📚 Year 10 OCR Statistics: Unit Test Mock Paper Analysis | Year 10 OCR 统计：单元测试模拟卷解析
This article provides a step-by-step walkthrough of a typical Year 10 OCR Statistics unit test, covering key topics from the GCSE (9‑1) specification. Each section analyses a mock question, highlights common mistakes, and offers clear revision tips. By working through these examples, you will strengthen your understanding of data handling, probability, and statistical diagrams.
本文是对一份典型的 Year 10 OCR 统计单元测试的逐步解析，涵盖 GCSE（9‑1）大纲的核心主题。每一节分析一个模拟题，指出常见错误，并提供清晰的复习建议。通过这些例题，你将加深对数据处理、概率和统计图表的理解。
1. Types of Data and Sampling Methods | 数据类型与抽样方法
Question 1 asks students to classify data as qualitative or quantitative, and to suggest a suitable sampling method for a school survey. Qualitative data are non‑numerical (e.g. favourite colour), while quantitative data are numerical and can be discrete (counted) or continuous (measured). A common error is confusing discrete and continuous – for example, shoe size is discrete even though it is a number, because it takes only specific values. For a survey of 200 students out of 1200, a stratified sample by year group ensures proportional representation and reduces bias better than a simple random sample.
第 1 题要求学生将数据分为定性或定量，并为一项学校调查建议合适的抽样方法。定性数据是非数值的（例如最喜欢的颜色），而定量数据是数值的，可以是离散的（计数）或连续的（测量）。常见错误是混淆离散与连续——例如，鞋码虽然是数字但它是离散的，因为它只取特定值。对于从 1200 名学生中抽取 200 人的调查，按年级分层抽样能确保比例代表性，比简单随机抽样更能减少偏差。
2. Mean, Median, Mode, and Range | 平均数、中位数、众数与极差
This question presents a small dataset: 12, 15, 18, 18, 21, 24, 30. The mean is (12+15+18+18+21+24+30) ÷ 7 = 138 ÷ 7 = 19.7 (1 d.p.). The median is the 4th value: 18. The mode is 18. The range is 30 – 12 = 18. A common slip is forgetting to order the data for the median or dividing by the wrong number for the mean. When comparing two datasets, use the mean and range together: a higher mean indicates a higher average, while a smaller range suggests more consistency.
本题给出一个小数据集：12, 15, 18, 18, 21, 24, 30。平均数是 (12+15+18+18+21+24+30) ÷ 7 = 138 ÷ 7 = 19.7（保留一位小数）。中位数是第 4 个值：18。众数是 18。极差是 30 – 12 = 18。常见的失误是求中位数时忘记排序，或求平均数时除以错误的个数。比较两个数据集时，应同时使用平均数和极差：平均数更高表示平均水平更高，而极差更小则说明更稳定。
3. Bar Charts and Pie Charts | 条形图与饼图
The mock paper gives a frequency table of favourite sports among 360 students. To draw a pie chart, calculate the angle for each sector: (frequency ÷ 360) × 360° = frequency × 1°. So a frequency of 90 gives an angle of 90°. For a dual bar chart comparing boys and girls, both axes must be clearly labelled, bars evenly spaced, and a key included. A common deduction is for missing axis titles or inconsistent scale.
模拟卷给出 360 名学生最喜欢运动的频数表。要绘制饼图，需要计算每个扇形的角度：(频数 ÷ 360) × 360° = 频数 × 1°。因此频数为 90 对应的角度就是 90°。绘制比较男女生的复式条形图时，两轴必须清楚标注，条形间距均匀，并附上图例。常见的扣分点是缺少轴标题或比例不一致。
4. Cumulative Frequency and Box Plots | 累积频率与箱线图
A grouped frequency table of test scores is used to construct a cumulative frequency graph. Add a cumulative frequency column by adding each frequency to the previous total. Plot the upper class boundary against cumulative frequency, join points with a smooth curve. From the graph, estimate the median (50th percentile) and the interquartile range (IQR = UQ – LQ). To draw a box plot, use the five‑number summary: lowest value, LQ, median, UQ, highest value. The box plot reveals skewness – if the median is closer to the left of the box, the data is positively skewed.
根据测试成绩的分组频数表构建累积频率图。添加累积频数列，将每个频数与之前的总和相加。以上组界为横坐标、累积频数为纵坐标描点，并用平滑曲线连接。从图中估计中位数（第 50 百分位数）和四分位距（IQR = UQ – LQ）。绘制箱线图时，需要五数总结：最小值、下四分位数、中位数、上四分位数、最大值。箱线图能显示偏态——如果中位数靠近箱体左侧，数据呈正偏态。
5. Basic Probability and Sample Spaces | 基础概率与样本空间
A question asks for the probability of rolling a prime number on a fair six‑sided die. Primes are 2, 3, 5, so probability = 3/6 = 1/2. Always express probabilities as fractions in simplest form. When listing outcomes for two events, a sample space diagram (grid) helps visualise all combinations. For mutually exclusive events, P(A or B) = P(A) + P(B). If events are independent, P(A and B) = P(A) × P(B). A common mistake is adding probabilities when they are not mutually exclusive; then you must subtract P(A and B).
题目要求计算掷一枚公平六面骰子得到质数的概率。质数为 2, 3, 5，因此概率 = 3/6 = 1/2。概率始终要用最简分数表示。列举两个事件的结果时，样本空间图（网格）有助于展示所有组合。对于互斥事件，P(A 或 B) = P(A) + P(B)。如果事件独立，P(A 与 B) = P(A) × P(B)。常见错误是在事件不互斥时直接相加；此时必须减去 P(A 与 B)。
6. Relative Frequency and Expected Value | 相对频数与期望值
An experiment spins a biased spinner 200 times. The relative frequency of landing on ‘blue’ is 0.32. The expected number of blues in 500 spins is 0.32 × 500 = 160. Relative frequency is used to estimate probability when outcomes are not equally likely. The more trials, the more reliable the estimate – this is the law of large numbers. In a fairground game with a prize worth £2 and a probability of winning 0.1, the expected winnings per play are £0.20. If the ticket costs 50p, the expected loss is 30p per play.
实验旋转一个偏斜转盘 200 次。落在“蓝色”的相对频数为 0.32。在 500 次旋转中，期望的蓝色次数为 0.32 × 500 = 160。当结果不是等可能时，用相对频数估计概率。试验次数越多，估计越可靠——这是大数定律。在一个奖品价值 £2、获胜概率为 0.1 的游乐场游戏中，每次游戏的期望收益为 £0.20。如果票价是 50 便士，那么每次游戏的期望亏损是 30 便士。
7. Tree Diagrams and Conditional Probability | 树状图与条件概率
A bag contains 4 red and 6 blue counters. Two counters are drawn without replacement. The tree diagram shows first pick probabilities: 4/10 red, 6/10 blue. For the second pick, branches adjust because the total counters reduce. The probability of drawing two reds is (4/10) × (3/9) = 12/90 = 2/15. For at least one blue, it is often quicker to calculate 1 – P(both red). Conditional probability is shown by the changing denominators on the second branches. Always check that the probabilities on each set of branches sum to 1.
袋中有 4 个红色和 6 个蓝色筹码，不放回地抽取两个。树状图显示第一次抽取概率：红 4/10，蓝 6/10。第二次抽取的分支会调整，因为总筹码数减少。抽出两个红色的概率是 (4/10) × (3/9) = 12/90 = 2/15。求至少一个蓝色时，通常用 1 – P（两个红）更快。条件概率体现在第二层分支的分母变化上。始终检查每组分支的概率之和是否为 1。
8. Scatter Graphs and Correlation | 散点图与相关性
A table gives hours of revision and exam marks for ten students. Plot each pair on a scatter graph with revision time on the x‑axis. The points show a positive correlation – as revision hours increase, marks tend to rise. Describe correlation in terms of type (positive/negative), strength (strong/moderate/weak), and form (linear). A line of best fit drawn by eye should pass through the mean point (x̄, ȳ) and have roughly equal points on each side. Use the line to estimate a mark for a given revision time, stating it is an interpolation or extrapolation. Do not extend the line beyond the data range for predictions – this is unreliable extrapolation.
表格给出 10 名学生的复习时间和考试成绩。将每对数据绘制在散点图上，x 轴为复习时间。点呈现正相关——复习时间越长，分数往往越高。描述相关性时应说明方向（正/负）、强度（强/中度/弱）和形状（线性）。通过目测绘制的最佳拟合直线应经过均值点 (x̄, ȳ)，且两侧点数大致相等。用这条线估计给定复习时间对应的分数，并说明是内插还是外推。不要将直线延伸到数据范围之外作预测——这是不可靠的外推。
9. Time Series and Moving Averages | 时间序列与移动平均
Quarterly sales data over three years are given. Plot the time series graph and notice seasonal fluctuations. A four‑point moving average smooths out these variations to reveal the underlying trend. Calculate the first moving average by averaging the first four values, then drop the earliest and add the next to get the second. Plot these averages against the mid‑time points of each group. The trend line shows a gradual increase in sales. Seasonal variation can then be estimated by subtracting the trend from the actual values. This helps make future predictions.
给出三年内每季度的销售额数据。绘制时间序列图并注意季节性波动。四点移动平均能平滑这些波动，揭示潜在趋势。计算第一个移动平均时，将前四个值平均，然后去掉最早的一个并加入下一个，依次进行。将这些平均值画在每组的中间时间点。趋势线显示销售额逐渐增长。然后，可以通过实际值减去趋势值来估计季节变动。这有助于作出未来预测。
10. Index Numbers and Rates of Change | 指数与变化率
A base year is selected with an index of 100. If the price of a bus fare rises from £1.50 to £1.80, the new index is (1.80 ÷ 1.50) × 100 = 120. The percentage increase is simply the index change: 20%. For weighted indices, multiply each item’s index by its weight, sum these, and divide by total weight. A common mistake is ignoring the base period or using raw values instead of index numbers. Chain base indices link each period to the previous one, showing period‑on‑period growth.
选定基年的指数为 100。如果公交车费从 £1.50 涨到 £1.80，新的指数为 (1.80 ÷ 1.50) × 100 = 120。百分比增幅就是指数的变化量：20%。对于加权指数，将每项指数乘以其权重，求和后除以总权重。常见错误是忽略基期或使用原始值而非指数。链基指数将每一期与上一期相连，显示逐期增长。
11. Choropleth Maps and Pictograms | 等值区域图与象形图
A mock question provides population density data for UK regions. A choropleth map uses shading to represent density classes – the darker the colour, the higher the density. It is essential to provide a key with clear boundaries. A pictogram showing car sales uses a car symbol to represent 1000 cars. For 3500 cars, 3.5 symbols are drawn. Inaccurate scaling is a typical error; ensure that the shape can be split clearly to represent part‑values. Pictograms make comparisons visually striking but are less precise than bar charts for small differences.
模拟题给出英国各地区的人口密度数据。等值区域图用阴影表示密度等级——颜色越深，密度越高。关键是要提供一个带有清晰边界的图例。显示汽车销量的象形图用一个汽车符号代表 1000 辆车。对于 3500 辆车，需画出 3.5 个符号。缩放不准是典型错误；确保图形可以清晰地分割以表示部分值。象形图使比较更具视觉冲击力，但对于微小差异不如条形图精确。
12. Statistical Enquiry and Evaluation | 统计调查与评估
The final section of a unit test often assesses the statistical enquiry cycle: plan, collect, process, discuss. A question might describe a flawed survey – e.g. a biased question like ‘Do you agree that homework is a waste of time?’ or a sample of only friends. Critique the method: question wording leading to response bias, small or unrepresentative sample, lack of randomisation. Suggest improvements such as using neutral language, stratified sampling, and piloting the questionnaire. When evaluating graphs, check for missing labels, distorted scales (not starting at zero), or 3D effects that mislead proportions.
单元测试的最后部分常考查统计调查周期：计划、收集、处理、讨论。题目可能描述一个有缺陷的调查——例如带有偏向性的问题“你是否同意家庭作业浪费时间？”或仅调查朋友的样本。批评其方法：问题措辞导致回答偏差，样本小或不具代表性，缺少随机化。建议改进，例如使用中性语言、分层抽样和试点调查问卷。评估图形时，检查缺少标签、比例失真的坐标轴（不从零开始）或误导比例的 3D 效果。
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