📚 Year 9 AQA Statistics: In-Depth Past Paper Analysis | Year 9 AQA 统计：历年真题深度解析
Welcome to our comprehensive revision guide for Year 9 AQA Statistics. This article breaks down real past paper questions, unpacking the key concepts and common pitfalls. By working through these topics, you will sharpen your data handling skills and learn exactly what examiners are looking for.
欢迎来到 Year 9 AQA 统计专项复习指南。本文精选历年真题进行深度剖析，逐一拆解核心概念与常见失分点。通过系统梳理这些题型，你将全面提升数据处理能力，精准把握评分标准。
1. Pictograms and Bar Charts | 象形图与条形图
Pictograms use symbols to represent frequencies, and bar charts display data as rectangular bars with heights proportional to the values. In AQA past papers, students often lose marks by misreading the key or forgetting to scale axes correctly.
象形图用符号代表频数，条形图则以矩形的高度呈现数据。在 AQA 历年真题中，学生常因误读图例或未正确标注坐标轴而丢分。
Example question (pictogram): A survey asked pupils about their favourite fruit. The key shows each icon = 3 pupils. Apple: 4 icons, Banana: 2 icons, Orange: 5 icons, Grape: 3 icons. (a) How many pupils chose Apple? (b) Which fruit was the least popular? (c) What is the difference between the number of pupils choosing Orange and Banana?
真题示例（象形图）：一项调查询问了学生最喜爱水果。图例显示每个图标 = 3 名学生。苹果：4 个图标，香蕉：2 个图标，橙子：5 个图标，葡萄：3 个图标。(a) 有多少名学生选择了苹果？(b) 哪种水果最不受欢迎？(c) 选择橙子和香蕉的学生人数相差多少？
Solution: (a) Apple: 4 × 3 = 12 pupils. (b) Banana has only 2 icons → 6 pupils, so it is the least popular. (c) Orange: 5 × 3 = 15, Banana: 6, difference = 15 – 6 = 9.
解析：(a) 苹果：4 × 3 = 12 名学生。(b) 香蕉仅有 2 个图标，即 6 名学生，因此最不受欢迎。(c) 橙子：5 × 3 = 15，香蕉：6，相差 15 − 6 = 9。
Bar charts commonly test reading off values and calculating totals or proportions. Always check the frequency scale – some questions use a broken scale or a different interval on the y-axis.
条形图常考察读取数值以及计算总数或比例。务必仔细检查频数轴刻度 — 有些题目会使用截断刻度或不同的 y 轴间距。
| Pet | Frequency |
|---|---|
| Dog | 14 |
| Cat | 9 |
| Fish | 5 |
| Other | 2 |
Using the table above, calculate the total number of pupils and the fraction who chose Cat.
利用上表，计算学生总人数和选择猫的学生所占分数。
Total = 14 + 9 + 5 + 2 = 30 pupils. Fraction for Cat = 9/30 = 3/10 (or 0.3).
总数 = 14 + 9 + 5 + 2 = 30 名学生。猫的比例 = 9/30 = 3/10（或 0.3）。
2. Interpreting Pie Charts | 解读饼图
Pie charts show proportions as angles at the centre of a circle. The total angle is 360°. Past paper questions often ask you to convert between angles and frequencies when the total frequency is known.
饼图用圆心角表示各部分的比例，整个圆周为 360°。历年真题往往要求你在已知总频数时，进行角度与频数的相互转换。
Typical exam task: A pie chart represents 180 students’ favourite subjects. Maths has an angle of 90°, English 120°, Science 60°, and the rest is Other. Calculate the number of students who prefer English, and find the angle for Other.
典型考题：某饼图展示了 180 名学生最喜爱的科目。数学的圆心角为 90°，英语 120°，科学 60°，其余为其他科目。求偏好英语的学生人数，并计算“其他”对应的圆心角。
Each degree corresponds to 180 ÷ 360 = 0.5 student. English students = 120 × 0.5 = 60. Total used angle = 90 + 120 + 60 = 270°, so Other angle = 360 − 270 = 90°.
每个度数对应 180 ÷ 360 = 0.5 名学生。英语人数 = 120 × 0.5 = 60。已用角度和 = 90 + 120 + 60 = 270°，因此“其他”角度 = 360 − 270 = 90°。
Watch out for questions that give you the frequency for one sector and ask you to find the total frequency: simply divide the frequency by the fraction of the angle. Always express your working clearly.
注意一类题目：给出某个扇区的频数，让你求总频数——只需用该频数除以该扇区角度占比即可。解题过程务必要写清楚。
Another common pitfall is reading angles with a protractor on a printed diagram; the exam may ask you to measure accurately and then calculate.
另一个常见易错点是需要用量角器测量印刷图上的角度；考试可能要求先精确测量，再进行计算。
3. Mean, Median, Mode and Range | 平均数、中位数、众数与极差
These four measures summarise data sets. The mean (x̄) is the sum of values divided by the number of values. The median is the middle value when data are ordered. The mode is the most frequent value, and the range is the difference between the largest and smallest.
这四个统计量用于概括数据集。平均数（x̄）是数值总和除以数据个数；中位数是排序后位于中间位置的数值；众数是出现次数最多的值；极差是最大值与最小值的差。
Past paper example: The heights (in cm) of 9 students: 145, 150, 152, 148, 150, 155, 147, 150, 149. Calculate the mean, median, mode and range.
真题示例：9 名学生身高（单位：cm）：145, 150, 152, 148, 150, 155, 147, 150, 149。计算平均数、中位数、众数和极差。
Sum = 145+150+152+148+150+155+147+150+149 = 1346. Mean = 1346 ÷ 9 ≈ 149.6 cm. Ordering: 145, 147, 148, 149, 150, 150, 150, 152, 155. Median = 150 cm. Mode = 150 cm (appears 3 times). Range = 155 − 145 = 10 cm.
总和 = 145+150+152+148+150+155+147+150+149 = 1346。平均数 = 1346 ÷ 9 ≈ 149.6 cm。升序排列：145, 147, 148, 149, 150, 150, 150, 152, 155。中位数 = 150 cm。众数 = 150 cm（出现三次）。极差 = 155 − 145 = 10 cm。
Examiners often ask which average best represents the data, linking to the presence of outliers. If an extreme value exists, the median is more robust than the mean.
考官常会问哪一个平均数最能代表数据，这需要结合异常值来判断。若存在极端值，中位数比平均数更具稳健性。
When data are given in a frequency table, remember to multiply each value by its frequency before summing. The mean is then total sum ÷ total frequency.
当数据以频数表形式给出时，要先让每个值乘以各自的频数再求和。平均数 = 总和 ÷ 总频数。
4. Stem-and-Leaf Diagrams | 茎叶图
Stem-and-leaf plots organise numerical data while preserving the original values. The ‘stem’ represents the leading digit(s), and the ‘leaf’ is the final digit. AQA exams test your ability to read and construct these plots, then find the median and range.
茎叶图既能整理数值数据，又能保留原始数值。“茎”代表前导数字，“叶”则是最后一个数字。AQA 考试会考查绘制与解读茎叶图，进而求中位数和极差。
Question: The stem-and-leaf diagram shows marks out of 50: Stem 1 | leaf: 4 7 8, Stem 2 | leaf: 0 3 5 5 9, Stem 3 | leaf: 1 2 6, Stem 4 | leaf: 0 5 7 (key: 1|4 means 14). (a) List all marks. (b) Find the median mark. (c) What is the range?
题目：茎叶图展示了满分 50 分的成绩：茎 1 | 叶：4 7 8，茎 2 | 叶：0 3 5 5 9，茎 3 | 叶：1 2 6，茎 4 | 叶：0 5 7（图例：1|4 表示 14）。(a) 列出所有分数。(b) 求分数的中位数。(c) 极差是多少？
Marks in order: 14, 17, 18, 20, 23, 25, 25, 29, 31, 32, 36, 40, 45, 47. Total 14 marks. Median = average of 7th and 8th = (25+29)/2 = 27. Range = 47 − 14 = 33.
分数顺序：14, 17, 18, 20, 23, 25, 25, 29, 31, 32, 36, 40, 45, 47。共 14 个数据。中位数 = 第 7 和第 8 个数的均值 = (25+29)/2 = 27。极差 = 47 − 14 = 33。
Some past papers ask you to compare two stem-and-leaf diagrams on the same stem. Focus on shape, spread, and median when making comparisons.
部分真题会要求比较同一茎上的两组茎叶图。进行比较时，需关注分布形态、离散程度和中位数。
5. Scatter Graphs and Correlation | 散点图与相关性
Scatter graphs display the relationship between two variables. A positive correlation means as one variable increases, the other tends to increase. AQA questions often involve drawing a line of best fit and using it for estimation.
散点图用于呈现两个变量之间的关系。正相关指一个变量增大时，另一个变量也趋于增大。AQA 考题常要求绘制最佳拟合线并进行估值。
Exam-style task: A table shows hours of revision (x) and test score (y): (2, 40), (4, 55), (6, 62), (8, 75), (10, 88). (a) Plot the points. (b) Describe the correlation. (c) Draw a line of best fit and estimate the score for 5 hours of revision.
考试风格任务：表格给出复习小时数 (x) 与测验分数 (y)：(2, 40), (4, 55), (6, 62), (8, 75), (10, 88)。(a) 描点绘图。(b) 描述相关性。(c) 画出最佳拟合线，并估计复习 5 小时的分数。
The points show a strong positive correlation. A line of best fit should pass through the ‘middle’ of the points. Reading from the line at x = 5 gives approximately y = 58. Estimates must be shown with vertical and horizontal dashed lines on the graph.
这些点显示出强正相关。最佳拟合线应当穿过点群的“中心”。从线上 x = 5 处读取，约 y = 58。估值时必须在图上用垂直和水平虚线标示出来。
Common mistakes: forcing the line through the origin when it does not fit the data, and making predictions outside the given range (extrapolation) without caution.
常见错误：强行让拟合线经过原点而忽视数据特征，以及在使用外推法时超出给定数据范围的预测而没有说明其风险。
6. Probability Basics | 概率基础
Probability is expressed as a fraction, decimal or percentage between 0 and 1. The probability of an event A is P(A) = (number of favourable outcomes) / (total number of outcomes). Past papers often combine probability with listing sample spaces.
概率可用 0 到 1 之间的分数、小数或百分数表示。事件 A 的概率 P(A) = 有利结果数 / 总结果数。历年真题常将概率与样本空间列举相结合。
Example: A fair six-sided die is rolled. (a) What is P(rolling a factor of 6)? (b) What is P(rolling a number less than 3)? The factors of 6 are 1, 2, 3, 6 → 4 outcomes, so probability = 4/6 = 2/3. Numbers less than 3 are 1, 2 → 2 outcomes, probability = 2/6 = 1/3.
例题：掷一颗均匀六面骰子。(a) P(掷出 6 的因数) 是多少？(b) P(掷出小于 3 的数) 是多少？6 的因数为 1, 2, 3, 6，有 4 个结果，概率 = 4/6 = 2/3。小于 3 的数为 1 和 2，有 2 个结果，概率 = 2/6 = 1/3。
When using two-stage events, e.g. spinning two spinners or drawing two counters, systematic listing or probability trees help avoid missing outcomes.
当涉及两阶段事件（如旋转两个转盘或抽取两个圆片）时，使用系统列表或概率树可以避免遗漏结果。
Questions on relative frequency often require you to calculate how many times an event is expected to happen: expected frequency = probability × number of trials.
相对频率类题目常要求计算期望发生次数：期望频数 = 概率 × 试验次数。
7. Two-Way Tables | 双向表
A two-way table summarises categorical data with two variables. You must be able to complete missing cells using row and column totals, then find probabilities based on the table.
双向表以两个分类变量汇总数据。你需要利用行/列合计补全缺失单元格，再根据表格计算概率。
| Boys | Girls | Total | |
|---|---|---|---|
| Like Tennis | 12 | 8 | 20 |
| Do not like | 5 | 15 | 20 |
| Total | 17 | 23 | 40 |
From the table above, find (a) P(selecting a girl who likes tennis) and (b) P(selecting a boy given they don’t like tennis).
根据上表，求 (a) P（选到喜欢网球的女生）和 (b) P（已知某人讨厌网球，此人为男生的概率）。
(a) There are 8 girls who like tennis out of 40 pupils → 8/40 = 1/5. (b) Total ‘do
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