📚 Year 9 WJEC Statistics: Mock Test Paper Walkthrough | 九年级 WJEC 统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 9 WJEC Statistics, covering key topics such as types of data, representation, measures of central tendency, spread, probability, and scatter graphs. Each section presents a typical exam-style question followed by its worked solution, highlighting essential techniques and common pitfalls to boost your confidence for the real assessment.
本文详细解析一份针对九年级 WJEC 统计课程的单元测试模拟卷，涵盖数据类型、数据表示、集中趋势量数、离散程度、概率和散点图等核心主题。每个小节给出一道典型考题及其完整解答过程，并强调关键技巧与常见错误，帮助你在真实考试中树立信心。
1. Types of Data and Sampling | 数据类型与抽样方法
Question 1: State whether each of the following data sources is primary or secondary. (a) A student measures the heights of plants in a biology experiment. (b) A journalist uses a government report on employment rates. (c) A school council collects opinions from pupils about lunch menus using a questionnaire.
题目 1：判断以下数据来源属于一手数据还是二手数据。(a) 学生在生物实验中测量植物的高度。(b) 记者引用政府关于就业率的报告。(c) 学校议会通过问卷收集学生对午餐菜单的意见。
Solution: Primary data is collected directly by the researcher for a specific purpose. Secondary data is obtained from existing sources that someone else gathered, often for a different reason. (a) Primary – the student gathers the measurements first‑hand. (b) Secondary – the government report was compiled by officials, not the journalist. (c) Primary – the school council is carrying out the survey itself.
解答：一手数据由研究者本人为特定目的直接收集；二手数据来自他人已收集的既有资料，通常出于别的目的。(a) 一手数据——学生亲自获取测量值。(b) 二手数据——政府报告由官员编制，非记者本人采集。(c) 一手数据——学校议会主动进行调查。
Common error: assuming that any official source is primary; always ask ‘Who collected the original information?’
常见错误：误以为任何官方来源都是一手数据；请始终追问“原始信息是由谁收集的？”
2. Frequency Tables and Bar Charts | 频率表与条形图
Question 2: 25 students recorded the number of books they read last month. The results are summarised in the frequency table below. Draw a bar chart to represent the data and state the mode.
题目 2：25 名学生记录了上月阅读的书籍数量。结果汇总于下表中。绘制条形图表示数据，并指出众数。
| Number of books (x) | Frequency (f) |
|---|---|
| 0 | 4 |
| 1 | 7 |
| 2 | 8 |
| 3 | 4 |
| 4 | 2 |
Solution: s between them, and the vertical scale should start at 0. The mode is the value with the highest frequency, which is 2 books (frequency 8). To find the total number of students, sum the frequencies: 4+7+8+4+2 = 25, which matches the sample size.
解答：条形图中，类别（书籍数量）位于横轴，频率位于纵轴。条形宽度须相等，之间留有空隙，纵轴刻度从 0 开始。众数是频率最高的数值，即 2 本书（频率 8）。计算学生总数时，将频数相加：4+7+8+4+2 = 25，与样本量吻合。
We can also compute the mean number of books read: Mean = (0×4 + 1×7 + 2×8 + 3×4 + 4×2) ÷ 25 = (0+7+16+12+8) ÷ 25 = 43 ÷ 25 = 1.72
我们也可以计算阅读书籍的平均数：平均数 = (0×4 + 1×7 + 2×8 + 3×4 + 4×2) ÷ 25 = (0+7+16+12+8) ÷ 25 = 43 ÷ 25 = 1.72
Common mistake: forgetting to multiply each value by its frequency before summing; simply adding all frequencies would give 5 values, not 25.
常见错误：在求和前忘记将每个数值与其频数相乘；如果只将频数简单相加，只能得到 5 个数据点而非 25 项。
3. Pie Charts and Angles | 饼图与角度计算
Question 3: The table shows how 30 employees travel to work. Calculate the angle for each sector to draw a pie chart.
题目 3：下表显示了 30 名员工的通勤方式。计算绘制饼图时每个扇区的角度。
| Method | Frequency |
|---|---|
| Car | 15 |
| Bus | 9 |
| Walk | 4 |
| Cycle | 2 |
Solution: ÷ 30) × 360° = 48°. Cycle: (2 ÷ 30) × 360° = 24°. Check: 180° + 108° + 48° + 24° = 360°, confirming the angles sum to a full circle. When drawing, use a protractor to measure each angle from the same radius.
解答：总频率为 30。饼图中每个扇区角度 = (频数 ÷ 总数) × 360°。汽车：(15 ÷ 30) × 360° = 180°。公交车：(9 ÷ 30) × 360° = 108°。步行：(4 ÷ 30) × 360° = 48°。自行车：(2 ÷ 30) × 360° = 24°。验证：180° + 108° + 48° + 24° = 360°，确认各角度之和为一个整圆。绘图时，用半圆规从同一条半径开始量取每个角度。
Common error: dividing by total frequency incorrectly or misplacing the decimal point when multiplying. Always express the fraction in its simplest form before multiplying by 360° to reduce mistakes.
常见错误：除以总数时出错，或相乘时小数点错位。建议先化简约分，再乘以 360°，以减少失误。
4. Stem-and-Leaf Diagrams | 茎叶图
Question 4: The ages of 10 members at a sports club are: 13, 21, 24, 8, 35, 30, 19, 22, 28, 31. Construct an ordered stem‑and‑leaf diagram, and find the median and range of the ages.
题目 4：某体育俱乐部 10 位会员的年龄为：13, 21, 24, 8, 35, 30, 19, 22, 28, 31。绘制有序茎叶图，并求年龄的中位数和极差。
Solution: as : Use the tens digit as the stem and the units digit as the leaf. First sort the data: 8, 13, 19, 21, 22, 24, 28, 30, 31, 35. Include a key: 1|3 = 13 years. The ordered diagram is:
解答：以十位数为茎，个位数为叶。先将数据排序：8, 13, 19, 21, 22, 24, 28, 30, 31, 35。附上说明：1|3 = 13 岁。有序茎叶图如下：
0 | 8
1 | 3 9
2 | 1 2 4 8
3 | 0 1 5
e 5th value is 22 and the 6th is 24, so median = (22 + 24) ÷ 2 = 23 years. Range = maximum – minimum = 35 – 8 = 27 years.
中位数位置：(10 + 1) ÷ 2 = 5.5，即第 5 和第 6 个值的平均。有序序列中第 5 位为 22，第 6 位为 24，故中位数 = (22 + 24) ÷ 2 = 23 岁。极差 = 最大值 – 最小值 = 35 – 8 = 27 岁。
Common error: forgetting to order the leaves or omitting the key. The diagram must have a clear key so that anyone can interpret the stems and leaves.
常见错误：忘记将叶排序，或遗漏说明。茎叶图必须有清晰的说明，以便他人解读茎和叶的含义。
5. Mean, Median, Mode and Range | 平均数、中位数、众数与极差
Question 5: The table below shows the number of pets owned by 30 households. Calculate the mode, median, mean and range.
题目 5：下表显示了 30 户家庭拥有的宠物数量。计算众数、中位数、平均数与极差。
| Number of pets (x) | Frequency (f) |
|---|---|
| 0 | 5 |
| 1 | 10 |
| 2 | 8 |
| 3 | 4 |
| 4 | 3 |
Solution: he 15th and 16th values are both 1 pet; median = 1. Mean: see calculation below. Range = 4 – 0 = 4 pets.
解答：众数 = 1 只宠物（最高频数 10）。中位数：总频数 n = 30，位置 (30+1)/2 = 15.5。累加频数：0 → 5, 1 → 15，故第 15 和第 16 个数据值均为 1；中位数 = 1。平均数见下方计算。极差 = 4 – 0 = 4 只宠物。
= (0×5 + 1×10 + 2×8 + 3×4 + 4×3) ÷ 30 = (0+10+16+12+12) ÷ 30 = 50 ÷ 30 = 1.67 (to 2 d.p.)
平均数 = (0×5 + 1×10 + 2×8 + 3×4 + 4×3) ÷ 30 = (0+10+16+12+12) ÷ 30 = 50 ÷ 30 ≈ 1.
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