📚 Year 10 OCR Statistics: Unit Test Mock Paper Analysis | 十年级 OCR 统计单元测试模拟卷解析
This article provides a comprehensive walkthrough of a Year 10 OCR Statistics unit test mock paper, covering key topics such as data types, sampling, charts, averages, spread, stem-and-leaf diagrams, box plots, scatter graphs, and probability. Each question is analysed with step-by-step explanations, common pitfalls, and exam tips to help you achieve top marks.
本文全面解析一份十年级 OCR 统计单元测试模拟卷，涵盖数据类型、抽样方法、图表、平均数、离散程度、茎叶图、箱线图、散点图与概率等核心考点。每道题目均提供逐步详解、常见错误及应试技巧，助你拿高分。
1. Mock Paper Overview | 模拟卷概览
The mock paper consists of 6 questions worth a total of 50 marks, to be completed in 45 minutes. It tests Unit 1 topics: collecting data, representing data, and probability basics. A calculator is allowed. You are expected to show all workings clearly.
本模拟卷共 6 道题，满分 50 分，限时 45 分钟，涵盖第一单元：数据收集、数据表示与基础概率。允许使用计算器，需清晰展示解题过程。
| Question | Topic | Marks |
|---|---|---|
| 1 | Data Types & Sampling | 8 |
| 2 | Pie Charts | 10 |
| 3 | Averages & Range | 9 |
| 4 | Stem-and-Leaf Diagrams | 8 |
| 5 | Box Plots & IQR | 8 |
| 6 | Scatter Graphs & Correlation | 7 |
Each question targets specific exam skills, from justifying sampling choices to interpreting distributions. Read on for full mark schemes and examiner guidance.
每道题目均针对特定考试技能，从论证抽样选择到解释分布特征。继续阅读，获取完整评分方案与考官指导。
2. Question 1: Data Types & Sampling | 题目1：数据类型与抽样方法
(a) Identify the sampling method used when every 10th student is chosen from an alphabetical register.
(a) 指出从字母顺序名单中每10人抽取一人所使用的抽样方法。
This is systematic sampling. A starting point is chosen at random within the first 10 names, and thereafter every 10th student is selected at a regular interval.
这是系统抽样。在前10个名字中随机选择一个起点，然后按固定间隔每隔10名抽取一名学生。
(b) State whether homework time (in minutes) is qualitative, quantitative discrete, or quantitative continuous.
(b) 判断以分钟计量的作业时间属于定性、定量离散还是定量连续变量。
Homework time is quantitative continuous, because it is measured on a scale and can take any value within a range, not just whole numbers.
作业时间属于定量连续变量，因为它是通过测量得到的，可以在一定范围内取任意数值，不限于整数。
(c) Give one advantage and one disadvantage of systematic sampling in this context.
(c) 在该情境下，给出系统抽样的一个优点和一个缺点。
Advantage: It is simple and quick to carry out, especially with a large population list. Disadvantage: If the list contains a hidden repeating pattern that coincides with the sampling interval, the sample may become biased and not representative.
优点：操作简便快速，尤其适合大型名单。缺点：若名单存在与抽样间隔吻合的隐藏周期模式，样本可能产生偏差，缺乏代表性。
(d) Describe how to obtain a stratified sample of 50 students from a Year 10 of 200, where 120 are boys and 80 are girls.
(d) 描述如何从200名十年级学生（120名男生，80名女生）中抽取50人的分层样本。
0/200) × 50 = 20. Then we randomly select 30 boys from the list of all boys, and 20 girls from the list of all girls. This ensures both genders are represented in the sample according to their true proportion in the population.
采用分层抽样时，我们按各层人数占比计算需抽取的数量。男生：(120/200) × 50 = 30 人；女生：(80/200) × 50 = 20 人。接着从全体男生名册中随机抽取30人，从全体女生名册中随机抽取20人。这样能确保样本中男女比例与总体一致，代表性更强。
3. Question 2: Pie Chart Construction | 题目2：饼图绘制
A survey asked 40 students their favourite soft drink. The results were: Cola 15, Lemonade 12, Orange 8, Water 5. (a) Calculate the angle for each sector. (b) Draw an accurate pie chart. (c) State the mode. (d) Explain why you cannot reliably say this represents the whole school’s preference.
一项调查询问了40名学生最喜欢的软饮料，结果如下：可乐15人，柠檬水12人，橙汁8人，水5人。(a) 计算每个扇区的角度。(b) 绘制准确的饼图。(c) 指出众数。(d) 解释为何不能据此可靠推断全校学生的偏好。
(a) Calculating angles:
(a) 计算角度：
requency = 15 + 12 + 8 + 5 = 40. The formula for each angle is (frequency / total) × 360°.
总频数 = 15 + 12 + 8 + 5 = 40。每个扇区角度公式为 (频数 / 总频数) × 360°。
| Drink | Frequency | Calculation | Angle (°) |
|---|---|---|---|
| Cola | 15 | (15/40) × 360 | 135 |
| Lemonade | 12 | (12/40) × 360 | 108 |
| Orange | 8 | (8/40) × 360 | 72 |
| Water | 5 | (5/40) × 360 | 45 |
Always check the angles sum to 360°: 135 + 108 + 72 + 45 = 360°.
务必检查总角度为360°：135 + 108 + 72 + 45 = 360°。
(b) Drawing the pie chart: Use compasses to draw a circle. Mark a radius, then use a protractor to measure each central angle consecutively. Label each sector clearly or provide a key. Ensure your sectors are accurately drawn to the nearest degree.
(b) 绘制饼图：使用圆规画圆，标记一条半径，然后用量角器依次量取每个圆心角。清晰标注各扇区或提供图例，确保角度精确到度。
(c) Mode: The mode is Cola as it has the highest frequency (15).
(c) 众数：众数是可乐，因其频数最高（15）。
(d) Representativeness: The sample of 40 students may only come from one class or year group, or could be volunteers. Unless the sample is random and sufficiently large, it will not reflect the views of the entire school population, which may have different preferences across different age groups.
(d) 代表性：40人的样本可能仅来自一个班级或年级，或是志愿者样本。除非样本随机且具有足够规模，否则无法反映全校学生的观点，不同年龄段可能有截然不同的偏好。
4. Question 3: Mean, Median, Mode and Range | 题目3：均值、中位数、众数与极差
The table shows the number of siblings for 30 Year 10 students.
下表显示了30名十年级学生的兄弟姐妹数。
| Number of siblings (x) | Frequency (f) |
|---|---|
| 0 | 5 |
| 1 | 8 |
| 2 | 10 |
| 3 | 4 |
| 4 | 3 |
(a) Calculate the mean number of siblings.
(a) 计算兄弟姐妹数的均值。
First add an fx column and find Σf and Σfx.
首先添加 fx 列，计算 Σf 与 Σfx。
| x | f | fx |
|---|---|---|
| 0 | 5 | 0 |
| 1 | 8 | 8 |
| 2 | 10 | 20 |
| 3 | 4 | 12 |
| 4 | 3 | 12 |
| Totals | Σf = 30 | Σfx = 52 |
Mean = Σfx / Σf = 52 / 30 = 1.733..., which rounds to 1.73 siblings (3 s.f.).
均值 = Σfx / Σf = 52 / 30 = 1.733...，四舍五入保留三位有效数字为 1.73 个兄弟姐妹。
(b) Find the median number of siblings.
(b) 求兄弟姐妹数的中位数。
tive frequency: 0s: 5; 1s: 5+8=13; 2s: 13+10=23. Both the 15th and 16th values fall in the '2 siblings' category, so median = 2.
30个数据的中位数位置为 (30+1)/2 = 15.5，取第15和第16个值的平均。累计频数：0个：5；1个：5+8=13；2个：13+10=23。第15和第16个值均属于“2个兄弟姐妹”，因此中位数为2。
(c) Write down the mode.
(c) 写出众数。
The highest frequency is 10, corresponding to 2 siblings. Mode = 2.
最高频数为10，对应2个兄弟姐妹，众数 = 2。
(d) Determine the range.
(d) 计算极差。
Range = highest value – lowest value = 4 – 0 = 4 siblings.
极差 = 最大值 – 最小值 = 4 – 0 = 4 个兄弟姐妹。
5. Question 4: Stem-and
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