📚 Year 9 WJEC Statistics: Unit Test Mock Paper Walkthrough | 九年级 WJEC 统计：单元模拟卷解析
This article provides a full walkthrough of a mock unit test for Year 9 WJEC Statistics. It is designed to help students review key topics, understand common question formats, and practise exam techniques. The mock paper covers data collection, charts, averages, probability, and more, all aligned to the WJEC KS3 specification.
本文详细解析了一份九年级 WJEC 统计单元测试模拟卷。旨在帮助学生复习重要知识点，熟悉常见题型，并练习考试技巧。模拟卷涵盖数据收集、图表、平均数、概率等内容，完全贴合 WJEC KS3 大纲要求。
1. Mock Paper Overview | 模拟卷概况
The mock test consists of eight questions, each targeting a specific skill area. A total of 50 marks are available, and students should aim to complete it within 45 minutes. The questions progress from basic data handling to more applied probability scenarios. Below we break down each question and provide step-by-step solutions, along with examiner tips.
模拟卷包含八道题，每道题针对一项特定技能。满分 50 分，建议学生在 45 分钟内完成。题目从基础数据处理逐步过渡到应用性概率情境。下面我们将逐一解析每道题，给出分步解答和考官技巧。
2. Question 1 – Collecting Data and Types of Data | 数据收集与数据类型
This question carries 6 marks and tests the understanding of data collection methods, data classification, and the census vs sample debate. Part (a) asks for a suitable method to investigate lunch preferences. A questionnaire is the most efficient way to gather responses from a large year group. An interview or observation could also be accepted, but a well-designed questionnaire allows for easy collation of categorical data like favourite dishes.
本题 6 分，考查数据收集方法、数据分类以及对普查与抽样的理解。第 (a) 小题要求提出一种调查午餐偏好的方法。问卷是收集大量学生回答最有效的方式。访谈或观察也可接受，但设计良好的问卷便于整理最喜爱菜肴等类别数据。
Part (b) requires one advantage and one disadvantage of a questionnaire. An advantage is that it can be distributed to many students simultaneously, saving time. However, a disadvantage is that some students may not return the questionnaire or may give insincere answers, leading to non-response bias. It is always good to link the advantage to the context – here, getting a wide range of lunch views quickly matches the school’s need.
第 (b) 小题要求给出问卷的一个优点和一个缺点。优点是能够同时发给大量学生，节省时间。但缺点是部分学生可能不交回问卷或给出不真实答案，导致无回应偏误。将优点与情境结合总是有益的——此处快速获得广泛的午餐意见符合学校需求。
For part (c), ‘favourite main dish’ is clearly categorical, as it names a category. ‘Number of portions consumed’ is numerical discrete because it counts whole portions. A ‘rating from 1 to 5’ is technically ordinal, but in Year 9 WJEC it is treated as numerical discrete. Always check that students do not mistakenly label ratings as categorical just because numbers are used as codes.
第 (c) 小题中，‘最喜爱的主菜’显然是类别数据，因为它描述类别。‘消费的份数’是数值离散型，因为它计算整份数。‘1 到 5 的评分’在技术上是有序的，但在九年级 WJEC 中被视为数值离散型。务必提醒学生不要仅仅因为数字用作编码，就将评分误标为类别数据。
Part (d) asks why a census of all Year 9 students might be better than a sample. A census includes every member of the population, so the results are completely representative of Year 9 lunch preferences. There is no sampling error, and the school can be confident in the accuracy of the data. However, a census is more time-consuming and resource-intensive, but the question only asks for a reason why it is ‘better’ in terms of data quality.
第 (d) 小题询问为什么对全体九年级学生进行普查可能比抽样更好。普查涵盖总体中的每位成员，因此结果能完全代表九年级午餐偏好。不存在抽样误差，学校对数据准确性充满信心。虽然普查耗时耗力，但题目仅要求从数据质量角度说明为何‘更好’。
3. Question 2 – Bar Chart and Frequency Table | 条形图与频数表
Question 2 (8 marks) presents a frequency table of favourite sports among Year 9 students: Football 15, Netball 10, Rugby 8, Other 7. Students must complete the table by calculating the total (40) and identify the mode. The mode is Football, as it has the highest frequency. It is essential to stress that the mode is the category name, not the frequency number.
第二题 (8 分) 给出九年级学生最喜爱运动的频数表：足球 15、篮网球 10、橄榄球 8、其他 7。学生需要计算总数 (40) 并找出众数。众数是足球，因其频数最高。必须强调众数是类别名称，而非频数数值。
Students are then asked to draw a bar chart on a provided grid. A common mistake is to forget to label axes or to make bars unequal in width. The horizontal axis should show the four sport categories, and the vertical axis should be scaled from 0 to 16 in even steps. Bars must be separated by equal gaps, and a title should be added, such as ‘Favourite Sports of Year 9 Students’.
随后要求学生用提供的网格绘制条形图。常见错误是忘记标注坐标轴，或条形宽度不一致。横轴应显示四类运动，纵轴应从 0 到 16 均匀标度。条形之间需有相等间隔，并添加标题，如‘九年级学生最喜爱运动’。
Finally, they calculate the fraction of students who chose Rugby. 8 out of 40 simplifies to 1/5. A simple probability reminder: always express the fraction in its simplest form. Some students might write 8/40 and lose a mark for not simplifying.
最后计算选择橄榄球的学生比例。8/40 化简为 1/5。简单概率提醒：务必以最简分数表示。部分学生可能会写 8/40，因未化简而丢分。
4. Question 3 – Pie Charts and Angle Calculations | 饼图与角度计算
In Question 3 (7 marks), an animal shelter recorded the types of animals rescued: 14 dogs, 10 cats, 6 rabbits, and 4 others. The total number of animals is 34. To draw a pie chart, students must first find the angle for each sector using the formula (frequency/total) × 360 degrees. For dogs: (14/34) × 360 ≈ 148.2°, which can be rounded to 148°. For cats: (10/34) × 360 ≈ 105.9° → 106°. Rabbits: (6/34) × 360 ≈ 63.5° → 64°. Others: (4/34) × 360 ≈ 42.4° → 42°. The sum of angles should be 360°.
第三题 (7 分) 中，某动物收容所记录了救助动物种类：狗 14 只、猫 10 只、兔子 6 只、其他 4 只。动物总数为 34。绘制饼图前，学生需先使用公式 (频数/总数) × 360 度计算每个扇区的角度。狗：(14/34) × 360 ≈ 148.2°，可四舍五入为 148°；猫：(10/34) × 360 ≈ 105.9° → 106°；兔子：(6/34) × 360 ≈ 63.5° → 64°；其他：(4/34) × 360 ≈ 42.4° → 42°。角度之和应为 360°。
When drawing, a protractor is used to measure each angle accurately. Labelling each sector or providing a key is essential. A follow-up part asks: if the number of rabbits doubled to 12, explain how the angle would change. The new total becomes 40, so the rabbit angle becomes (12/40) × 360 = 108°. This shows that multiplying the frequency does not simply multiply the angle; the total also changes, making proportional reasoning crucial.
绘图时需用量角器精确量取各角度。必须为每个扇区添加标签或图例。后续小题询问：如果兔子数量翻倍至 12 只，解释角度将如何变化。新的总数变为 40，因此兔子角度为 (12/40) × 360 = 108°。这表明频数翻倍并不简单地使角度翻倍，总数也会变化，比例推理至关重要。
5. Question 4 – Mean, Median, Mode, and Range | 平均数、中位数、众数和极差
Question 4 (8 marks) gives test scores: 45, 52, 61, 58, 45, 64, 45. The first task is to calculate the mean. Sum the values: 45+52+61+58+45+64+45 = 370. Divide by 7 to get a mean of 52.857…, which rounds to 52.9 or 53 depending on context. For WJEC, often one decimal place is acceptable. The mode is 45, as it appears three times. The median is found by ordering the data: 45, 45, 45, 52, 58, 61, 64. The middle value (4th) is 52. The range is maximum – minimum = 64 – 45 = 19.
第四题 (8 分) 给出测试分数：45、52、61、58、45、64、45。首先计算平均数。求和：45+52+61+58+45+64+45 = 370。除以 7 得到平均数 52.857…，根据上下文可四舍五入为 52.9 或 53。WJEC 通常接受一位小数。众数是 45，因为它出现了三次。将数据排序：45、45、45、52、58、61、64。中位数（第 4 个）为 52。极差 = 最大值 – 最小值 = 64 – 45 = 19。
The question then asks which average best represents the data. Since the mean (52.9) is very close to the median (52), both are representative, but the presence of repeated low scores makes the mean slightly lower. In this case, the median is robust to the repeated 45s. However, because the data has no extreme outliers, the mean is also acceptable. A strong answer explains that the median is not pulled down by multiple low scores, while the mean considers all values.
随后题目询问哪个平均数最能代表该数据集。由于平均数 (52.9) 与中位数 (52) 非常接近，两者均有代表性，但重复的低分使平均数略低。此时中位数对重复的 45 具有稳健性。但由于数据没有极端异常值，平均数也可接受。一个有力的答案会解释中位数不会被多个低分拉低，而平均数考虑了所有数值。
6. Question 5 – Scatter Graphs and Correlation | 散点图与相关性
This 6-mark question provides a table linking outdoor temperature (in °C) with ice cream sales (in units). The temperatures: 14, 17, 20, 23, 26, 29. Corresponding sales: 22, 30, 38, 42, 50, 55. Students plot the points on a given grid. The completed scatter graph shows a clear positive correlation – as temperature increases, ice cream sales tend to rise. Words like ‘strong’, ‘positive’, and ‘linear’ can be used to describe it.
本题 6 分，给出室外温度 (°C) 与冰淇淋销量 (单位) 的表格。温度：14、17、20、23、26、29；对应销量：22、30、38、42、50、55。学生在网格上描点。完成的散点图显示出明显的正相关——温度升高，冰淇淋销量也趋于上升。可用‘强’、‘正’、‘线性’等词描述。
Part (c) expects students to use the trend to predict sales at 22°C. By drawing a line of best fit (a straight line passing through the general centre of points), they can interpolate a sales value of approximately 41 units. It is important to show working by drawing dashed lines on the graph. Part (d) warns against extrapolation for a temperature like 35°C, as the relationship may change beyond the data range – melting issues or stock limits could flatten sales, so predictions become unreliable.
第 (c) 小题要求学生利用趋势预测 22°C 时的销量。通过绘制最佳拟合线（大致穿过点群中心的直线），可内插出约 41 单位的销量。必须在图上用虚线显示工作过程。第 (d) 小题提醒不要外推预测 35°C 时的销量，因为关系在数据范围外可能改变——冰淇淋融化或库存限制可能使销量趋于平缓，因此预测变得不可靠。
7. Question 6 – Basic Probability and Tree Diagrams | 基础概率与树状图
Question 6 (9 marks) involves a bag containing 5 red balls, 3 blue balls, and 2 green balls. Total balls = 10. Part (a) asks for P(red) = 5/10 = 1/2, P(blue) = 3/10, P(green) = 2/10 = 1/5. Always simplify fractions where possible. Part (b) introduces the scenario of picking two balls without replacement. A tree diagram is required to show all possible outcomes for the two picks.
第六题 (9 分) 涉及一个装有 5 个红球、3 个蓝球和 2 个绿球的袋子。总球数 = 10。第 (a) 小题求 P(红) = 5/10 = 1/2，P(蓝) = 3/10，P(绿) = 2/10 = 1/5。务必尽可能化简分数。第 (b) 小题引入不放回地连续取两次球的场景，需要树状图展示两次取球的所有可能结果。
The tree begins with three branches for the first pick, labelled with probabilities. For the second pick, because the ball is not replaced, the total decreases to 9, and the number of that colour reduces by one. For example, after picking a red, the probabilities for the second pick become: 4/9 red, 3/9 blue, 2/9 green. Students must label all branches clearly. The probability of two reds is calculated by multiplying along the branch: (5/10) × (4/9) = 20/90 = 2/9.
树状图从第一次抽取的三个分支开始，标注概率。对于第二次抽取，由于球不放回，总数减为 9，且该颜色的数量减少 1。例如，第一次抽到红球后，第二次抽球概率变为：红 4/9、蓝 3/9、绿 2/9。学生必须清晰标注所有分支。两次都抽到红球的概率通过将分支上的概率相乘计算：(5/10) × (4/9) = 20/90 = 2/9。
8. Question 7 – Sample Space and Combined Events | 样本空间与组合事件
This 5-mark probability question uses three tosses of a fair coin. Part (a) asks to list the sample space systematically. Using a table or tree, students should obtain 8 equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Part (b) requires the probability of exactly two heads. The favourable outcomes are HHT, HTH, THH – three out of eight, so P(exactly 2 heads) = 3/8.
这道 5 分的概率题涉及抛一枚标准硬币三次。第 (a) 小题要求系统列出样本空间。利用表格或树状图，学生应得到 8 种等可能结果：HHH, HHT, HTH, HTT, THH, THT, TTH, TTT。第 (b) 小题求恰好两次正面的概率。有利结果为 HHT, HTH, THH —— 8 种中的 3 种，因此 P(恰好两次正面) = 3/8。
Part (c) asks for the probability of at least one tail. It is quicker to use the complement rule: 1 – P(no tails) = 1 – P(HHH) = 1 – 1/8 = 7/8. Alternatively, students could count all outcomes except HHH and verify they get 7. A common mistake is to misinterpret ‘at least one tail’ as exactly one tail, which would be wrong. Careful reading of the command word is essential.
第 (c) 小题求至少一次反面的概率。使用补集规则更快：1 – P(无反面) = 1 – P(HHH) = 1 – 1/8 = 7/8。学生也可数出除 HHH 外的所有结果并验证得到 7。常见错误是将‘至少一次反面’误解为恰好一次反面，这将导致错误。仔细读懂指令词至关重要。
9. Question 8 – Grouped Frequency Table and Estimated Mean | 分组频数表与估算平均数
Question 8 (9 marks) provides grouped data on the time (minutes) students spent on homework: 0 ≤ t < 10 (frequency 3), 10 ≤ t < 20 (7), 20 ≤ t < 30 (6), 30 ≤ t < 40 (4). Total frequency is 20. To estimate the mean, we first find the midpoint of each interval: 5, 15, 25, 35. Multiply each midpoint by its frequency: 5×3=15, 15×7=105, 25×6=150, 35×4=140. Summing gives Σfx = 410. Estimated mean = 410/20 = 20.5 minutes.
第八题 (9 分) 提供学生做作业时长 (分钟) 的分组数据：0 ≤ t < 10 (频数 3)、10 ≤ t < 20 (7)、20 ≤ t < 30 (6)、30 ≤ t < 40 (4)。总频数为 20。为估算平均数，首先找出各区间的中点：5、15、25、35。将各中点乘以对应频数：5×3=15，15×7=105，25×6=150，35×4=140。求和得 Σfx = 410。估算平均数 = 410/20 = 20.5 分钟。
The median group is found by locating the position of the (20+1)/2 = 10.5th value. Cumulative frequencies: 3, 10, 16, 20. The 10.5th value lies in the second interval (10 ≤ t < 20). Part (d) asks why the estimated mean might differ from the true mean. Because we use midpoints to represent all values in each interval, the actual data are unlikely to be symmetrically spread around the midpoint. Some students may have spent more time near the upper end, which would cause the true mean to be higher. The estimate is only an approximation.
中位数组通过定位第 (20+1)/2 = 10.5 个值来找到。累积频数：3、10、16、20。第 10.5 个值落在第二个区间 (10 ≤ t < 20)。第 (d) 小题询问为什么估算平均数可能与真实平均数不同。因为我们用中点代表区间内的所有值，而实际数据不太可能围绕中点均匀分布。有些学生可能用时偏向上限，导致真实平均数更高。估算值仅为近似值。
10. Common Mistakes and Examiner Tips | 常见错误与考官提示
Across the mock paper, several pattern of errors appear. In data collection questions, students often confuse categorical and numerical data. Remember, if a number is used merely as a label (e.g., jersey number 7), it is categorical. Ratings that express an ordered scale are treated as numerical. For bar charts, forgetting to start the vertical axis at 0 can distort comparisons. Always check that the scale runs from 0, and bars are separate.
整份模拟卷中，出现几类常见错误。在数据收集题中，学生常混淆类别数据和数值数据。记住，若数字仅用作标签（如 7 号球衣），则为类别数据。表达有序尺度的评分被视为数值数据。对于条形图，纵轴未从 0 开始可能扭曲比较。务必确保比例从 0 开始，且条形分开。
When calculating the mean from grouped data, using the wrong midpoint is a frequent slip. Double-check the interval is correctly interpreted, especially if it says 0– (meaning 0 ≤ t < 10). In probability without replacement, always adjust the denominator and numerator for the second event. The most common failure is to treat the draws as independent, leading to fractions like (5/10)² for two reds, which ignores the change in total. Draw the tree and label carefully.
计算分组数据平均数时，使用错误的中点是一个常见失误。仔细检查区间是否解释正确，特别是当它写为 0– (表示 0 ≤ t < 10)。在不放回概率中，务必调整第二次事件的分子和分母。最常见的错误是将抽取视为独立事件，导致用 (5/10)² 计算两次红球概率，忽略了总数的变化。画出树状图并仔细标注。
Lastly, when interpreting scatter graphs, do not assume correlation implies causation. Just because temperature and ice cream sales are correlated does not mean temperature causes sales directly; other factors like day of the week could influence both. Use cautious language such as ‘there is a positive relationship’ rather than ‘temperature makes sales go up’.
最后，解读散点图时，不要假定相关性意味着因果关系。温度与冰淇淋销量相关，并不意味着温度直接导致销量变化；其他因素如星期几也可能同时影响两者。使用谨慎的语言，如‘存在正相关关系’，而非‘温度使销量上升’。
11. Final Summary and Revision Tips | 总结与复习
Published by TutorHao | Year 9 统计 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply