📚 Year 11 OCR Statistics Unit Test Mock Paper Walkthrough | Year 11 OCR 统计单元测试模拟卷解析
This walkthrough provides a full mock unit test for Year 11 OCR Statistics, covering key topics from the specification. Every question is presented with a step-by-step solution, highlighting the reasoning and calculations you need to master for the actual exam. Use this resource to test your knowledge, refine your skills and gain confidence in tackling real exam-style problems.
本文为 Year 11 OCR 统计学提供一套完整的单元测试模拟卷，涵盖考试大纲的核心主题。每道题目均配有逐步解析，突出实际考试中必须掌握的推理与计算过程。利用这份资料检验自己的知识掌握情况、打磨解题技巧，并增强应对真实考题的信心。
1. Sampling Methods | 抽样方法
A headteacher wants to investigate students’ views on the new canteen menu. The school has 600 students in Years 10–11 and 400 students in Years 12–13. Describe how to take a stratified sample of 100 students, and give one advantage of this method over simple random sampling.
一位校长想调查学生对食堂新菜单的看法。学校有 600 名 10–11 年级学生和 400 名 12–13 年级学生。请描述如何抽取一个包含 100 名学生的分层样本，并说明这种方法相比简单随机抽样的一个优点。
First, calculate the proportion of students in each stratum. Total population = 600 + 400 = 1000. Years 10–11 proportion = 600/1000 = 0.6, so sample size = 0.6 × 100 = 60. Years 12–13 proportion = 400/1000 = 0.4, so sample size = 0.4 × 100 = 40.
首先，计算每一层的学生比例。总体人数 = 600 + 400 = 1000。10–11 年级的比例 = 600/1000 = 0.6，因此样本量为 0.6 × 100 = 60。12–13 年级的比例 = 400/1000 = 0.4，因此样本量为 0.4 × 100 = 40。
Next, within each year group, use simple random sampling (e.g. a random number generator) to select the required number of students.
接着，在每个年级组内部使用简单随机抽样（例如随机数生成器）选出所需数量的学生。
One advantage of stratified sampling is that it guarantees every subgroup is represented in the correct proportion, reducing sampling bias and giving more reliable results for the whole school population.
分层抽样的一个优点是，它能保证每个子群按正确比例被代表，从而降低抽样偏差，并为整个学校群体提供更可靠的结果。
2. Histograms and Cumulative Frequency | 直方图与累积频率
The table shows the reaction times of 50 students in a computer test. Reaction time, t (milliseconds): 0 ≤ t < 50, frequency 5; 50 ≤ t < 100, frequency 12; 100 ≤ t < 150, frequency 18; 150 ≤ t < 200, frequency 10; 200 ≤ t < 300, frequency 5. (a) Draw a histogram to represent the data. (b) Use a cumulative frequency graph to estimate the median reaction time.
下表显示了 50 名学生在一次电脑测试中的反应时间。反应时间 t（毫秒）：0 ≤ t < 50，频数 5；50 ≤ t < 100，频数 12；100 ≤ t < 150，频数 18；150 ≤ t < 200，频数 10；200 ≤ t < 300，频数 5。(a) 绘制直方图表示数据。(b) 利用累积频率图估计反应时间的中位数。
For a histogram, we must use frequency density because the class widths are unequal. Frequency density = frequency ÷ class width. Compute: 0–50: width 50, density 5/50 = 0.1; 50–100: width 50, density 12/50 = 0.24; 100–150: width 50, density 18/50 = 0.36; 150–200: width 50, density 10/50 = 0.2; 200–300: width 100, density 5/100 = 0.05. Draw bars with these densities on the vertical axis and time on the horizontal axis.
绘制直方图时，由于组距不相等，必须使用频数密度。频数密度 = 频数 ÷ 组距。计算：0–50：组距 50，密度 5/50 = 0.1；50–100：组距 50，密度 12/50 = 0.24；100–150：组距 50，密度 18/50 = 0.36；150–200：组距 50，密度 10/50 = 0.2；200–300：组距 100，密度 5/100 = 0.05。在纵轴上以这些密度、横轴上以时间绘制直条。
Cumulative frequencies: 5, 17, 35, 45, 50. The median position is the 25.5th value (50/2 = 25.5). This lies in the 100 ≤ t < 150 interval. Use linear interpolation: median = 100 + ((25.5 – 17) / (35 – 17)) × 50 = 100 + (8.5 / 18) × 50 ≈ 100 + 23.6 = 123.6 ms.
累积频数：5、17、35、45、50。中位数的位置是第 25.5 个数值（50/2 = 25.5）。它落在 100 ≤ t < 150 这一组。利用线性插值：中位数 = 100 + ((25.5 – 17) / (35 – 17)) × 50 = 100 + (8.5 / 18) × 50 ≈ 100 + 23.6 = 123.6 毫秒。
3. Measures of Central Tendency and Spread | 集中趋势与离散程度
The marks of 8 students in a quiz are: 8, 12, 15, 18, 20, 22, 22, 30. Find the mean, median, lower quartile, upper quartile, interquartile range and determine if there are any outliers.
8 名学生在一次小测验中的分数为：8, 12, 15, 18, 20, 22, 22, 30。求平均数、中位数、下四分位数、上四分位数、四分位距，并判断是否存在离群值。
Sum = 8+12+15+18+20+22+22+30 = 147. Mean = 147/8 = 18.375. Ordered list remains the same. For an even number of values, median = average of the 4th and 5th: (18+20)/2 = 19. The lower half (8,12,15,18) has median (12+15)/2 = 13.5, so Q₁ = 13.5. The upper half (20,22,22,30) has median (22+22)/2 = 22, so Q₃ = 22.
总和 = 8+12+15+18+20+22+22+30 = 147。平均数 = 147/8 = 18.375。按顺序排列同上。数据个数为偶数时，中位数 = 第 4 和第 5 个数的平均数：(18+20)/2 = 19。下半部分数据 (8,12,15,18) 的中位数为 (12+15)/2 = 13.5，因此 Q₁ = 13.5。上半部分数据 (20,22,22,30) 的中位数为 (22+22)/2 = 22，因此 Q₃ = 22。
IQR = Q₃ – Q₁ = 22 – 13.5 = 8.5. Outlier boundaries: lower fence = Q₁ – 1.5 × IQR = 13.5 – 12.75 = 0.75; upper fence = Q₃ + 1.5 × IQR = 22 + 12.75 = 34.75. No marks fall outside 0.75 to 34.75, so there are no outliers.
四分位距 IQR = Q₃ – Q₁ = 22 – 13.5 = 8.5。离群值判定界限：下界 = Q₁ – 1.5 × IQR = 13.5 – 12.75 = 0.75；上界 = Q₃ + 1.5 × IQR = 22 + 12.75 = 34.75。没有分数落在 0.75 到 34.75 之外，因此无离群值。
4. Scatter Graphs and Correlation | 散点图与相关性
The table shows hours of revision (x) and test score (y) for 8 students. x: 2, 3, 5, 6, 8, 9, 10, 12; y: 45, 50, 55, 60, 65, 70, 75, 85. Calculate Spearman’s rank correlation coefficient and interpret the result.
下表展示了 8 名学生的复习时间（x，小时）与测试成绩（y）。x：2, 3, 5, 6, 8, 9, 10, 12；y：45, 50, 55, 60, 65, 70, 75, 85。计算斯皮尔曼等级相关系数并解释结果。
Rank both x and y separately in ascending order. Ranks for x: 1,2,3,4,5,6,7,8. Ranks for y: 1,2,3,4,5,6,7,8. Differences d: all zero. Σd² = 0. Using the formula rₛ = 1 – (6Σd²) / (n(n² – 1)), with n = 8: rₛ = 1 – (6×0) / (8×(64–1)) = 1 – 0 = 1. This indicates a perfect positive monotonic correlation: as revision time increases, test score consistently increases.
分别对 x 和 y 按升序赋予等级。x 的等级：1,2,3,4,5,6,7,8。y 的等级：1,2,3,4,5,6,7,8。差值 d 均为零。Σd² = 0。代入公式 rₛ = 1 – (6Σd²) / (n(n² – 1))，n = 8：rₛ = 1 – (6×0) / (8×(64–1)) = 1 – 0 = 1。这表明存在完全正单调相关：复习时间越长，测试成绩持续提高。
If there were tied ranks, we would use mid‑ranks and adjust the calculation, but here the data yield a perfect correlation, meaning one variable can be used to predict the other with perfect rank accuracy.
如有并列等级，需使用中间等级并调整计算，但本题数据呈现完全相关，意味着一个变量可用于以完美的等级准确性预测另一个变量。
5. Probability Basics | 概率基础
Two fair six‑sided dice are rolled. Find: (a) the probability that at least one die shows a 6; (b) the probability that the sum of the two numbers is less than 5.
掷两枚公平的六面骰子。求：(a) 至少有一枚骰子显示 6 的概率；(b) 两数之和小于 5 的概率。
Total outcomes = 6 × 6 = 36. For (a), it is easier to use the complement rule. P(no 6) = (5/6) × (5/6) = 25/36. So P(at least one 6) = 1 – 25/36 = 11/36. You could also count the 11 favourable outcomes: (6,1) to (6,6) and (1,6) to (5,6).
总结果数 = 6 × 6 = 36。对于 (a)，采用互补规则更简便。P(无 6) = (5/6) × (5/6) = 25/36。因此 P(至少一个 6) = 1 – 25/36 = 11/36。也可以数出 11 种有利结果：(6,1) 至 (6,6) 以及 (1,6) 至 (5,6)。
For (b), list pairs whose sum < 5: (1,1), (1,2), (1,3), (2,1), (2,2), (3,1). There are 6 favourable outcomes, so probability = 6/36 = 1/6.
对于 (b)，列出和小于 5 的数对：(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)。共 6 种有利结果，因此概率 = 6/36 = 1/6。
6. Conditional Probability | 条件概率
In a survey of 50 students about extending lunch break, 30 are male and 20 female. 18 males support the extension, and 10 females support it. (a) Complete the two‑way table. (b) Find the probability that a randomly chosen student is female and does not support the extension. (c) Given that a student is male, find the probability that he supports the extension.
在一项关于延长午餐时间的调查中，50 名学生中有 30 名男生和 20 名女生。18 名男生支持延长，10 名女生支持延长。(a) 完成双向表。(b) 求随机选到的学生是
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