📚 Year 12 OCR Statistics: Unit Test Mock Paper Walkthrough | OCR Year 12 统计单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for the Year 12 OCR Statistics syllabus. The paper covers key topics: statistical sampling, data presentation and interpretation, measures of central tendency and dispersion, probability, the binomial distribution, and hypothesis testing. Each section below addresses a specific part of the mock test, presenting both the solution and an explanation in English, followed by its Chinese translation. All mathematical notation uses standard Unicode characters, ensuring clarity whether you are revising or simulating exam conditions.
本文详细解析了一份针对 OCR Year 12 统计课程的单元测试模拟卷。试卷涵盖统计抽样、数据呈现与解读、集中趋势和离散度量、概率、二项分布以及假设检验等核心主题。以下每一节针对模拟卷中的特定部分，先给出英文解答与讲解，随后附上对应的中文翻译。所有数学符号均采用标准 Unicode 字符，方便学生在复习或模拟考试时清晰理解。
1. Simple Random Sample | 简单随机抽样
A simple random sample of size n is one where every possible sample of size n has an equal chance of being selected from the population. For example, using a random number generator to pick 30 students from a year group of 200 ensures each possible group of 30 is equally likely. This method eliminates selection bias but requires a complete sampling frame.
简单随机抽样是指从总体中抽取容量为 n 的样本时，每一个可能的容量为 n 的样本被选中的概率都相等。例如，使用随机数生成器从 200 名学生的年级组中抽取 30 人，确保任一组 30 人的组合具有同等的机会。该方法消除了选择偏倚，但需要有完整的抽样框。
2. Stratified Sampling | 分层抽样
In stratified sampling, the population is divided into mutually exclusive strata (e.g., by gender or year group), and a simple random sample is taken from each stratum. The number sampled from each stratum is proportional to its size. This guarantees representation from all subgroups and can improve the precision of estimates when strata are homogeneous internally.
在分层抽样中，总体被划分为互不相交的层（如按性别或年级），然后从每一层中分别进行简单随机抽样。每一层的抽样数目与其在总体中所占的比例保持一致。这保证了所有子群体都被代表，并且当层内同质性较高时，可以提高估计的精度。
3. Cumulative Frequency Table | 累积频数表
Given a grouped frequency table of test scores, cumulative frequency is found by adding the frequencies up to the end of each class. For instance, if the intervals 0–10, 10–20, 20–30 have frequencies 5, 8, 12, the cumulative frequencies are 5, 13, 25. This table allows quick estimation of medians and percentiles when plotting a cumulative frequency curve.
给定一个分组的考试成绩频数表，累积频数通过将截至每一组上限的频数相加得到。例如，区间 0–10、10–20、20–30 的频数分别为 5、8、12，则累积频数为 5、13、25。该表格便于在绘制累积频数曲线时快速估算中位数和百分位数。
4. Box Plot and Skewness | 箱线图与偏态
A box plot displays the minimum, lower quartile (Q₁), median (Q₂), upper quartile (Q₃), and maximum. The interquartile range (IQR = Q₃ − Q₁) measures spread. Skewness is judged by comparing the whiskers and the position of the median: if Q₃ − Q₂ > Q₂ − Q₁, the data are positively skewed; if the opposite, negatively skewed. In our mock data, Q₁ = 34, Q₂ = 48, Q₃ = 62, so Q₃ − Q₂ = 14 > Q₂ − Q₁ = 14? Actually equal here, indicating roughly symmetric, but with an upper whisker longer than the lower, slight positive skew might still be evident from the full five-number summary.
箱线图显示了最小值、下四分位数 (Q₁)、中位数 (Q₂)、上四分位数 (Q₃) 和最大值。四分位距 (IQR = Q₃ − Q₁) 用于度量离散程度。偏态通过比较须的长度和中位数的位置来判断：若 Q₃ − Q₂ > Q₂ − Q₁，数据呈正偏态；反之则为负偏态。在我们的模拟数据中，Q₁ = 34，Q₂ = 48，Q₃ = 62，因此 Q₃ − Q₂ = 14，Q₂ − Q₁ = 14，两者相等时表明大致对称，但上须比下须更长，从完整的五数总结中仍可能看出轻微的正偏态。
5. Standard Deviation | 标准差
The standard deviation measures the average distance of data points from the mean. For a sample, it is calculated using the formula:
s = √[ Σ(x − x̄)² / (n − 1) ]
Given the values 12, 15, 18, 20, 25, we first find x̄ = 18. Then Σ(x − x̄)² = (12−18)² + (15−18)² + (18−18)² + (20−18)² + (25−18)² = 36 + 9 + 0 + 4 + 49 = 98. With n = 5, s = √(98/4) = √24.5 ≈ 4.95 (3 s.f.).
标准差衡量数据点与均值之间的平均距离。对于样本，其计算公式为：
s = √[ Σ(x − x̄)² / (n − 1) ]
给定数值 12、15、18、20、25，首先求得均值 x̄ = 18。然后计算 Σ(x − x̄)² = (12−18)² + (15−18)² + (18−18)² + (20−18)² + (25−18)² = 36 + 9 + 0 + 4 + 49 = 98。由于 n = 5，s = √(98/4) = √24.5 ≈ 4.95（保留三位有效数字）。
6. Comparing Data Sets | 比较数据集
When comparing two groups, we look at both a measure of location (mean or median) and a measure of spread (standard deviation or IQR). In the mock test, Class A had a mean of 65 and s.d. of 8, while Class B had a mean of 65 and s.d. of 15. Although the centres are identical, Class B shows much greater variability, meaning its students’ scores are more spread out around the same average. Thus, Class A performed more consistently.
在比较两组数据时，需要同时考察集中趋势的度量（均值或中位数）和离散程度的度量（标准差或四分位距）。模拟卷中，A 班的均值为 65，标准差为 8；B 班的均值为 65，标准差为 15。尽管中心位置相同，B 班的变异程度大得多，意味着其学生成绩围绕同一均值的分布更分散。因此，A 班的表现更稳定。
7. Tree Diagrams | 树形图
Tree diagrams help visualise multi‑stage probability experiments. Suppose a bag contains 4 red and 6 blue discs, and two discs are drawn without replacement. The first branch shows P(Red) = 4/10 and P(Blue) = 6/10. For the second draw, the probabilities change: if a red was taken first, P(Red second) = 3/9, P(Blue second) = 6/9. Multiplying along branches gives probabilities like P(RR) = (4/10)×(3/9) = 12/90 = 2/15.
树形图有助于将多阶段概率实验可视化。假设一个袋子里有 4 个红色和 6 个蓝色圆盘，每次抽取后不放回。第一层分支显示 P(红) = 4/10，P(蓝) = 6/10。第二次抽取时概率发生变化：若第一次抽到红色，则 P(红第二个) = 3/9，P(蓝第二个) = 6/9。沿分支相乘即可得到诸如 P(RR) = (4/10)×(3/9) = 12/90 = 2/15 的概率。
8. Conditional Probability Calculation | 条件概率计算
Using the same tree, suppose we want P(Blue second | Red first). This is simply the conditional probability on the branch: 6/9 = 2/3. The formula P(A|B) = P(A ∩ B) / P(B) confirms this: P(Blue second ∩ Red first) = (4/10)×(6/9) = 24/90; P(Red first) = 4/10; therefore (24/90) ÷ (4/10) = 6/9. This illustrates how the formula works in practice.
仍以同一树形图为例，假设我们需要求 P(蓝第二个 | 红第一个)。这直接就是该分支上的条件概率：6/9 = 2/3。公式 P(A|B) = P(A ∩ B) / P(B) 可以验证：P(蓝第二个 ∩ 红第一个) = (4/10)×(6/9) = 24/90；P(红第一个) = 4/10；因此 (24/90) ÷ (4/10) = 6/9。这展示了公式的实际应用。
9. Binomial Distribution: Setting up and Calculating | 二项分布：设定与计算
A random variable X follows a binomial distribution if there are a fixed number n of independent trials, each with two outcomes (success/failure) and a constant probability of success p. In the mock question, a spinner lands on a ‘win’ sector with p = 0.2 and is spun 10 times. Thus X ~ B(10, 0.2). The probability of exactly 3 wins is P(X = 3) = ₁₀C₃ × (0.2)³ × (0.8)⁷ = 120 × 0.008 × 0.2097152 ≈ 0.2013 (4 d.p.).
若随机变量 X 满足：试验次数 n 固定、各次试验独立、每次试验只有两个结果（成功/失败）且成功概率 p 不变，则 X 服从二项分布。在模拟题中，一个转盘停在“获胜”区域的概率 p = 0.2，共旋转 10 次。因此 X ~ B(10, 0.2)。恰好获胜 3 次的概率为 P(X = 3) = ₁₀C₃ × (0.2)³ × (0.8)⁷ = 120 × 0.008 × 0.2097152 ≈ 0.2013（保留四位小数）。
10. Binomial Distribution: Using Tables | 二项分布：使用表格
Binomial cumulative probability tables provide P(X ≤ k) for various n and p. For X ~ B(10, 0.2), the table shows P(X ≤ 3) = 0.8791. To find P(X ≥ 4), use the complement: 1 − P(X ≤ 3) = 1 − 0.8791 = 0.1209. Tables are especially useful for hypothesis testing where tail probabilities are needed.
二项分布累积概率表给出了不同 n 和 p 下 P(X ≤ k) 的值。对于 X ~ B(10, 0.2)，查表得 P(X ≤ 3) = 0.8791。要求 P(X ≥ 4) 则利用互补事件：1 − P(X ≤ 3) = 1 − 0.8791 = 0.1209。在假设检验中需要尾部概率时，表格尤其有用。
11. Hypothesis Testing: Hypotheses and Critical Region | 假设检验：假设与临界域
In hypothesis testing for a binomial proportion, the null hypothesis H₀ states p = p₀, while the alternative H₁ can be one‑tail (p < p₀ or p > p₀) or two‑tail (p ≠ p₀). Suppose a manufacturer claims a defect rate is at most 5% (p = 0.05), and we test 20 items. Let X ~ B(20, 0.05). The critical region at a 5% significance level for a one‑tail test H₁: p > 0.05 consists of the smallest k such that P(X ≥ k) ≤ 0.05. From tables, P(X ≥ 3) = 1 − P(X ≤ 2) ≈ 1 − 0.9245 = 0.0755 > 0.05, while P(X ≥ 4) = 1 − 0.9841 = 0.0159 ≤ 0.05, so the critical region is X ≥ 4.
在对二项比例进行假设检验时，原假设 H₀ 声明 p = p₀，备择假设 H₁ 可以是单侧（p < p₀ 或 p > p₀）或双侧（p ≠ p₀）。假设某厂商声称缺陷率不超过 5%（p = 0.05），我们测试 20 件产品。令 X ~ B(20, 0.05)。在 5% 显著性水平下进行单侧检验 H₁: p > 0.05，临界域是满足 P(X ≥ k) ≤ 0.05 的最小 k 值。查表，P(X ≥ 3) = 1 − P(X ≤ 2) ≈ 1 − 0.9245 = 0.0755 > 0.05，而 P(X ≥ 4) = 1 − 0.9841 = 0.0159 ≤ 0.05，故临界域为 X ≥ 4。
12. Hypothesis Testing: Conclusion and Interpretation | 假设检验：结论与解释
If the observed number of defects in the sample is, say, 5, then the result falls inside the critical region. We reject H₀ at the 5% significance level and conclude there is sufficient evidence that the defect rate exceeds 5%. If the observed number were 2, we would not reject H₀. Always phrase the conclusion in the context of the problem: “There is/is not enough evidence at the 5% level to suggest that the defect rate is greater than 5%.”
若样本中观察到的缺陷数为 5，则该结果落入临界域。我们在 5% 显著性水平下拒绝 H₀，并得出结论：有充分证据表明缺陷率超过 5%。若观察值为 2，则不拒绝 H₀。结论始终要结合问题背景进行表述：“在 5% 的水平上，有/没有足够证据表明缺陷率大于 5%。”
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