18。B 组分数为：13, 13, 14, 15, 15。请分别计算每组分数的平均数和极差。哪一组的成绩更稳定？
Group A mean: (10+12+14+16+18) ÷ 5 = 70 ÷ 5 = 14. Range: 18 – 10 = 8. Group B mean: (13+13+14+15+15) ÷ 5 = 70 ÷ 5 = 14. Range: 15 – 13 = 2. Both groups have the same mean, but Group B has a much smaller range, indicating its scores are more consistent.
A 组平均数：(10+12+14+16+18) ÷ 5 = 70 ÷ 5 = 14。极差：18 – 10 = 8。B 组平均数：(13+13+14+15+15) ÷ 5 = 70 ÷ 5 = 14。极差：15 – 13 = 2。两组平均数相同，但 B 组的极差小得多，这表明 B 组的成绩更加稳定。
9. Interpreting a Line Graph | 解读折线图
A line graph shows the monthly rainfall (in mm) from January to June: Jan 50, Feb 45, Mar 55, Apr 60, May 70, Jun 65. Describe the trend from January to May. Between which two consecutive months was the greatest increase in rainfall?
一幅折线图显示了一月至六月的月降雨量（单位：毫米）：一月 50，二月 45，三月 55，四月 60，五月 70，六月 65。请描述从一月到五月的趋势。在哪两个连续月份之间降雨量增长最大？
From January to May, the general trend is upward, though there is a slight dip in February. The greatest increase occurs between April (60 mm) and May (70 mm), with a rise of 10 mm. Other increases are 5 mm or less.
从一月到五月，整体趋势是上升的，尽管二月略有下降。降雨量增长最大的是四月（60 毫米）到五月（70 毫米），增长了 10 毫米。其他相邻月份的增长都不超过 5 毫米。
10. Choosing the Right Average | 选择合适的平均数
A small company has five employees with annual salaries: £18,000, £20,000, £22,000, £24,000, and £26,000. The owner earns £120,000. If the owner’s salary is included, which measure of average — mean or median — better represents the typical salary? Explain why.
一家小公司有五名员工，年薪分别为：£18,000, £20,000, £22,000, £24,000, £26,000。老板的年薪为 £120,000。如果计入老板的薪水，平均数和中位数哪个更能代表典型的薪水？请解释原因。
Including the owner, the mean is (£18,000+£20,000+£22,000+£24,000+£26,000+£120,000) ÷ 6 = £230,000 ÷ 6 ≈ £38,333, which is much higher than most salaries. The median remains at £23,000 (middle of £22,000 and £24,000). The median is the better measure because it is not affected by the extreme value of £120,000 and gives a more realistic idea of a typical salary.
计入老板的薪水后，平均数为 (£18,000+£20,000+£22,000+£24,000+£26,000+£120,000) ÷ 6 = £230,000 ÷ 6 ≈ £38,333，远高于多数人的薪水。中位数则保持在 £23,000（£22,000 和 £24,000 的中间值）。中位数是更合适的度量，因为它不受极端值 £120,000 的影响，能更真实地反映典型的薪水水平。
Published by TutorHao | Statistics Revision Series | aleveler.com
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