Year 7 Cambridge Statistics: Unit Test Mock Paper Walkthrough | 剑桥 7 年级统计：单元测试模拟卷解析

📚 Year 7 Cambridge Statistics: Unit Test Mock Paper Walkthrough | 剑桥 7 年级统计：单元测试模拟卷解析

This walkthrough covers a full Unit Test mock paper for Year 7 Cambridge Statistics. We explain each question step by step, building your skills in data collection, frequency tables, bar charts, pie charts, measures of centre, and basic probability. Use this article to review key concepts and see exactly how marks are awarded.

本文全面解析一份剑桥 7 年级统计单元测试模拟卷，逐步讲解每一道题目，帮助你巩固数据收集、频率表、条形图、饼图、平均数、中位数和概率等核心技能。通过这份解析，你可以清晰理解评分标准和解题思路。

1. Classifying Data | 数据分类

Question: A survey collected the following information from students: favourite subject, number of pens in their pencil case, time spent on homework (in minutes), and preferred lunch option. Classify each variable as qualitative or quantitative. For any quantitative variable, state whether it is discrete or continuous.

题目：一项调查收集了学生的以下信息：最喜欢的科目、铅笔盒里笔的数量、作业时间（分钟）和午餐选择。将每个变量分为定性变量或定量变量。如有定量变量，请说明是离散的还是连续的。

Answer and explanation: ‘Favourite subject’ is qualitative because it describes a category or quality, not a number. ‘Number of pens’ is quantitative discrete – it can be counted in whole numbers (you cannot have 2.3 pens). ‘Time spent on homework’ is quantitative continuous – it can take any value within a range (e.g. 34.5 minutes). ‘Preferred lunch option’ is qualitative, just like favourite subject.

答案与解析：“最喜欢的科目”是定性变量，因为它描述的是类别或属性，而非数值。“笔的数量”是离散定量变量——它只能用整数计数（不可能有 2.3 支笔）。“作业时间”是连续定量变量——它可以在一个范围内取任意值（如 34.5 分钟）。“午餐选择”是定性变量，与最喜欢的科目同理。

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2. Constructing a Frequency Table | 构建频率表

Question: The quiz scores of 20 students are recorded below. Construct a frequency table showing each score and its frequency. Scores: 7, 8, 6, 9, 7, 7, 8, 10, 6, 7, 8, 9, 7, 8, 10, 7, 6, 9, 8, 7.

题目：记录了 20 名学生的测验分数。请构建一个频率表，列出每个分数及其频数。分数如下：7、8、6、9、7、7、8、10、6、7、8、9、7、8、10、7、6、9、8、7。

Method: First, list all possible scores from the data set: 6, 7, 8, 9, 10. Then count how many times each score appears. Score 6 appears 3 times, score 7 appears 7 times, score 8 appears 5 times, score 9 appears 3 times, and score 10 appears 2 times. The frequency table should have two columns: Score and Frequency.

方法：首先列出数据集中所有可能的分数：6、7、8、9、10。然后数出每个分数出现的次数。6 分出现 3 次，7 分出现 7 次，8 分出现 5 次，9 分出现 3 次，10 分出现 2 次。频率表应包含两列：分数与频数。

Final frequency table:

最终频率表：

Score	Frequency
6	3
7	7
8	5
9	3
10	2

Always check that the sum of frequencies equals the total number of data values: 3 + 7 + 5 + 3 + 2 = 20. This confirms the table is correct.

请务必检查频数总和等于数据总数：3 + 7 + 5 + 3 + 2 = 20。这可以确认表格无误。

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3. Interpreting a Bar Chart | 解读条形图

Question: A bar chart shows the number of pets owned by students in a class. The bars represent: 0 pets – 5 students, 1 pet – 12 students, 2 pets – 8 students, 3 pets – 3 students, 4 pets – 2 students. Use the chart to find: a) total number of students surveyed; b) the modal number of pets; c) how many more students have 1 pet than have 2 pets.

题目：一幅条形图显示了班级中每位学生拥有宠物的数量。柱形表示：0 只宠物——5 名学生，1 只宠物——12 名学生，2 只宠物——8 名学生，3 只宠物——3 名学生，4 只宠物——2 名学生。利用图表求：a) 受访学生总数；b) 宠物数量的众数；c) 养 1 只宠物的学生比养 2 只宠物的学生多多少人。

Solution: a) Total students = 5 + 12 + 8 + 3 + 2 = 30. b) The mode is the category with the highest frequency; here, the tallest bar is for 1 pet (12 students), so modal number of pets is 1. c) Students with 1 pet = 12, with 2 pets = 8. The difference is 12 − 8 = 4. So 4 more students have 1 pet than 2 pets.

解答：a) 学生总数 = 5 + 12 + 8 + 3 + 2 = 30。b) 众数是频数最高的类别；图中最高柱对应 1 只宠物（12 名学生），因此宠物数量的众数为 1。c) 养 1 只宠物的学生有 12 人，养 2 只宠物的学生有 8 人。相差 12 − 8 = 4。因此养 1 只宠物的学生比养 2 只宠物的多 4 人。

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4. Drawing a Bar Chart Step by Step | 绘制条形图的步骤

Question: Draw a bar chart for the data: Favourite colour – Red: 9, Blue: 15, Green: 6, Yellow: 10. Include axes labels, a title, and equal bar widths with gaps.

题目：为以下数据绘制条形图：最喜欢的颜色——红色：9，蓝色：15，绿色：6，黄色：10。要求标出坐标轴标签、标题，并保证柱宽一致且柱间留有间隔。

Explanation: On graph paper, draw a horizontal axis (x-axis) and a vertical axis (y-axis). Label the x-axis ‘Favourite colour’ and the y-axis ‘Frequency’. Write the title ‘Favourite Colour of Students’ above the graph. Choose a sensible scale for the y-axis: since the maximum frequency is 15, you could use 1 cm = 2 units, making the axis go up to 16. Draw bars for each colour with equal width (e.g. 1 cm) and equal gaps (e.g. 0.5 cm). The height for Red is 9, Blue is 15, Green is 6, Yellow is 10. Finally, check that each bar is labelled with its colour underneath.

解析：在方格纸上画出横轴（x 轴）和纵轴（y 轴）。x 轴标注“最喜欢的颜色”，y 轴标注“频数”。图形上方写上标题“学生最喜欢的颜色”。为 y 轴选择一个合理的刻度：最大频数为 15，可以用 1 cm 代表 2 个单位，坐标轴最高到 16。每种颜色画一个等宽的柱形（如 1 cm 宽），柱间留等距空隙（如 0.5 cm）。红色的柱高为 9，蓝色为 15，绿色为 6，黄色为 10。最后检查每个柱形下方是否标有相应的颜色名称。

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5. Calculating Pie Chart Angles | 计算饼图的角度

Question: The favourite fruit of 30 students is recorded: Apple: 15, Banana: 10, Cherry: 5. Calculate the angle of each sector for a pie chart.

题目：30 名学生最喜欢的水果记录如下：苹果：15，香蕉：10，樱桃：5。计算饼图中每个扇形的圆心角。

Method: The total frequency is 15 + 10 + 5 = 30. The full circle is 360°. The angle for each sector is found by using the formula:

方法：总频数为 15 + 10 + 5 = 30。整圆为 360°。每个扇形的角度由以下公式求得：

Angle = (Frequency ÷ Total Frequency) × 360°

For Apple: (15 ÷ 30) × 360° = 0.5 × 360° = 180°. For Banana: (10 ÷ 30) × 360° = ⅓ × 360° = 120°. For Cherry: (5 ÷ 30) × 360° = ⅙ × 360° = 60°. Check: 180° + 120° + 60° = 360°. The pie chart can now be drawn using a protractor to measure these angles.

苹果：(15 ÷ 30) × 360° = 0.5 × 360° = 180°。香蕉：(10 ÷ 30) × 360° = ⅓ × 360° = 120°。樱桃：(5 ÷ 30) × 360° = ⅙ × 360° = 60°。验证：180° + 120° + 60° = 360°。现在可以用量角器测量这些角度来绘制饼图。

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6. Finding the Mean | 计算平均数

Question: The heights (in cm) of 5 students are: 150, 155, 160, 165, 150. Calculate the mean height.

题目：5 名学生的身高（厘米）分别为：150、155、160、165、150。计算平均身高。

Solution: The mean is found by adding all values together and then dividing by the number of values. Sum of heights = 150 + 155 + 160 + 165 + 150 = 780 cm. Number of students = 5. Therefore, mean = 780 ÷ 5 = 156 cm.

解答：平均数通过将所有数值相加再除以数值的个数得出。身高总和 = 150 + 155 + 160 + 165 + 150 = 780 cm。学生人数 = 5。因此，平均身高 = 780 ÷ 5 = 156 cm。

Mean = (Sum of all values) ÷ (Number of values)

Always include the unit (cm) in your final answer when the data has a unit. The mean height is 156 cm, which represents a typical height for this group.

当数据带有单位时，最终答案务必包含单位（cm）。平均身高为 156 cm，这代表了这组学生的典型身高。

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7. Finding the Median | 求中位数

Question: The ages of 7 children are: 12, 13, 11, 12, 14, 15, 12. Find the median age.

题目：7 个孩子的年龄分别为：12、13、11、12、14、15、12。求出年龄的中位数。

Step-by-step: First, arrange the data in ascending order: 11, 12, 12, 12, 13, 14, 15. There are 7 numbers, so the median is the (7 + 1) ÷ 2 = 4th value in the ordered list. The 4th value is 12. Therefore, the median age is 12 years.

分步解答：首先，将数据按升序排列：11、12、12、12、13、14、15。共有 7 个数，因此中位数是排序后第 (7 + 1) ÷ 2 = 4 个值。第 4 个值是 12。因此，年龄的中位数为 12 岁。

If the data set had an even number of values, you would take the mean of the two middle numbers. Here, with an odd count, the median is simply the middle value. The median is useful because it is not affected by extreme values.

如果数据集的数值个数为偶数，则需要取中间两个数的平均数。本例中数值个数为奇数，中位数就是正中间的那个值。中位数很有用，因为它不受极端值的影响。

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8. Identifying the Mode | 找出众数

Question: Shoe sizes of 8 students are recorded: 5, 6, 5, 7, 6, 5, 8, 6. Find the mode(s).

题目：8 名学生的鞋码记录如下：5、6、5、7、6、5、8、6。求众数。

Explanation: The mode is the value that occurs most frequently. Create a small frequency tally: size 5 appears 3 times, size 6 appears 3 times, size 7 appears 1 time, size 8 appears 1 time. Both size 5 and size 6 have the highest frequency of 3. Therefore, this data set has two modes – it is bimodal. The modes are 5 and 6.

解析：众数是出现次数最多的值。做一个简单的频数统计：5 码出现 3 次，6 码出现 3 次，7 码出现 1 次，8 码出现 1 次。5 码和 6 码都出现了最高频数 3。因此，这个数据集有两个众数——它是双众数的。众数为 5 和 6。

A data set can have one mode, more than one mode, or no mode if all values occur equally. Always state all modes clearly in your answer.

一个数据集可以有一个众数、多个众数，或者如果所有值出现的次数相同则没有众数。在回答中务必清晰列出所有众数。

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9. Calculating the Range | 计算极差

Question: The daily maximum temperatures (in °C) for a week were: 23, 25, 20, 22, 27. Find the range.

题目：一周每日最高气温（°C）记录为：23、25、20、22、27。求极差。

Method: The range measures the spread of the data. It is calculated by subtracting the smallest value from the largest value. First, identify the maximum temperature: 27°C. Minimum temperature: 20°C. Range = 27 − 20 = 7°C.

方法：极差衡量数据的分散程度，由最大值减去最小值计算得出。首先找出最高气温：27°C，最低气温：20°C。极差 = 27 − 20 = 7°C。

Range = Maximum value − Minimum value

The range tells us that the temperatures varied by 7 degrees over the week. A larger range indicates more variability. Remember to include the unit in your answer when applicable.

极差告诉我们这周的气温变化幅度为 7 度。极差越大表示变化程度越大。如适用，请记得在答案中写上单位。

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10. Introduction to Probability | 概率入门

Question: A bag contains 3 red marbles, 2 blue marbles, and 5 green marbles. One marble is picked at random. Find the probability that it is (a) blue, (b) not green, (c) yellow. Express each answer as a fraction in simplest form.

题目：一个袋子里装有 3 个红色弹珠、2 个蓝色弹珠和 5 个绿色弹珠。随机取出一个弹珠。求下列概率：(a) 蓝色，(b) 不是绿色，(c) 黄色。每个答案用最简分数表示。

Solution: Total number of marbles = 3 + 2 + 5 = 10. (a) Probability of blue = number of blue marbles ÷ total = 2 ÷ 10 = 1/5. (b) ‘Not green’ means the marble is red or blue. Number of marbles that are not green = 3 + 2 = 5, so probability = 5 ÷ 10 = 1/2. (c) There are no yellow marbles in the bag, so the number of favourable outcomes is 0. Probability = 0 ÷ 10 = 0. This is an impossible event.

解答：弹珠总数 = 3 + 2 + 5 = 10。(a) 蓝色概率 = 蓝色弹珠数量 ÷ 总数 = 2 ÷ 10 = 1/5。(b)“不是绿色”意味着弹珠是红色或蓝色。非绿色弹珠数量 = 3 + 2 = 5，因此概率 = 5 ÷ 10 = 1/2。(c) 袋中没有黄色弹珠，因此有利结果数为 0。概率 = 0 ÷ 10 = 0。这是一个不可能事件。

Probability = (Number of favourable outcomes) ÷ (Total number of possible outcomes)

Probabilities can be written as fractions, decimals, or percentages, always between 0 (impossible) and 1 (certain). In this question, answers as simplified fractions are expected. Always check your fraction is in its simplest form.

概率可以写成分数、小数或百分数，总是在 0（不可能）到 1（必然）之间。本题要求以最简分数书写答案。请务必检查你的分数是否已化为最简形式。

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Published by TutorHao | Statistics Revision Series | aleveler.com

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