Year 7 CIE Statistics: Unit Test Mock Paper Analysis | 7年级CIE统计：单元测试模拟卷解析

📚 Year 7 CIE Statistics: Unit Test Mock Paper Analysis | 7年级CIE统计：单元测试模拟卷解析

This mock paper analysis walks you through a complete Year 7 CIE Statistics unit test, question by question. Each section shows a typical exam-style question, explains the key concepts, and provides a full model solution. By working through these worked examples, you will sharpen your skills in data handling, chart construction, averages, and probability – all essential for success in Lower Secondary Checkpoint and beyond.

这份模拟卷解析将带你逐步完成一套完整的 7 年级 CIE 统计单元测试，逐题精讲。每个小节展示一道典型考题，解释核心概念，并提供完整的标准答案。通过这些例题的详细讲解，你将提升数据处理、图表绘制、平均数计算和概率分析的技能，为 Lower Secondary Checkpoint 及后续学习打下扎实基础。

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1. Question 1: Identifying Data Types | 第1题：识别数据类型

Question 1 on the mock paper lists four variables: ‘favourite subject’, ‘height in centimetres’, ‘number of cars in a household’ and ‘type of mobile phone’. Students must classify each one as categorical or numerical, and for numerical variables state whether they are discrete or continuous.

模拟卷的第1题列出了四个变量：”最喜欢的科目”、”身高（厘米）”、”家庭拥有汽车的数量” 和 “手机类型”。学生需要将每一个变量分类为类别数据或数值数据，并指出数值数据是离散型的还是连续型的。

Part (a): The variable ‘favourite subject’ is categorical because it describes a non-numerical quality. Even though subjects can be coded with numbers, they represent categories, not measurements.

第 (a) 部分：”最喜欢的科目” 是类别数据，因为它描述的是非数值的性质。尽管科目可以用数字编码，但数字代表的仍是类别，而不是测量值。

Part (b): ‘Height in centimetres’ is numerical continuous. Height is measured on a continuous scale and can take any value within a range, such as 142.5 cm.

第 (b) 部分：”身高（厘米）” 是数值连续型数据。身高是在一个连续尺度上测量得到的，可以在某个区间内取任意值，例如 142.5 厘米。

Part (c): ‘Number of cars in a household’ is numerical discrete. It is obtained by counting whole cars; you cannot have 1.5 cars in a household.

第 (c) 部分：”家庭拥有汽车的数量” 是数值离散型数据。它是通过计数整辆汽车得到的，一个家庭不可能拥有 1.5 辆汽车。

Part (d): ‘Type of mobile phone’ is categorical. The brand or model acts as a label and does not give a numerical measurement.

第 (d) 部分：”手机类型” 是类别数据。品牌或型号起到标签的作用，并不能提供数值测量。

Common pitfall: Some students confuse numerical codes (like jersey numbers) with numerical data. Always ask: does the number represent a count or a measurement? If it is just a label, treat it as categorical.

常见误区：一些学生容易将数值编码（例如球衣号码）与数值数据混淆。请始终问自己：这个数字代表的是计数还是测量？如果它只是一个标签，那么就把它当作类别数据来处理。

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2. Question 2: Tally Charts and Frequency Tables | 第2题：划记表与频数表

Question 2 gives the raw data of 30 students’ favourite fruits: Apple, Banana, Apple, Orange, Apple, Banana, Apple, Apple, Orange, Banana, Banana, Apple, Grape, Orange, Apple, Banana, Apple, Grape, Orange, Banana, Banana, Apple, Grape, Apple, Banana, Orange, Apple, Banana, Orange, Apple. The task is to complete a frequency table using tally marks.

第2题给出了 30 名学生最喜爱水果的原始数据：Apple, Banana, Apple, Orange, … 任务是使用划记法完成一张频数表。

Begin by listing the categories ‘Apple’, ‘Banana’, ‘Orange’, ‘Grape’ in the first column. For each data point, draw one tally mark in the appropriate row. After four vertical lines, the fifth mark is drawn diagonally across to form a ‘gate’ of five – this makes counting by fives much quicker.

首先在第一列列出类别 “Apple”、”Banana”、”Orange”、”Grape”。每遇到一个数据点，就在相应的行中画一道划记。画满四条竖线后，第五道斜线穿过前四条线形成一个 “五柱门” —— 这样方便五个五个地快速计数。

After completing the tally, count the groups: Apple shows 12 marks (two gates of five plus two extra), Banana 9, Orange 6 and Grape 3. Double-check that the total adds up to 30. The frequency column should always sum to the total number of data values.

完成划记后，统计各组数目：Apple 出现了 12 次（两个五柱门再加两条），Banana 9 次，Orange 6 次，Grape 3 次。一定要重新核对总数是否等于 30。频数列的总和必须始终等于数据的总个数。

An exam tip: label your frequency column with ‘Frequency’ and add a brief title such as ‘Favourite Fruits of 30 Students’. Tidy presentation can earn method marks.

考试技巧：给你的频数列标上 “Frequency” 的标题，并添加一个简短的标题，例如 “30 名学生最喜爱的水果”。整洁的作答能够获得方法分。

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3. Question 3: Drawing and Reading a Bar Chart | 第3题：绘制与解读条形图

Question 3 provides the frequency table from Question 2 and asks you to: (a) draw a vertical bar chart, (b) state which fruit is the most popular, and (c) work out how many more students chose Apple than Orange.

第3题给出了第2题中的频数表，要求：(a) 绘制一幅垂直条形图，(b) 说明哪种水果最受欢迎，(c) 计算出选择 Apple 的学生比选择 Orange 的多多少人。

For the bar chart, use graph paper and a sharp pencil. Draw and label the horizontal axis with the fruit categories and the vertical axis with frequency. Choose a sensible scale – here each large square could represent 2 students, so the highest bar (Apple, frequency 12) reaches 6 large squares. Bars must be of equal width with equal gaps between them.

绘制条形图时，使用方格纸和削尖的铅笔。画出并标记横轴（水果类别）和纵轴（频数）。选择一个合理的尺度——在这里，每个大格可以代表 2 名学生，因此最高的条形（Apple，频数 12）应达到 6 个大格。条形必须等宽，条与条之间保持相等的间距。

Part (b): The most popular fruit is Apple because it has the highest frequency of 12. Part (c): Number choosing Apple = 12, Orange = 6. The difference is 12 − 6 = 6, so 6 more students chose Apple over Orange.

第 (b) 部分：最受欢迎的水果是 Apple，因为它具有最高的频数 12。第 (c) 部分：选择 Apple 的人数为 12，Orange 为 6。两者之差为 12 − 6 = 6，因此选择 Apple 的学生比 Orange 多 6 人。

Always check that your bar heights match the frequencies exactly and that you do not join the bars or draw them touching – this distinguishes a bar chart from a histogram, which will be introduced in later years.

务必检查条形的高度是否精确对应频数，并且不要将条形连起来或让它们相互接触——这一点将条形图与今后将要学习的直方图区分开。

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4. Question 4: Pie Chart Construction | 第4题：饼图的绘制

Question 4 uses the same data of 30 students. It asks you to calculate the angle for each sector and then construct a pie chart to represent the data. The central angle formula is used to convert each frequency into degrees.

第4题使用同样的 30 名学生的数据。题目要求计算每个扇区的角度，然后绘制一幅饼图来表示数据。使用中心角公式将每个频数转换为度数。

Angle = (frequency ÷ total) × 360°

For Apple: frequency = 12, total = 30, so angle = (12 ÷ 30) × 360° = 144°. For Banana: (9 ÷ 30) × 360° = 108°. Orange: (6 ÷ 30) × 360° = 72°. Grape: (3 ÷ 30) × 360° = 36°. Check that the four angles sum to 144° + 108° + 72° + 36° = 360°, which confirms the calculation is correct.

Apple：频数 = 12，总数 = 30，因此角度 = (12 ÷ 30) × 360° = 144°。Banana：(9 ÷ 30) × 360° = 108°。Orange：(6 ÷ 30) × 360° = 72°。Grape：(3 ÷ 30) × 360° = 36°。检查四个角度之和是否为 144° + 108° + 72° + 36° = 360°，这可以确认计算正确。

To draw the pie chart, first draw a circle using compasses. Draw a vertical radius to serve as the starting line. Place the protractor on the centre and measure the first angle (144°) anticlockwise. Draw the next radius. Repeat for each sector. Label each sector with the fruit name and either the frequency or the percentage. Add a clear title.

绘制饼图时，先用圆规画一个圆。画一条竖直的半径作为起始线。将量角器放在圆心处，逆时针量取第一个角度（144°）。画出下一条半径。对每个扇区重复此步骤。在每个扇区上标注水果名称，以及频数或百分比。最后加上清晰的标题。

If you are asked to calculate percentages, remember that (frequency ÷ total) × 100 converts a frequency directly into a percentage. For Apple this is (12 ÷ 30) × 100 = 40%. This serves as a quick plausibility check – the largest sector should correspond to the largest percentage.

如果被要求计算百分比，记住 (频数 ÷ 总数) × 100 可直接将频数转换为百分比。以 Apple 为例，(12 ÷ 30) × 100 = 40%。这可以作为一种快速的合理性检查——最大的扇区应当对应最大的百分比。

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5. Question 5: Mean, Median, Mode and Range | 第5题：平均数、中位数、众数和极差

Question 5 presents the following set of data: 12, 15, 10, 18, 14, 15, 13. Students are asked to calculate the mean, median, mode and range, showing all steps.

第5题给出了以下一组数据：12, 15, 10, 18, 14, 15, 13。要求学生计算平均数、中位数、众数和极差，并展示所有步骤。

Mean: Add all values: 12 + 15 + 10 + 18 + 14 + 15 + 13 = 97. Count how many values there are: 7. Divide the sum by the count: 97 ÷ 7 = 13.857… This can be rounded to 13.9 or left as 13.9 (1 d.p.).

平均数：将所有数值相加：12 + 15 + 10 + 18 + 14 + 15 + 13 = 97。数出数值的个数：7。用总和除以个数：97 ÷ 7 = 13.857… 可以四舍五入到 13.9（保留一位小数）。

Median: First, rewrite the list in order: 10, 12, 13, 14, 15, 15, 18. There are 7 numbers, so the median is the 4th value. The 4th value is 14, so median = 14.

中位数：首先，将数据按顺序重排：10, 12, 13, 14, 15, 15, 18。共有 7 个数，因此中位数是第 4 个数值。第 4 个数值是 14，所以中位数 = 14。

Mode: Look for the value that appears most often. 15 appears twice, while all others appear once. Therefore, the mode is 15. If no value repeated, the data set would have no mode.

众数：寻找出现最频繁的数值。15 出现了两次，其他数值均只出现一次。因此，众数是 15。如果没有数值重复，则该数据组没有众数。

Range: Range = largest value − smallest value = 18 − 10 = 8. The range gives a simple measure of how spread out the data are.

极差：极差 = 最大值 − 最小值 = 18 − 10 = 8。极差提供了一种衡量数据分散程度的简单方式。

Examiners look for full working: the ordered list for the median, the sum for the mean, and the two values used for the range. A final statement summarising all four statistics often earns a communication mark.

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Published by TutorHao | Year 7 统计 Revision Series | aleveler.com

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