📚 Year 9 OCR Statistics: Unit Test Mock Paper Walkthrough | 九年级 OCR 统计单元测试模拟卷解析
Welcome to this detailed walkthrough of a Year 9 OCR Statistics unit test mock paper. We will go through ten carefully designed questions that cover key topics: data types, data collection, charts, averages, cumulative frequency, and probability. Each question is followed by a full solution and examiner-style commentary to strengthen your understanding.
欢迎阅读这份九年级 OCR 统计单元测试模拟卷的详细解析。我们将逐一讲解十道精心设计的题目，覆盖数据类型、数据收集、图表、平均数、累积频率和概率等关键主题。每道题均附有完整解答和类似考官的点评，以加深你的理解。
1. Identifying Data Types | 识别数据类型
Question 1 (3 marks): Classify each of the following as qualitative, discrete quantitative or continuous quantitative: (a) Eye colour, (b) Number of books read in one month, (c) Time taken to swim 50 metres.
第1题（3分）：将下列各项分类为定性数据、离散定量数据或连续定量数据：(a) 眼睛颜色，(b) 一个月内阅读的书籍数量，(c) 游泳50米所用时间。
Answer: (a) Qualitative; (b) Discrete quantitative; (c) Continuous quantitative.
答案：(a) 定性数据；(b) 离散定量数据；(c) 连续定量数据。
Explanation: Eye colour is a descriptive category, so it is qualitative. The number of books is a count that can only be a whole number, making it discrete quantitative. Time is measured on a continuous scale and can take any value, so it is continuous quantitative.
解析：眼睛颜色是描述性类别，因此是定性数据。书籍数量是计数，只能取整数值，因此是离散定量数据。时间在连续尺度上测量，可以取任意数值，因此是连续定量数据。
2. Primary vs Secondary Data | 原始数据与二手数据
Question 2 (2 marks): A student wants to investigate Year 9 favourite lunch options. She designs a questionnaire. Explain why a questionnaire is more suitable than observation in this context, and state whether the collected data is primary or secondary.
第2题（2分）：一名学生想调查九年级最喜欢的午餐选择。她设计了一份问卷。解释为什么在此情景下问卷比观察更合适，并说明收集的数据是原始数据还是二手数据。
Answer: A questionnaire allows her to ask students directly about their opinions, which cannot be observed. The data is primary because it is collected first-hand by the investigator for the specific purpose.
答案：问卷能让她直接询问学生的意见，而意见无法通过观察获得。数据是原始数据，因为它由研究者出于特定目的亲手收集。
Examiner note: Remember, primary data is gathered yourself; secondary data is obtained from existing sources such as the internet or publications. Questionnaires are ideal for gathering opinions and preferences.
考官点评：记住，原始数据是自己收集的；二手数据来自已有来源，如互联网或出版物。问卷是收集意见和偏好的理想方式。
3. Frequency Tables and Bar Charts | 频数表与条形图
Question 3 (4 marks): The tally chart below shows the number of pets owned by 20 students. Complete the frequency column and draw a bar chart on the grid provided (grid omitted, describe how to plot).
第3题（4分）：下面的计数表显示了20名学生拥有的宠物数量。完成频率栏，并在提供的网格上绘制条形图（网格略去，描述如何绘制）。
| Number of pets | Tally | Frequency |
|---|---|---|
| 0 | |||| | 4 |
| 1 | |||| | | 6 |
| 2 | ||| | 3 |
| 3 | || | 2 |
| 4 | | | 1 |
Answer: The frequencies are shown above. To draw a bar chart, label the horizontal axis with 'Number of pets' (0 to 4) and the vertical axis with 'Frequency' (scale 0 to 6). Draw bars of equal width, with heights corresponding to each frequency: 4, 6, 3, 2 and 1. Leave gaps between bars because the data is discrete.
答案：频率如上表所示。绘制条形图时，横轴标记为“宠物数量”（0至4），纵轴标记为“频率”（刻度0至6）。绘制等宽的长条，高度分别对应频率4、6、3、2和1。条形之间留有空隙，因为数据是离散的。
4. Pie Charts | 饼图
Question 4 (4 marks): Using the pet ownership data from Question 3, calculate the angle for each sector of a pie chart. Which category represents the largest proportion?
第4题（4分）：利用第3题的宠物拥有数据，计算饼图中各扇区的角度。哪个类别占比最大？
60° = 54°. 3 pets: (2/20)×360° = 36°. 4 pets: (1/20)×360° = 18°. The largest proportion is '1 pet' with 108°.
答案：总频数 = 20。角度 = (频数 / 20) × 360°。0只宠物: (4/20)×360° = 72°；1只宠物: (6/20)×360° = 108°；2只宠物: 54°；3只宠物: 36°；4只宠物: 18°。最大占比是“1只宠物”，对应108°。
sum of your angles is 360°. Here, 72+108+54+36+18 = 360°, which confirms the calculations.
饼图能清晰显示比例。务必检查角度之和为360°。此题72+108+54+36+18 = 360°，验证了计算。
5. Stem-and-Leaf Diagrams | 茎叶图
Question 5 (3 marks): The stem-and-leaf diagram below shows the test scores of 13 students. Key: 1|2 means 12. Find the median score.
第5题（3分）：下面的茎叶图显示了13名学生的测验成绩。图例：1|2 表示12。求中位数成绩。
Stem 1: 2, 5, 7 | Stem 2: 0, 3, 3, 8, 9 | Stem 3: 1, 4, 4, 6 | Stem 4: 0
茎：1 叶：2,5,7；茎：2 叶：0,3,3,8,9；茎：3 叶：1,4,4,6；茎：4 叶：0
Answer: Order the data: 12, 15, 17, 20, 23, 23, 28, 29, 31, 34, 34, 36, 40. There are 13 values. The median is the 7th value, which is 28.
答案：将数据排序：12, 15, 17, 20, 23, 23, 28, 29, 31, 34, 34, 36, 40。共13个数值。中位数是第7个值，即28。
= 7. Always arrange the leaves in ascending order from the stem before finding the median.
对于奇数个数据点，中位数位置 = (n+1)/2 = (13+1)/2 = 7。在找中位数之前，始终要将每个茎上的叶子从小到大排列。
6. Measures of Central Tendency and Range | 集中趋势与范围
Question 6 (5 marks): Using the same test scores from Question 5, calculate the mean, mode, median, and range.
第6题（5分）：使用第5题相同的测验成绩，计算平均数、众数、中位数和范围。
Scores: 12, 15, 17, 20, 23, 23, 28, 29, 31, 34, 34, 36, 40.
成绩：12, 15, 17, 20, 23, 23, 28, 29, 31, 34, 34, 36, 40。
3 ≈ 26.3 (to 3 s.f.). The modes are 23 and 34 (bimodal). Median = 28 (from Q5). Range = 40 – 12 = 28.
答案：总和 = 342。平均数 = 342 ÷ 13 ≈ 26.3（保留三位有效数字）。众数为23和34（双峰）。中位数 = 28（来自第5题）。范围 = 40 – 12 = 28。
Tip: The mean is sensitive to extreme
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