📚 Year 9 CIE Statistics: Unit Test Mock Paper Walkthrough | 九年级CIE统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test paper for Year 9 CIE Statistics. The paper covers core topics including measures of central tendency, graphical representation of data, probability, cumulative frequency, and box plots. Each question is presented with a step-by-step solution in both English and Chinese, making it an ideal revision resource for students aiming to secure top marks.
本文为一份针对九年级 CIE 统计课程的单元测试模拟卷提供逐题解析。试卷涵盖集中趋势度量、数据图形表示、概率、累积频率和箱线图等核心主题。每道题均配有中英双语的逐步解答过程，是帮助学生冲刺高分的理想复习资料。
1. Mean, Median, Mode and Range | 平均数、中位数、众数与极差
Question: Find the mean, median, mode and range of the following data set: 12, 15, 14, 10, 15, 16, 13, 15, 11.
题目：求以下数据集的平均数、中位数、众数和极差：12, 15, 14, 10, 15, 16, 13, 15, 11。
15 + 14 + 10 + 15 + 16 + 13 + 15 + 11 = 121. Number of values = 9. Mean = 121 ÷ 9 ≈ 13.44.
解答：首先计算平均数。总和 = 12 + 15 + 14 + 10 + 15 + 16 + 13 + 15 + 11 = 121。数据个数 = 9。平均数 = 121 ÷ 9 ≈ 13.44。
To find the median, arrange the data in ascending order: 10, 11, 12, 13, 14, 15, 15, 15, 16. The median is the middle value, which is the 5th value: 14.
求中位数时，将数据按升序排列：10, 11, 12, 13, 14, 15, 15, 15, 16。中位数为中间的数值，即第 5 个数值：14。
The mode is the value that appears most frequently. Here, 15 appears three times, so the mode is 15.
众数是出现频率最高的值。此处 15 出现了三次，因此众数为 15。
Range = maximum – minimum = 16 – 10 = 6.
极差 = 最大值 – 最小值 = 16 – 10 = 6。
2. Mean from a Frequency Table | 从频率表计算平均数
Question: The table below shows the scores of students in a quiz. Calculate the mean score.
题目：下表显示了学生在一次测验中的得分。计算平均得分。
| Score (x) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency (f) | 2 | 5 | 8 | 4 | 1 |
Solution: Extend the table to include fx (score × frequency).
解答：扩展表格，加入 fx（得分 × 频数）。
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f | 2 | 5 | 8 | 4 | 1 |
| fx | 2 | 10 | 24 | 16 | 5 |
4 + 1 = 20. Sum of fx (Σfx) = 2 + 10 + 24 + 16 + 5 = 57. Mean = Σfx ÷ Σf = 57 ÷ 20 = 2.85.
频数之和 (Σf) = 2 + 5 + 8 + 4 + 1 = 20。fx 之和 (Σfx) = 2 + 10 + 24 + 16 + 5 = 57。平均数 = Σfx ÷ Σf = 57 ÷ 20 = 2.85。
Thus, the mean score is 2.85.
因此，平均得分为 2.85。
3. Estimated Mean from Grouped Data | 分组数据的估算平均数
Question: The heights of 25 students are recorded in the grouped frequency table below. Estimate the mean height.
题目：25 名学生的身高记录如下分组频率表所示。估算平均身高。
| Height (h cm) | Frequency |
|---|---|
| 140 ≤ h < 150 | 4 |
| 150 ≤ h < 160 | 10 |
| 160 ≤ h < 170 | 8 |
| 170 ≤ h < 180 | 3 |
Solution: To estimate the mean, we use the midpoint of each class interval.
解答：为估算平均数，我们使用每组的组中点。
| Height | Midpoint (x) | f | f × x |
|---|---|---|---|
| 140-150 | 145 | 4 | 580 |
| 150-160 | 155 | 10 | 1550 |
| 160-170 | 165 | 8 | 1320 |
| 170-180 | 175 | 3 | 525 |
25. Σ(f × midpoint) = 580 + 1550 + 1320 + 525 = 3975. Estimated mean = 3975 ÷ 25 = 159 cm.
Σf = 4 + 10 + 8 + 3 = 25。Σ(f × 中点) = 580 + 1550 + 1320 + 525 = 3975。估算平均身高 = 3975 ÷ 25 = 159 cm。
Therefore, the estimated mean height of the students is 159 cm.
因此，学生的估算平均身高为 159 厘米。
4. Pie Charts | 饼图
Question: A survey asked 50 people to choose their favourite colour. The results are: Red (10), Blue (15), Green (5), Yellow (20). Calculate the angle for each sector and describe how to draw the pie chart.
题目：一项调查询问了 50 人他们最喜欢的颜色。结果为：红色 (10)，蓝色 (15)，绿色 (5)，黄色 (20)。计算每个扇形的角度，并描述如何绘制饼图。
mber of responses = 10 + 15 + 5 + 20 = 50. Each sector angle = (frequency ÷ total) × 360°.
解答：总回答数 = 10 + 15 + 5 + 20 = 50。每个扇形的角度 = (频数 ÷ 总数) × 360°。
Red: (10/50) × 360° = 72°
红色：(10/50) × 360° = 72°
Blue: (15/50) × 360° = 108°
蓝色：(15/50) × 360° = 108°
Green: (5/50) × 360° = 36°
绿色：(5/50) × 360° = 36°
Yellow: (20/50) × 360° = 144°
黄色：(20/50) × 360° = 144°
To draw the chart, draw a circle, use a protractor to mark each angle from the centre, and label each sector clearly.
绘制时，先画一个圆，使用量角器从中心标出每个角度，并清楚地标记每个扇形。
5. Bar Charts and Comparative Analysis | 条形图与比较分析
Question: The table shows the grade distribution for two classes in a mathematics test. Draw a suitable dual bar chart and compare the overall performance.
题目：下表显示了两个班级在一次数学测试中的成绩分布。绘制合适的双条形图，并比较整体表现。
| Grade | A | B | C | D | E |
|---|---|---|---|---|---|
| Class X | 2 | 5 | 8 | 4 | 1 |
| Class Y | 3 | 6 | 7 | 3 | 1 |
Solution: Draw axes; horizontal axis for grades, vertical axis for frequency. Use two bars per grade (side by side) in different colours to represent Class X and Class Y. Labels and a legend are essential.
解答：绘制坐标轴；横轴为等级，纵轴为频数。每个等级使用两个不同颜色的条形（并排）分别代表 X 班和 Y 班。需标注图例和轴标签。
Comparing medians: Class X mode is grade C (8 students), and Class Y mode is also C (7). However, Class Y has slightly more A and B grades, indicating a marginally better performance. The distributions are similar.
比较中位数：X 班的众数为 C 等（8 人），Y 班众数也为 C 等（7 人）。然而，Y 班取得 A 和 B 的人数略多，这表明整体表现稍好。两个分布相似。
6. Scatter Plots and Correlation | 散点图与相关性
Question: The data shows hours of study and test scores for 8 students. Plot a scatter diagram and describe the relationship.
题目：数据展示了 8 名学生的学习时间与测试分数的关系。绘制散点图并描述其关系。
| Study (h) | 1 | 2 | 2.5 | 3 | 4 | 5 | 5.5 | 6 |
|---|---|---|---|---|---|---|---|---|
| Score (%) | 45 | 50 | 55 | 60 | 70 | 75 | 78 | 85 |
Solution: Plot each pair (study hours, score) on a graph with study time on the x-axis and score on the y-axis. The points rise from left to right, showing a clear upward trend. This indicates a positive correlation – as study time increases, test scores tend to increase. The correlation appears linear and moderately strong.
解答：将每对数据（学习时间，分数）绘制在图上，x 轴为学习时间，y 轴为分数。点从左到右上升，呈现明显的上升趋势。这表明正相关——随着学习时间增加，测试分数倾向于提高。相关关系呈线性，强度中等偏强。
7. Basic Probability | 基础概率
Question: A bag contains 3 red balls, 5 blue balls and 2 green balls. One ball is chosen at random. Find: (a) P(red); (b) P(not blue); (c) P(red or green).
题目：一个袋子里有 3 个红球、5 个蓝球和 2 个绿球。随机抽取一个球。求：(a) P(红球)；(b) P(非蓝球)；(c) P(红球或绿球)。
Solution: Total number of balls = 3 + 5 + 2 = 10.
解答：球的总数 = 3 + 5 + 2 = 10。
(a) P(red) = number of red balls / total = 3/10 = 0.3.
(a) P(红) = 红球数量 / 总数 = 3/10 = 0.3。
(b) P(not blue) = 1 – P(blue) = 1 – (5/10) = 0.5, or (red+green)/total = (3+2)/10 = 0.5.
(b) P(非蓝) = 1 – P(蓝) = 1 – (5/10) = 0.5，或 (红+绿)/总数 = (3+2)/10 = 0.5。
(c) P(red or green) = (3+2)/10 = 5/10 = 0.5.
(c) P(红或绿) = (3+2)/10 = 5/10 = 0.5。
8. Cumulative Frequency and Quartiles | 累积频率与四分位数
Question: The time taken (in minutes) for 40 students to complete a puzzle is grouped. Construct a cumulative frequency table and estimate the median and interquartile range.
题目：40 名学生完成拼图所用的时间（分钟）分组如下。构建累积频率表，并估算中位数和四分位距。
| Time (min) | Frequency |
|---|---|
| 0 ≤ t < 5 | 4 |
| 5 ≤ t < 10 | 10 |
| 10 ≤ t < 15 | 12 |
| 15 ≤ t < 20 | 8 |
| 20 ≤ t < 25 | 6 |
Solution: Build the cumulative frequency column.
解答
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