KS3 AQA Statistics: Unit Test Mock Paper Walkthrough | KS3 AQA 统计：单元测试模拟卷解析

📚 KS3 AQA Statistics: Unit Test Mock Paper Walkthrough | KS3 AQA 统计：单元测试模拟卷解析

Mock papers are one of the best ways to prepare for your KS3 Statistics assessment. They help you get familiar with the style of questions, practise applying your skills under timed conditions, and identify any topics you need to revisit. In this walkthrough, we will go through a typical AQA unit test mock paper step by step, explaining the thinking behind each answer and the key statistical concepts being tested. Whether you are aiming for a high score or just want to build confidence, this detailed review will support your revision.

模拟卷是备考 KS3 统计评估的最好方式之一。它们帮助你熟悉题型，在限时条件下练习运用所学技能，并找出需要复习的知识点。在这份解析中，我们将逐步梳理一套典型的 AQA 单元测试模拟卷，解释每道题背后的思路和所考查的关键统计概念。无论你的目标是取得高分还是想增强信心，这份详细讲解都会为你的复习提供支持。

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1. Data Collection and Question Types | 数据收集与问题类型

The first section of the mock paper looks at designing a questionnaire. You are asked: “A school wants to find out how students travel to school. Write one closed question and one open question that could be used in a survey.”

模拟卷的第一部分考查问卷设计。题目问：“一所学校想了解学生的上学交通方式。写出一个可用于调查的封闭式问题和一个开放式问题。”

A closed question gives respondents a limited set of choices. For example, “How do you usually travel to school?” with options: walk, bus, car, bicycle, other. This type of question is easy to analyse because responses fall into clear categories.

封闭式问题给出一组有限的选项。例如：“你通常怎样上学？”选项：步行、公共汽车、私家车、自行车、其他。这类问题易于分析，因为回答可以归入明确的类别。

An open question allows people to answer in their own words, providing more detail. For example, “What do you think could improve your journey to school?” This might reveal ideas the researcher had not considered, but it is harder to summarise.

开放式问题允许人们用自己的话回答，提供更多细节。例如：“你认为可以怎样改善你上学的路程？”这可能会揭示研究者没考虑到的想法，但更难总结。

In the mark scheme, you would also be expected to comment on avoiding leading or biased wording, such as “Don’t you agree that walking is the healthiest way to get to school?” which pushes respondents toward a particular answer. Good questionnaire design keeps questions neutral and clear.

在评分标准中，考生还需要评论避免引导性或带有偏见的措辞，例如“你是否认为步行是最健康的上学方式？”这样的问题会推动受访者选择特定答案。好的问卷设计保持问题中立和清晰。

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2. Sampling Methods | 抽样方法

This question presents a scenario: “A library wants to know which new books to buy. The librarian asks the first 30 people who walk in on Monday morning. Explain one advantage and one disadvantage of this sampling method.”

本题给出一个情景：“一家图书馆想知道应该购买哪些新书。图书管理员问了周一早上最早进来的30个人。解释这种抽样方法的一个优点和一个缺点。”

The method described is convenience sampling. An advantage is that it is quick and easy to carry out – the librarian does not need to select a random sample or wait for returns from a longer survey. It can give rough ideas at a low cost.

描述的方法是便利抽样。一个优点是快捷简便——图书管理员不需要选取随机样本或等待更长时间的调查反馈。它能以较低的代价获得初步想法。

A disadvantage is that the sample may not be representative. People visiting on Monday morning might be retired adults or parents with young children, and their preferences could differ from those of evening or weekend users. This can introduce bias, making the results less reliable for the whole library community.

一个缺点是样本可能不具代表性。周一早上来的可能是退休人士或带小孩的家长，他们的偏好可能与晚间或周末借阅者不同。这可能引入偏差，使得结果对整个图书馆读者的可靠性降低。

To improve the study, a random sample of library members could be chosen, or a stratified sample to ensure different age groups are included proportionally. For KS3, you need to know that random sampling gives everyone an equal chance of being selected, which reduces bias.

为了改进研究，可以随机选择图书馆会员，或采用分层抽样确保不同年龄组按比例包含。对于 KS3，你需要知道随机抽样给予每个人均等的入选机会，从而减少偏差。

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3. Stem-and-Leaf Diagrams | 茎叶图

The mock paper includes a stem-and-leaf diagram showing the marks (out of 50) scored by a class on a maths test. The diagram is presented as follows: stem (tens) and leaf (units) – 1 | 5 7 8, 2 | 0 1 3 4 6, 3 | 2 5 8 9, 4 | 1 2 5. A key tells you that 1|5 represents 15.

模拟卷中有一个茎叶图，显示了一个班级在数学测验中的分数（满分 50）。茎叶图如下：茎（十位）叶（个位）——1 | 5 7 8, 2 | 0 1 3 4 6, 3 | 2 5 8 9, 4 | 1 2 5。图例说明 1|5 代表 15。

One question asks you to find the median. Count the total number of leaves: there are 3+5+4+3 = 15 values. Ordered from smallest to largest, the 8th value will be the median. Counting through the stem-and-leaf, we have 15,17,18,20,21,23,24,26 – so the 8th value is 26. Therefore, the median mark is 26 out of 50.

一道问题要求找到中位数。数一数叶子的总数：3+5+4+3 = 15 个数值。从小到大排列，第 8 个值就是中位数。依次数茎叶图：15,17,18,20,21,23,24,26 ——所以第 8 个值是 26。因此中位数分数是 26 分（满分 50）。

Next, find the range. The smallest value is 15, and the largest is 45. Range = largest – smallest = 45 – 15 = 30 marks. The range tells us how spread out the scores are.

接着求极差。最小值是 15，最大值是 45。极差 = 最大值 – 最小值 = 45 – 15 = 30 分。极差告诉我们分数的分散程度。

Another common task is to identify the mode (the most frequent value). In this diagram, no leaf appears more than twice, but you might spot that 25 appears once (leaf 5 on stem 2) and 32 appears once. Actually, check carefully: stem 3 has leaves 2,5,8,9 – so 32,35,38,39. There is no repeated leaf, so there is no mode, or you can say the data set has no mode. In KS3, it is acceptable to state that there is no mode if no value occurs more than once.

另一个常见任务是找到众数（最常见的值）。在这个茎叶图中，没有重复出现超过一次的叶子。你可以说这组数据没有众数。在 KS3 中，如果没有数值出现超过一次，可以说明没有众数。

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4. Mean, Median, Mode and Range | 平均值、中位数、众数和极差

A table in the mock paper records the number of books read by 20 students in one month. The data are summarised: 3, 5, 2, 7, 5, 4, 6, 3, 5, 8, 2, 5, 3, 4, 6, 5, 4, 7, 3, 5. You are asked to calculate the mean number of books read, and then explain when the median might be more useful than the mean.

模拟卷中的一个表格记录了 20 名学生在一个月内阅读的书籍数量。数据汇总为：3, 5, 2, 7, 5, 4, 6, 3, 5, 8, 2, 5, 3, 4, 6, 5, 4, 7, 3, 5。要求计算所读书籍的平均数，然后解释什么时候中位数可能比平均数更有用。

To find the mean, add all the values: first total the frequencies. Sum = 3+5+2+7+5+4+6+3+5+8+2+5+3+4+6+5+4+7+3+5 = 92. There are 20 students. Mean = 92 ÷ 20 = 4.6 books. The mean tells us the average number of books read, but it is affected by the high value of 8.

求平均数，先把所有数值加起来：总和 = 92。有 20 名学生。平均数 = 92 ÷ 20 = 4.6 本书。平均数告诉我们平均读书数量，但它会受到较高值 8 的影响。

Now find the median. Order the data: 2,2,3,3,3,3,4,4,4,5,5,5,5,5,5,6,6,7,7,8. The median is the middle value; since n=20, it is the average of the 10th and 11th values. The 10th value is 5, and the 11th is also 5, so median = 5 books. The mode is 5 as well, because 5 occurs most frequently (six times). The range = 8 – 2 = 6 books.

现在找中位数。排序：2,2,3,3,3,3,4,4,4,5,5,5,5,5,5,6,6,7,7,8。中位数是中间值；因为 n=20，取第 10 和 11 个值的平均数。第 10 个值是 5，第 11 个也是 5，所以中位数为 5 本书。众数也是 5，因为 5 出现最频繁（6 次）。极差 = 8 – 2 = 6 本书。

The median is more useful than the mean when there are extreme values (outliers). The value 8 is a bit higher than the rest, pulling the mean up to 4.6, while the median is a robust 5. If we had an extremely high value, say 30 books, the mean would shift dramatically, but the median would stay much closer to the typical number of books read. The median is not influenced by outliers.

当存在极端值（离群值）时，中位数比平均数更有用。数值 8 略高于其他数据，把平均数拉高到 4.6，而中位数保持在稳健的 5。如果有一个极高值，比如 30 本书，平均数会大幅偏移，而中位数则会保持更接近典型的读书数量。中位数不受离群值影响。

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5. Interpreting Bar Charts and Pie Charts | 解读条形图和饼图

A dual bar chart is presented in the mock paper comparing the favourite subjects of Year 7 and Year 8 students. The chart shows that more students in both years prefer PE and Art, but Year 8 has a noticeably higher number choosing Science than Year 7. You are asked to make two comparisons and use numbers from the chart to support them.

模拟卷中有一个复式条形图，对比了七年级和八年级学生最喜欢的科目。图表显示，两个年级都有更多学生喜欢体育和美术，但八年级选择科学的人数明显多于七年级。问题要求进行两个比较，并用图表中的数字支持比较。

First comparison: In Year 7, 35 students chose PE as their favourite subject, while in Year 8, 40 students chose PE. This shows that PE is slightly more popular among Year 8 students by 5 students.

第一个比较：七年级有 35 名学生选择体育作为最喜欢的科目，八年级有 40 名。这说明体育在八年级学生中略受欢迎，多了 5 人。

Second comparison: For Science, Year 7 had 20 students, whereas Year 8 had 38 students – a difference of 18 students. This suggests that interest in Science grows significantly between Year 7 and Year 8. When interpreting charts, always refer to specific numbers or scales, not just “it is higher”.

第二个比较：在科学方面，七年级有 20 人，而八年级有 38 人——相差 18 人。这表明从七年级到八年级，对科学的兴趣大幅增长。解读图表时，要始终引用具体的数字或比例，而不只是说“它更高”。

A pie chart might also appear, showing the proportion of votes for different school lunch options. If the sector for “pizza” has an angle of 120 degrees, you know it represents 120/360 = 1/3 of the total votes. If there were 240 students surveyed, the number choosing pizza would be 1/3 × 240 = 80 students. Understanding this relationship between angle, fraction and quantity is essential.

饼图也可能出现，显示不同学校午餐选项的投票比例。如果“比萨”的扇区角度为 120 度，你知道它代表总投票的 120/360 = 1/3。如果调查了 240 名学生，那么选择比萨的人数是 1/3 × 240 = 80 名。理解角度、分数和数量之间的关系至关重要。

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6. Drawing Pie Charts (Calculating Angles) | 绘制饼图（计算角度）

This question provides a frequency table of pets owned by 60 families: Dog – 24 families, Cat – 15, Fish – 9, Other – 12. You must calculate the angle for each sector in a pie chart.

本题给出一个 60 户家庭拥有宠物的频数表：狗 – 24 家，猫 – 15，鱼 – 9，其他 – 12。你需要计算饼图中每个扇区的角度。

The formula is: sector angle = (frequency ÷ total frequency) × 360°. For dog, (24 ÷ 60) × 360° = 0.4 × 360° = 144°. For cat, (15 ÷ 60) × 360° = 0.25 × 360° = 90°. For fish, (9 ÷ 60) × 360° = 0.15 × 360° = 54°. For other, (12 ÷ 60) × 360° = 0.2 × 360° = 72°. Check the sum: 144° + 90° + 54° + 72° = 360°, which is exact.

公式为：扇区角度 =（频数 ÷ 总频数）× 360°。狗：(24 ÷ 60) × 360° = 0.4 × 360° = 144°。猫：(15 ÷ 60) × 360° = 0.25 × 360° = 90°。鱼：(9 ÷ 60) × 360° = 0.15 × 360° = 54°。其他：(12 ÷ 60) × 360° = 0.2 × 360° = 72°。检查总和：144° + 90° + 54° + 72° = 360°，完全吻合。

After calculating the angles, you would be asked to draw the pie chart accurately, labelling each sector clearly. Even if the drawing is not perfectly to scale, the angles should be realistic. In an exam, you may not need to draw the whole chart; sometimes just completing one missing sector or calculating the angles is enough.

计算好角度后，你会被要求精确绘制饼图，并清晰地给每个扇区加标签。即使绘图不十分精确，角度也应切合实际。在考试中，你可能不需要画出整个图表；有时候只需完成一个缺失的扇区或计算角度即可。

A common error is forgetting to multiply by 360° or using the wrong total. Always double-check that your fractions add up to 1 (24/60+15/60+9/60+12/60=60/60=1) before converting to degrees.

一个常见错误是忘了乘以 360° 或使用了错误的总数。在转换为度数之前，务必检查分数加起来是否为 1（24/60+15/60+9/60+12/60=60/60=1）。

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7. Scatter Graphs and Correlation | 散点图与相关性

A scatter graph in the mock paper plots the number of hours spent revising against the test score (out of 50) for 15 students. The points generally rise from bottom left to top right, forming a pattern. You are asked to describe the correlation and then estimate the score for a student who revised for 6 hours, using a line of best fit.

模拟卷中的散点图描绘了 15 名学生复习小时数与考试分数（满分 50）的关系。点阵大致从左下至右上上升，形成一个模式。问题要求描述相关性，并利用最佳拟合线估计复习了 6 小时的学生的分数。

There is a positive correlation between revision time and test score: as one variable increases, the other tends to increase. The correlation looks fairly strong because the points are quite close to a straight line. There are no obvious outliers.

复习时间与考试分数之间存在正相关：当一个变量增加时，另一个也趋向增加。相关性看起来相当强，因为点很靠近一条直线。没有明显的离群点。

To estimate the score for 6 hours, draw a line of best fit that goes through the middle of the points, having roughly the same number of points above and below. Extend the line to x=6, then read across to the y-axis. Suppose the line gives a value of about 42. So, a student revising 6 hours is predicted to score around 42 out of 50. The line of best fit is an estimate, so your answer may vary within a reasonable range.

要估计复习 6 小时的分数，可以画一条最佳拟合线，使其穿过点的中心，两侧的点数大致相等。将线延伸至 x=6，然后读取 y 轴数值。假设这条线给出的值大约为 42。因此，复习 6 小时的学生预计得分约 42 分（满分 50）。最佳拟合线是估计值，所以你的答案在一个合理范围内波动即可。

If the question asks for the meaning of an outlier, you could say: “An outlier is a point that lies far away from the general trend.” For instance, a student who revised for 8 hours but only scored 20 would be an outlier, perhaps due to factors like illness or a misunderstanding of the material.

如果问题问离群值的含义，你可以说：“离群值是一个远离总体趋势的点。”例如，一个复习了 8 小时却只得了 20 分的学生就是离群值，可能因为生病或对内容理解有误等因素。

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8. Basic Probability | 基础概率

This section tests your understanding of probability on a scale from 0 to 1, including using fractions, decimals and percentages. A question states: “A bag contains 3 red, 2 blue and 5 green counters. One counter is taken at random. What is the probability it is blue?”

本部分考察你对 0 到 1 概率标度的理解，包括使用分数、小数和百分比。有一题：“一个袋子里有 3 个红色、2 个蓝色和 5 个绿色筹码。随机取出一个，它是蓝色的概率是多少？”

Total number of counters = 3+2+5 = 10. Number of blue counters = 2. Probability (blue) = number of favourable outcomes ÷ total number of possible outcomes = 2/10 = 1/5. This can also be written as 0.2 or 20%.

筹码总数 = 3+2+5 = 10。蓝色筹码数量 = 2。概率（蓝色） = 有利结果数 ÷ 可能结果总数 = 2/10 = 1/5。也可以写成 0.2 或 20%。

A follow-up question might ask: “What is the probability that the counter is not green?” There are 5 green counters, so not green = 3 red + 2 blue = 5 counters. P(not green) = 5/10 = 1/2. Alternatively, P(green) = 5/10, so P(not green) = 1 – P(green) = 1 – 0.5 = 0.5. This demonstrates the complementary rule.

后续问题可能会问：“取出筹码不是绿色的概率是多少？”有 5 个绿色筹码，那么非绿色 = 3 红 + 2 蓝 = 5 个。P（非绿）= 5/10 = 1/2。或者，P（绿）= 5/10，因此 P（非绿）= 1 – P（绿）= 1 – 0.5 = 0.5。这展示了互补规则。

You might also see questions about mutually exclusive events. For example, can a single counter be both red and blue? No, so the events are mutually exclusive, and we simply add probabilities: P(red or blue) = P(red)+P(blue)=3/10+2/10=5/10.

你还可能看到关于互斥事件的题目。例如，一个筹码可能同时是红色和蓝色吗？不能，所以这些事件是互斥的，我们只需将概率相加：P（红或蓝）= P（红）+P（蓝）= 3/10+2/10=5/10。

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9. Relative Frequency and Experimental Probability | 相对频率与实验概率

A question describes an experiment: “Charlie flips a biased coin 100 times and gets 65 heads. Based on this experiment, what is the relative frequency of tails?” This connects to experimental probability.

一道题描述了一个实验：“查理抛一枚偏置硬币 100 次，得到 65 次正面。基于这个实验，反面的相对频率是多少？”这联系到实验概率。

Number of tails = 100 – 65 = 35. Relative frequency of tails = number of tails ÷ total trials = 35 ÷ 100 = 0.35. The experimental probability of getting tails with this biased coin is estimated as 0.35, but it is not the theoretical probability. The more trials you do, the closer the relative frequency is likely to get to the true probability (the law of large numbers).

反面次数 = 100 – 65 = 35。反面相对频率 = 反面次数 ÷ 总试验次数 = 35 ÷ 100 = 0.35。用这枚偏置硬币得到反面的实验概率估计为 0.35，但它并非理论概率。试验次数越多，相对频率越可能接近真实概率（大数定律）。

If the coin were fair, you would expect about 50 heads. Because the result is 65, the coin seems biased towards heads. However, 100 tosses is not enormous, so there is still some randomness. You might be asked: “If Charlie tossed the coin 1000 times, would you expect the relative frequency of heads

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