📚 Year 9 Cambridge Statistics: Mock Unit Test Walkthrough | 剑桥九年级统计：单元测试模拟卷解析
This mock test walkthrough is designed to help Year 9 students consolidate their understanding of core statistical concepts in the Cambridge curriculum. Covering data collection, graphical representation, measures of central tendency, spread, and basic probability, each question is unpacked with clear working steps and explanations. Whether you are preparing for a classroom unit test or simply revising, this session will build your confidence in handling real exam-style questions.
这份模拟卷解析旨在帮助九年级学生巩固剑桥课程中的核心统计概念。内容涵盖数据收集、图形表示、集中趋势量数、离散程度和基础概率，每道题都配有详细解题步骤与解释。不论你是在为单元测验做准备，还是日常复习，本次讲解都将提升你应对真实考试题型的信心。
1. Overview of the Mock Test | 模拟卷概述
The mock paper consists of six compulsory questions, carrying a total of 50 marks. The questions progress from simple data classification and frequency tables to pie chart calculations, summary statistics, box plots, and probability problems. Students are allowed 45 minutes and should present all working clearly. The structure mirrors typical Cambridge progression tests, ensuring familiarity with the command words and expected presentation.
本模拟卷共六道必答题，总分 50 分。题目难度逐步递进：从简单的数据分类和频数表，到饼图计算、概括统计量、箱线图，再到概率问题。建议用时 45 分钟，所有解题步骤需书写清晰。试卷结构参考了典型的剑桥阶段测试，有助于学生熟悉指令词和作答规范。
| Question | Topic | Marks |
|---|---|---|
| Q1 | Data Types & Collection Methods | 6 |
| Q2 | Frequency Table & Bar Chart | 8 |
| Q3 | Pie Chart Calculations | 8 |
| Q4 | Mean, Median, Mode & Range | 10 |
| Q5 | Quartiles & Box Plot | 10 |
| Q6 | Basic Probability | 8 |
2. Question 1: Data Types & Collection Methods | 第1题：数据类型与收集方法
In this question, students are given a list of variables: ‘favourite colour’, ‘height in cm’, ‘number of pets’, ‘time taken to run 100 m’, and ‘rating of a film (1-5)’. They must classify each as qualitative or quantitative, and for quantitative ones, further distinguish between discrete and continuous data. The second part asks students to suggest a suitable data collection method for a survey on favourite colours among classmates.
本题给出了一系列变量：“最喜欢的颜色”“身高（厘米）”“宠物数量”“100 米跑步用时”“电影评分（1-5）”。要求学生判断每个变量是定性还是定量数据，对于定量数据还需进一步区分离散与连续。第二部分要求为调查班级同学最喜欢的颜色推荐合适的数据收集方法。
The variable ‘favourite colour’ is qualitative (categorical) because it describes a quality or characteristic. ‘Height’ is quantitative and continuous – it can be measured on a scale and take any value within a range. ‘Number of pets’ is quantitative but discrete, as it can only take whole numbers. ‘Time taken to run 100 m’ is quantitative continuous; time is measured and can include decimals. ‘Rating of a film (1-5)’ is often treated as discrete quantitative data in statistics, although some might see it as ordered categorical. In a Cambridge context, it is safest to label it as quantitative discrete because the values are limited integers.
变量“最喜欢的颜色”是定性（分类）数据，因为它描述一种性质或特征。“身高”属于定量连续数据——可在一定范围内取任意值。“宠物数量”是定量但属离散数据，因为只能是整数。“100 米跑步用时”是定量连续数据，时间可以包含小数。“电影评分（1-5）”在统计中常被视为定量离散数据，尽管有人可能认为它是定序分类数据。在剑桥考试中，最稳妥的做法是将其归为定量离散数据，因为取值是有限的整数。
For the data collection method, a questionnaire or a quick show-of-hands survey would be suitable. A questionnaire allows responses to be recorded systematically, while keeping the survey anonymous can encourage honest answers. Observing or experimenting would not be appropriate for this type of opinion-based data.
至于数据收集方法，问卷或快速举手调查都很合适。问卷能够系统记录回答，匿名填写还能鼓励诚实作答。观察或实验不适合这类基于观点的数据。
Key rule: Qualitative → categorical; Quantitative discrete → counts; Quantitative continuous → measurements.
3. Question 2: Frequency Table & Bar Chart | 第2题：频率表与条形图
The question provides raw data of 25 students’ preferred sports: Football, Basketball, Football, Tennis, Basketball, Cricket, Football, Tennis, Basketball, Tennis, Cricket, Football, Cricket, Football, Basketball, Tennis, Basketball, Basketball, Cricket, Football, Football, Tennis, Cricket, Cricket, Football. Students construct a frequency table and then draw a bar chart, ensuring axes are labelled and bars have equal width with gaps.
本题给出了 25 名学生最喜爱运动的原始数据：足球、篮球、足球、网球、篮球、板球、足球、网球、篮球、网球、板球、足球、板球、足球、篮球、网球、篮球、篮球、板球、足球、足球、网球、板球、板球、足球。学生需要制作频数表，然后绘制条形图，确保轴标签完整，条形等宽且之间留有空隙。
To construct the frequency table, tally marks are used to count each occurrence. The resulting frequencies are: Football 8, Basketball 6, Tennis 5, Cricket 6. The total is 25, which should be checked. The table should have three columns: Sport, Tally, Frequency.
制作频数表时，使用画记符号统计每个项目的出现次数。最终频数为：足球 8，篮球 6，网球 5，板球 6。总数为 25，需要检查。表格应包含三列：运动项目、画记、频数。
When drawing the bar chart, the horizontal axis represents the sport categories, and the vertical axis shows frequency, using a scale that goes at least to 8. All bars must be of equal width, separated by gaps, and each bar must be the correct height corresponding to its frequency. A title such as ‘Favourite Sports of 25 Students’ completes the graph.
绘制条形图时，横轴表示运动项目类别，纵轴表示频数，刻度至少要到 8。所有条形宽度一致且保持间隔，每个条形的高度必须与对应的频数相符。最后加上如图表标题“25 名学生最喜爱的运动”。
4. Question 3: Pie Chart Calculations | 第3题：饼图计算
Students are given a table showing the favourite colours of 30 pupils: Red 12, Blue 8, Green 5, Yellow 5. The task is to calculate the angle for each sector and to construct an accurate pie chart. Marks are awarded for correct angle calculation and for drawing sectors to the nearest degree.
题目给出 30 名学生最喜爱颜色的表格：红色 12，蓝色 8，绿色 5，黄色 5。要求计算每个扇形的圆心角，并绘制准确的饼图。评分点包括角度计算正确以及扇形角度绘制的精度。
The central angle formula is applied:
Angle = (Frequency ÷ Total Frequency) × 360°
Red: (12 ÷ 30) × 360 = 144°; Blue: (8 ÷ 30) × 360 = 96°; Green: (5 ÷ 30) × 360 = 60°; Yellow: (5 ÷ 30) × 360 = 60°. The sum of angles must equal 360°, which serves as a check.
应用圆心角公式：
角度 = (频数 ÷ 总频数) × 360°
红色：(12 ÷ 30) × 360 = 144°；蓝色：(8 ÷ 30) × 360 = 96°；绿色：(5 ÷ 30) × 360 = 60°；黄色：(5 ÷ 30) × 360 = 60°。角度之和必须为 360°，这可以作为验算。
To draw the pie chart, start with a circle and draw a radius vertically. Use a protractor to measure 144° for Red, then from the new radius measure 96° for Blue, then 60° for Green, and the remainder automatically becomes Yellow. Label each sector with its colour and percentage if required (Red 40%, Blue 26.7%, Green 16.7%, Yellow 16.7%).
绘制饼图时，先画一个圆并画出竖直半径。用量角器先量出 144° 的红色扇形，接着从新半径再量 96° 的蓝色扇形，然后量出 60° 绿色，余下部分自然为黄色。按要求给每个扇形标注颜色和百分比（红 40%，蓝 26.7%，绿 16.7%，黄 16.7%）。
5. Question 4: Mean, Median, Mode & Range | 第4题：平均数、中位数、众数和极差
The data set provided is: 12, 15, 14, 18, 20, 15, 16, 22. Students must calculate the mean, median, mode, and range, showing full working. The question also asks for an interpretation of which average best represents the data.
给出的数据集为：12, 15, 14, 18, 20, 15, 16, 22。要求学生计算平均数、中位数、众数和极差，并展示完整过程。问题还要求解释哪一个平均数最能代表这组数据。
The mean is the sum divided by the number of values.
Mean = (Σx) ÷ n
Sum = 12 + 15 + 14 + 18 + 20 + 15 + 16 + 22 = 132. n = 8. Mean = 132 ÷ 8 = 16.5.
平均数是总和除以数据个数。
平均数 = Σx ÷ n
总和 = 12 + 15 + 14 + 18 + 20 + 15 + 16 + 22 = 132。n = 8。平均数 = 132 ÷ 8 = 16.5。
To find the median, first arrange the data in ascending order: 12, 14, 15, 15, 16, 18, 20, 22. Since there are 8 values (even), the median is the average of the 4th and 5th terms. (15 + 16) ÷ 2 = 15.5. The median is 15.5.
求中位数时，先将数据升序排列：12, 14, 15, 15, 16, 18, 20, 22。因为有 8 个值（偶数），中位数是第 4 项和第 5 项的平均数。(15 + 16) ÷ 2 = 15.5。中位数为 15.5。
The mode is the most frequent value; here 15 appears twice, all others once. Mode = 15.
众数是出现频次最高的值；这里 15 出现两次，其余均一次。众数 = 15。
The range is the difference between the maximum and minimum: Range = 22 − 12 = 10.
极差是最大值与最小值之差：极差 = 22 − 12 = 10。
The mean (16.5) is pulled up slightly by the value 22, whereas the median (15.5) and mode (15) are closer to the centre of the majority. In this case, the median or mode might better describe the typical value because the mean is affected by the relatively high outlier. However, with only 8 values, all measures provide useful insights.
平均数 16.5 被数值 22 略微拉高，而中位数 15.5 和众数 15 更接近多数数据的中心。在这个例子中，中位数或众数可能更能代表典型值，因为平均数受到相对较大值的影响。但仅凭 8 个数据，所有量数都能提供有用信息。
6. Question 5: Quartiles and Box Plot | 第5题：四分位数与箱线图
For the data set: 5, 7, 8, 10, 12, 14, 15, 18, 20, 22 (ten values), students determine the lower quartile (Q1), median (Q2), upper quartile (Q3), and then construct a box-and-whisker plot. They are also asked to identify any outliers using the 1.5 × IQR rule.
数据集：5, 7, 8, 10, 12, 14, 15, 18, 20, 22（10 个值），要求学生确定下四分位数 Q1、中位数 Q2、上四分位数 Q3，然后绘制箱线图。还需用 1.5 × IQR 规则识别异常值。
The data are already ordered. For n = 10, median position is the average of the 5th and 6th terms: (12 + 14) ÷ 2 = 13, so Q2 = 13.
数据已按序排列。n = 10，中位数位置在第 5 项与第 6 项的平均数：(12 + 14) ÷ 2 = 13，因此 Q2 = 13。
The lower half is the first five values: 5, 7, 8, 10, 12. The median of this half is the 3rd term: Q1 = 8.
下半部分为前五个值：5, 7, 8, 10, 12。该部分的中位数为第 3 项：Q1 = 8。
The upper half is the last five values: 14, 15, 18, 20, 22. Its median is the 3rd term: Q3 = 18.
上半部分为后五个值：14, 15, 18, 20, 22。其中位数为第 3 项：Q3 = 18。
Interquartile range IQR = Q3 − Q1 = 18 − 8 = 10. The outlier boundaries are: lower fence = Q1 − 1.5 × IQR = 8 − 15 = −7; upper fence = Q3 + 1.5 × IQR = 18 + 15 = 33. Since all data values lie between −7 and 33, there are no outliers.
四分位距 IQR = Q3 − Q1 = 18 − 8 = 10。异常值边界：下边界 = Q1 − 1.5 × IQR = 8 − 15 = −7；上边界 = Q3 + 1.5 × IQR = 18 + 15 = 33。所有数据值均介于 −7 与 33 之间，故无异常值。
To draw the box plot, a scale is drawn. The box extends from Q1 (8) to Q3 (18), with a line at the median (13). Whiskers extend from the box to the minimum (5) and maximum (22). A neat, labelled box plot is required.
绘制箱线图时，先画出数轴。箱体从 Q1 (8) 延伸至 Q3 (18)，在 Q2 (13) 处画中线。须线从箱体延伸至最小值 (5) 和最大值 (22)。要求绘制整洁、标注清晰的箱线图。
7. Question 6: Basic Probability | 第6题：基础概率
The scenario involves a fair six-sided die rolled once. Students find the probability of specific outcomes and combine events. An additional part uses a bag containing 3 red, 5 blue, and 2 green marbles to test complementary events and ‘or’ probability.
本题场景为掷一枚公平的六面骰子一次。学生需计算特定结果的概率，并进行事件组合。附加部分使用一个装有 3 个红球、5 个蓝球和 2 个绿球的袋子，考察互补事件和“或”概率。
For a fair die, the sample space is {1, 2, 3, 4, 5, 6}, each with probability 1/6. (a) P(rolling a 4) = 1/6. (b) P(rolling an even number) = number of even outcomes (2,4,6) / 6 = 3/6 = 1/2. (c) P(rolling a number greater than 4) = outcomes {5, 6} gives 2/6 = 1/3.
对于公平骰子，样本空间为 {1, 2, 3, 4, 5, 6}，每个结果的概率为 1/6。(a) P(掷出 4) = 1/6。(b) P(掷出偶数) = 偶数结果数 (2,4,6) / 6 = 3/6 = 1/2。(c) P(掷出大于 4 的数) = 结果集 {5, 6} 给出 2/6 = 1/3。
With the bag of marbles, total marbles = 3 + 5 + 2 = 10. P(red) = 3/10, P(blue) = 5/10 = 1/2, P(green) = 2/10 = 1/5. For complementary events: P(not blue) = 1 − P(blue) = 1 − 1/2 = 1/2, which is the same as P(red or green) = 3/10 + 2/10 = 5/10 = 1/2.
对于袋子里的球，总球数 = 3 + 5 + 2 = 10。P(红) = 3/10，P(蓝) = 5/10 = 1/2，P(绿) = 2/10 = 1/5。互补事件：P(非蓝) = 1 − P(蓝) = 1 − 1/2 = 1/2，这与 P(红或绿) = 3/10 + 2/10 = 5/10 = 1/2 一致。
For mutually exclusive events, students should recall that P(A or B) = P(A) + P(B). In this context, drawing a red or a blue marble: P(red or blue) = 3/10 + 5/10 = 8/10 = 4/5. The use of fractions in simplest form is expected.
对于互斥事件，学生应记住 P(A 或 B) = P(A) + P(B)。在此背景下，抽到红色或蓝色弹珠：P(红或蓝) = 3/10 + 5/10 = 8/10 = 4/5。答案应化为最简分数。
8. Common Mistakes and Exam Tips | 常见错误与备考建议
Many students lose marks by confusing discrete and continuous data – remember, if you can have decimals, it’s normally continuous. Also, in pie charts, forgetting to divide by total frequency before multiplying by 360 is a frequent error. For averages, always reorder the data for the median; don’t just pick middle positions in the original list.
许多学生因混淆离散和连续数据而丢分——记住，如果能出现小数，通常为连续数据。此外，在饼图中，经常出现的错误是忘记先除以总频数再乘以 360。计算平均数时，务必为求中位数而重新排序；不要在原列表中直接取中间位置。
In box plots, ensure you use the correct quartiles; mixing up Q1 and Q3 can entirely distort the box. For probability, always check that probabilities sum to 1 for complementary events, and express answers as simplified fractions unless otherwise specified.
在箱线图中，务必使用正确的四分位数；混淆 Q1 和 Q3 会完全歪曲图形。在概率中，要检查互补事件的概率之和是否为 1，除非另有说明，答案都应以最简分数形式呈现。
Regular practice with past paper questions, reading the command words carefully (e.g. ‘calculate’, ‘draw’, ‘explain’), and showing all working clearly will maximise your marks. Time management is key – spend proportionate time on questions with higher mark allocations.
定期练习历年真题，仔细阅读指令词（如“计算”“绘制”“解释”），并清晰地展示所有解题步骤，能够最大化你的得分。时间管理至关重要——在高分值题目上需要投入相应比例的时间。
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