📚 KS3 CIE Statistics: In-depth Analysis of Past Paper Questions | KS3 CIE 统计：历年真题深度解析
Past paper questions are the best way to prepare for KS3 CIE Statistics exams. They reveal how concepts are tested, the types of wording used, and the step-by-step thinking expected. This article selects ten representative real‑exam‑style questions covering bar charts, averages, pie charts, two‑way tables, scatter graphs, questionnaires, comparison, misleading graphs, probability experiments, and a mixed challenge. Each question is broken down with clear reasoning in English and Chinese, helping you build both statistical skills and exam confidence.
历年真题是备考 KS3 CIE 统计的最佳资料。它们能揭示概念的考查方式、常见措辞以及期望的逐步推理过程。本文精选十道具有代表性的真题风格问题，涵盖条形图、平均数、饼图、双向表、散点图、问卷设计、数据比较、误导性图表、概率实验以及一道综合题。每道题都会用中英文清晰的思路进行拆解，帮助你同时提升统计技能和考试信心。
1. Interpreting a Bar Chart | 解读条形图
A bar chart shows the number of books read by each student in one month. The vertical axis is labelled ‘Frequency’, with bars for 0, 1, 2, 3, 4 books. The heights are: 3 students read 0 books, 5 read 1 book, 8 read 2 books, 4 read 3 books, and 2 read 4 books. The question asks: (a) How many students are in the class? (b) What is the modal number of books read?
一幅条形图显示了一个月内每名学生阅读的书籍数量。纵轴标记为“频数”，横轴上的条形对应 0、1、2、3、4 本书。高度分别是：3 名学生读了 0 本，5 名读了 1 本，8 名读了 2 本，4 名读了 3 本，2 名读了 4 本。题目要求：(a) 班级里共有多少名学生？(b) 阅读量的众数是多少？
To find the total number of students, add all the frequencies: 3 + 5 + 8 + 4 + 2 = 22 students. The mode is the category with the highest frequency, which is 2 books with 8 students. Always check that you are reading the bar heights correctly – the scale might not start at zero, though here it does.
要计算学生总数，将所有频数相加：3 + 5 + 8 + 4 + 2 = 22 名学生。众数是频数最高的类别，即 2 本书，对应 8 名学生。务必检查你读取条形高度是否正确——有时刻度可能不从零开始，不过本题是从零开始的。
2. Calculating Mean, Median, Mode and Range | 计算平均数、中位数、众数和极差
A question gives the scores of 9 students in a spelling test: 5, 7, 6, 8, 5, 9, 7, 6, 10. It asks for the mean, median, mode, and range. These four measures summarise the data in different ways, and CIE often tests your ability to choose the best one for a given context.
一道题给出了 9 名学生在拼写测试中的分数：5, 7, 6, 8, 5, 9, 7, 6, 10。要求计算平均数、中位数、众数和极差。这四个统计量以不同的方式概括数据，CIE 经常考查你在特定情境下选择最合适统计量的能力。
First, sort the data: 5, 5, 6, 6, 7, 7, 8, 9, 10. Mean = sum ÷ number of values = (5+5+6+6+7+7+8+9+10) ÷ 9 = 63 ÷ 9 = 7. Median is the middle value in an ordered list: the 5th value is 7. Mode is the most frequent value: both 5, 6, and 7 appear twice, so there are three modes (multimodal). Range = maximum – minimum = 10 – 5 = 5.
首先将数据排序：5, 5, 6, 6, 7, 7, 8, 9, 10。平均数 = 总和 ÷ 数据个数 = (5+5+6+6+7+7+8+9+10) ÷ 9 = 63 ÷ 9 = 7。中位数是有序列表的中间值：第 5 个值是 7。众数是出现次数最多的值：5、6 和 7 都出现了两次，因此有三个众数（多众数）。极差 = 最大值 – 最小值 = 10 – 5 = 5。
When the question also asks which average best represents the data, we note that the mean = median = 7, and the data is fairly symmetrical, so either works well. The mode is less useful here because there are several. In real exams, always justify your choice briefly.
当题目还问哪个平均数最能代表数据时，我们注意到平均数与中位数都等于 7，数据分布较为对称，所以两者都合适。由于这里有多个众数，众数在此用处不大。在实际考试中，要简要说明你的选择理由。
3. Drawing and Interpreting a Pie Chart | 绘制与解读饼图
A survey of 60 students’ favourite fruits gives: Apple 20, Banana 15, Orange 10, Grapes 15. The question asks to calculate the angle for each sector and then draw a pie chart. It may also ask what fraction of students chose Orange.
一项针对 60 名学生最喜欢水果的调查结果如下：苹果 20 人，香蕉 15 人，橙子 10 人，葡萄 15 人。题目要求计算每个扇区的角度，然后画出饼图。可能还会问选择橙子的学生所占的比例是多少。
The total frequency is 60. The angle for a category = (frequency ÷ total) × 360°. Apple: (20/60) × 360° = 120°. Banana: (15/60) × 360° = 90°. Orange: (10/60) × 360° = 60°. Grapes: (15/60) × 360° = 90°. Always check that the angles sum to 360°: 120 + 90 + 60 + 90 = 360°. The fraction for Orange is 10/60, which simplifies to 1/6.
总频数为 60。某一类的角度 = (频数 ÷ 总数) × 360°。苹果：(20/60) × 360° = 120°。香蕉：(15/60) × 360° = 90°。橙子：(10/60) × 360° = 60°。葡萄：(15/60) × 360° = 90°。务必检查角度之和为 360°：120 + 90 + 60 + 90 = 360°。选择橙子的比例是 10/60，化简为 1/6。
When drawing, use a protractor and label each sector clearly. CIE often gives a partially drawn pie chart and asks you to complete it or interpret it. Remember that the size of the sector shows proportion, not the actual frequency unless the total is known.
绘图时，要使用量角器并清晰地标出每个扇区。CIE 经常给出一个部分绘制的饼图，让你补全或解读。请记住，扇区的大小反映的是比例，而非实际频数，除非已知总数。
4. Probability from Two-Way Tables | 双向表中的概率
A two-way table shows the number of boys and girls who prefer football or basketball. Boys: Football 12, Basketball 8. Girls: Football 6, Basketball 10. Questions: A student is chosen at random. Find (a) P(girl), (b) P(likes football), (c) P(boy and likes basketball).
一张双向表显示喜欢足球或篮球的男生和女生人数。男生：足球 12 人，篮球 8 人。女生：足球 6 人，篮球 10 人。问题：随机选择一名学生。求 (a) P(女生)，(b) P(喜欢足球)，(c) P(男生且喜欢篮球)。
First, complete the table with totals: Total boys = 20, total girls = 16, overall total = 36. Total football = 18, total basketball = 18. (a) P(girl) = 16/36 = 4/9. (b) P(football) = 18/36 = 1/2. (c) P(boy and basketball) is directly from the cell: 8/36 = 2/9. Always simplify fractions unless told otherwise.
首先，算出合计：男生总计 20 人，女生总计 16 人，全班总计 36 人。喜欢足球的共 18 人，喜欢篮球的共 18 人。(a) P(女生) = 16/36 = 4/9。(b) P(足球) = 18/36 = 1/2。(c) P(男生且篮球) 直接从对应单元格获取：8/36 = 2/9。除非另有要求，否则务必化简分数。
Some questions will ask for conditional probability, e.g., ‘Given that the student is a boy, find P(likes basketball).’ Here the denominator changes to the total boys (20), so answer = 8/20 = 2/5. Always check what the condition restricts.
有些题目会要求条件概率，例如：“已知该学生是男生，求他喜欢篮球的概率”。此时分母变为男生总数 20，所以答案 = 8/20 = 2/5。一定要看清条件限制了什么。
5. Scatter Graphs and Correlation | 散点图与相关性
A past paper shows a scatter graph of hours spent revising and test marks. Points generally rise from bottom-left to top-right. Questions: (a) Describe the correlation. (b) Draw a line of best fit. (c) Estimate the mark for a student who revised 6.5 hours.
一道真题展示了一幅关于复习时间与考试成绩的散点图。点大致从左下方向右上方上升。问题：(a) 描述相关性。(b) 画出最佳拟合线。(c) 估算一名复习了 6.5 小时的学生的成绩。
(a) There is a strong positive correlation – as revision hours increase, marks tend to increase. (b) Draw a straight line through the middle of the points, trying to have roughly equal numbers of points above and below. (c) Find 6.5 on the horizontal axis, go up to the line, then across to read the mark from the vertical axis. The estimate should be around 70–75 depending on the line drawn.
(a) 存在强正相关——随着复习时间增加，成绩也倾向于提高。(b) 穿过点群中间画一条直线，尽量使线上方和下方的点数大致相等。(c) 在横轴上找到 6.5，向上引至最佳拟合线，再水平引至纵轴读取成绩。根据所画的线，估算值大约在 70–75 之间。
Interpolation (estimating inside the data range) is generally reliable, but extrapolation (outside the range) can be unreliable. If asked about an 8‑hour revision, but data only goes to 7 hours, we must state that the estimate may be inaccurate because we don’t know if the trend continues.
内插法（在数据范围内估算）通常是可靠的，但外推法（超出数据范围）则可能不可靠。如果被问及复习 8 小时的情况，但数据只到 7 小时，我们必须说明该估算可能不准确，因为我们不知道趋势是否会继续。
6. Designing a Questionnaire | 设计调查问卷
A CIE question might present a flawed survey question and ask you to criticise it, e.g., ‘How much TV do you watch? A lot A little Sometimes.’ You must identify issues: overlapping response options, vague categories, lack of a time frame. Then write an improved question.
CIE 的题目可能会给出一个有缺陷的调查问题，要求你进行批评，例如：“你看多少电视？ 很多 很少 有时”。你必须指出问题：回答选项重叠、类别模糊、缺少时间范围。然后写出改进后的问题。
Common problems: the categories ‘A lot’, ‘A little’, ‘Sometimes’ mix frequency and amount, and they are subjective – different people interpret them differently. There is no time period (per day? per week?). An improved version: ‘On average, how many hours of TV do you watch per week? 0–5 hours 6–10 hours 11–15 hours 16–20 hours More than 20 hours’. This uses objective, non‑overlapping numerical ranges and a clear time frame.
常见问题：“很多”、“很少”、“有时”这些类别混合了频率和数量，而且主观性强——不同人的理解不同。并且没有时间期限（每天？每周？）。改进后的版本：“平均而言，你每周看多少小时电视？ 0–5 小时 6–10 小时 11–15 小时 16–20 小时 超过 20 小时”。这使用了客观、互不重叠的数值区间和明确的时间框架。
Always ensure response options are exhaustive (cover all possibilities) and mutually exclusive. Also avoid leading questions that suggest a ‘correct’ answer. In the exam, you must give a specific improved question, not just general advice.
始终确保回答选项是穷尽的（包含所有可能情况）且互斥的。同时避免引导性问题，即暗示存在“正确”答案的问题。在考试中，你必须给出具体的改进问题，而不仅仅是笼统的建议。
7. Comparing Data Sets Using Averages and Spread | 使用平均数与离散程度比较数据集
Two classes take the same maths test. Class A: mean = 62, median = 64, range = 45. Class B: mean = 65, median = 60, range = 25. The question asks: Compare the performance and consistency of the two classes.
两个班级参加相同的数学测试。A 班：平均数 = 62，中位数 = 64，极差 = 45。B 班：平均数 = 65，中位数 = 60，极差 = 25。题目要求：比较两个班级的表现和一致性。
Start with central tendency: Class B has a slightly higher mean (65 vs 62), so on average they performed better. However, Class A has a higher median (64 vs 60), suggesting that the typical student in A scored higher, but the mean is pulled down by low outliers. Compare spread: Class B’s range is much smaller (25 vs 45), so their scores are more consistent and less spread out. Class A’s large range indicates some very low or very high scores, which makes the mean less representative.
首先看集中趋势：B 班的平均数略高（65 对比 62），因此平均而言他们表现更好。然而，A 班的中位数更高（64 对比 60），这表明 A 班的典型学生得分更高，但平均数被低分拉低了。比较离散程度：B 班的极差小很多（25 对比 45），所以他们的成绩更一致，分布更集中。A 班较大的极差表明存在一些非常低或非常高的分数，这使得平均数的代表性较差。
For full marks, always quote the statistics directly and say what they mean in context. You can also comment on which average is more appropriate – with large range, the median is usually better for describing ‘typical’ performance.
要获得满分，务必直接引用统计数据，并结合情境说明其含义。你还可以评论哪个平均数更合适——当极差较大时，中位数通常更能描述“典型”表现。
8. Identifying Misleading Graphs | 识别误导性图表
An exam question shows two bar charts representing sales figures. In the first chart, the vertical axis starts at 100, making the difference between 120 and 130 look huge. The second chart starts at 0, where the same data show only a small difference. The task is to explain why the first graph is misleading.
一道考题展示了两个代表销售额的条形图。在第一个图中，纵轴从 100 开始，使得 120 和 130 之间的差异看起来巨大。第二个图从 0 开始，相同的数据显示的差异很小。任务是要解释为什么第一个图具有误导性。
When the vertical axis does not start at zero, the heights of bars are not proportional to the actual values, exaggerating differences. The viewer might wrongly think the difference is large. Also, irregular scale intervals, missing labels, or 3D effects can distort perception. A fair graph uses a scale starting from zero (unless there is a special need) and clearly labelled axes.
当纵轴不从零开始时，条形的高度与实际数值不成比例，从而夸大了差异。观看者可能会错误地认为差异很大。此外，不规则的刻度间距、缺少标签或三维效果都会扭曲认知。公正的图表应使用从零开始的刻度（除非有特殊需要）并清晰标注坐标轴。
In your answer, point out the specific manipulation: ‘The vertical axis starts at 100, not 0, so the bar for 130 is twice as tall as 120, giving a false impression.’ Remember that CIE often asks you to redraw the graph correctly.
在你的答案中，要指出具体的操纵手法：“纵轴从 100 开始，而不是从 0 开始，因此代表 130 的条形高度是 120 的两倍，给人以错误印象。”请记住，CIE 经常要求你正确地重新绘制图表。
9. Probability from Experiments and Expected Frequency | 实验概率与期望频数
A die is rolled 300 times. The results are: 1 → 45, 2 → 52, 3 → 48, 4 → 55, 5 → 50, 6 → 50. Question: (a) Find the experimental probability of rolling a 4. (b) Is the die likely to be fair? Give a reason. (c) If the die is fair, how many times would you expect to roll a 6 in 600 rolls?
一枚骰子被投掷 300 次。结果如下：1 → 45 次，2 → 52 次，3 → 48 次，4 → 55 次，5 → 50 次，6 → 50 次。问题：(a) 求投掷出 4 点的实验概率。(b) 这枚骰子可能是公平的吗？给出理由。(c) 如果骰子是公平的，在 600 次投掷中，你期望出现多少次 6 点？
(a) Experimental probability = frequency of 4 ÷ total rolls = 55/300 = 11/60 or about 0.183. (b) For a fair die, each number has probability 1/6 ≈ 0.167. The experimental probabilities are all close to 0.167 (e.g., 55/300 ≈ 0.183), so small variation is expected due to chance. The die seems reasonably fair. (c) Expected frequency = probability × number of trials = (1/6) × 600 = 100 times.
(a) 实验概率 = 出现 4 点的次数 ÷ 总投掷次数 = 55/300 = 11/60，约 0.183。(b) 对于公平的骰子，每个数字的概率为 1/6 ≈ 0.167。实验概率都接近 0.167（例如 55/300 ≈ 0.183），因此由于随机性的影响，存在小范围偏差是正常的。这枚骰子看起来基本上是公平的。(c) 期望频数 = 概率 × 试验次数 = (1/6) × 600 = 100 次。
This concept links theoretical and experimental probability. Remember: ‘expected frequency’ does not mean the outcome will be exactly that number; it’s the average in the long run. Use clear working steps: probability, then multiply by number of trials.
这个概念将理论概率与实验概率联系起来。记住：“期望频数”并不意味着结果一定恰好是那个数字，而是长期运行下的平均数。解题步骤要清晰：先求概率，再乘以试验次数。
10. Mixed Past Paper Challenge | 综合真题挑战
A combined question: Mrs. Smith records the time (in minutes) her students spent on homework and their test scores. Data for 10 students are given in a table. Part (a) asks to draw a scatter graph. Part (b) asks for the mean time and mean score. Part (c) asks to describe correlation and draw a line of best fit. Part (d) asks: ‘Estimate the score for a student who spends 25 minutes. Comment on the reliability.’
一道综合题：史密斯老师记录了她学生做家庭作业所花的时间（分钟）以及他们的测试成绩。给出了 10 名学生的数据表格。(a) 部分要求绘制散点图。(b) 部分要求计算平均时间和平均成绩。(c) 部分要求描述相关性并画出最佳拟合线。(d) 部分要求：“估算花 25 分钟做作业的学生的成绩。并对可靠性进行评论。”
Step‑by‑step: Plot each pair carefully, label axes ‘Time (min)’ and ‘Score’. Mean time = sum of times ÷ 10; mean score = sum of scores ÷ 10. The scatter points generally go upward, so there is positive correlation. Draw a line balancing points above and below. For 25 minutes, find the corresponding score on the line, e.g., about 78. However, if all the times in the data are between 10 and 30 minutes, 25 is within this range, so the estimate is reliable (interpolation). If 25 were beyond the max, say 35, the estimate would be unreliable (extrapolation) because we do not know if the pattern continues.
逐步解答：仔细描点，给坐标轴标注“时间（分钟）”和“成绩”。平均时间 = 时间总和 ÷ 10；平均成绩 = 成绩总和 ÷ 10。散点大致向上分布，因此存在正相关。画一条使上下点数大致平衡的直线。对于 25 分钟，在线上找到对应的成绩，例如大约 78 分。然而，如果数据中所有时间都在 10 到 30 分钟之间，那么 25 分钟位于此范围内，因此这个估算是可靠的（内插法）。如果 25 分钟超出了最大值，比如数据最大 20 分钟，问 35 分钟，那么估算就不可靠（外推法），因为我们不知道规律是否会延续。
Always structure mixed answers logically, showing all calculations, and use the context to judge reliability. In CIE mark schemes, a correct statement about interpolation/extrapolation often carries the final mark.
要逻辑清晰地组织综合解答，展示所有计算过程，并利用情境来判断可靠性。在 CIE 的评分标准中，关于内插法或外推法的正确陈述通常占最后的一分。
Published by TutorHao | Statistics 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