AS OCR Statistics: Unit Test Mock Paper Analysis | AS OCR 统计：单元测试模拟卷解析

📚 AS OCR Statistics: Unit Test Mock Paper Analysis | AS OCR 统计：单元测试模拟卷解析

Mock examinations are invaluable for reinforcing statistical concepts and honing exam technique. This analysis walkthrough dissects a typical AS OCR Unit Test paper, covering sampling methods, data presentation, probability, discrete random variables, binomial distribution, and hypothesis testing. Each section provides model solutions, key formulae, and examiner insights to help you avoid common pitfalls.

模拟考试是巩固统计概念和磨练考试技巧的宝贵工具。本次解析逐题剖析了一套典型的 AS OCR 单元测试卷，内容涵盖抽样方法、数据呈现、概率、离散随机变量、二项分布和假设检验。每个部分都提供了标准答案、关键公式和考官洞察，帮助你避开常见陷阱。

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1. Mock Paper Overview | 模拟卷概述

The mock paper is designed for a 1-hour session and carries a total of 60 marks. It consists of four themed sections: Sampling (10 marks), Data Presentation & Interpretation (15 marks), Probability & Discrete Random Variables (20 marks), and Binomial Distribution with Hypothesis Testing (15 marks). Each question is structured to test both procedural fluency and conceptual understanding, mirroring the style of actual OCR AS Statistics papers.

这份模拟卷设计为 1 小时内完成，总分 60 分。试卷包含四个主题部分：抽样（10 分）、数据呈现与解读（15 分）、概率与离散随机变量（20 分）以及二项分布与假设检验（15 分）。每道题既考查操作熟练度，又检验概念理解，完全模仿 OCR AS 统计真题的风格。

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2. Sampling Methods Questions | 抽样方法题解析

Question 1 presented a scenario where a school canteen manager wants to survey eating habits across Year 10, 11, and 12 students. Part (a) asked for a suitable sampling method and a justification. Part (b) required calculation of sample sizes if 80 students are to be selected from a population of 200, 150, and 130 per year group respectively.

第 1 题给定一个情境：学校食堂经理想调查 10 年级、11 年级和 12 年级学生的饮食习惯。(a) 小题要求给出合适的抽样方法并说明理由，(b) 小题要求计算若从总数分别为 200、150 和 130 的三个年级中抽取 80 名学生时各年级的样本量。

Stratified sampling is the most appropriate choice. It guarantees proportional representation from each year group, reducing sampling bias and ensuring that the views of smaller cohorts are not under-represented. Simple random sampling could, by chance, overlook a whole year group.

分层抽样是最合适的选择。它能保证每个年级按比例被抽到，从而减少抽样偏差，确保人数较少的年级意见不会被低估。简单随机抽样则可能偶然漏掉某个年级。

To allocate the sample sizes, we compute the stratum proportions. Total population N = 200 + 150 + 130 = 480. The Year 10 sample size = (200/480) × 80 ≈ 33.33 → 33 students. Year 11: (150/480) × 80 = 25 students exactly. Year 12: (130/480) × 80 ≈ 21.67 → 22 students. Note that rounding must preserve the total of 80, so we adjust accordingly.

计算各层样本量时，我们按层比例分配。总人数 N = 200 + 150 + 130 = 480。10 年级样本量 = (200/480) × 80 ≈ 33.33 → 33 人；11 年级：(150/480) × 80 = 25 人；12 年级：(130/480) × 80 ≈ 21.67 → 22 人。注意取整后需保持总和 80，因此适当调整。

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3. Data Presentation: Histograms and Cumulative Frequency | 数据呈现：直方图与累积频率

Question 2 provided a grouped frequency table for the time (in minutes) spent on homework by a sample of 60 students. The groups were 0–10, 10–20, 20–30, 30–50, and 50–80. Candidates had to draw a histogram and a cumulative frequency curve, then estimate the median and interquartile range.

第 2 题给出了一张分组频数表，记录了 60 名学生做家庭作业的时间（分钟）。分组为 0–10、10–20、20–30、30–50 和 50–80。考生需要绘制直方图和累积频率曲线，然后估算中位数和四分位距。

Time (min)	Frequency	Class width	Frequency density
0 – 10	6	10	0.6
10 – 20	14	10	1.4
20 – 30	18	10	1.8
30 – 50	16	20	0.8
50 – 80	6	30	0.2

For a histogram, the vertical axis must represent frequency density, not frequency, because the class widths are unequal. Frequency density is calculated as frequency ÷ class width. The bars should be drawn with no gaps, and area of each bar corresponds to frequency.

对于直方图，因为组距不相等，纵轴必须表示频率密度，而非频数。频率密度 = 频数 ÷ 组距。条形之间不应留空隙，每个条形的面积与频数成正比。

Cumulative frequencies are obtained by successively adding: 6, 20, 38, 54, 60. Plot these against the upper class boundaries (10, 20, 30, 50, 80) and join with a smooth curve. From the curve, the median (30th value) lies at about 27 minutes; the lower quartile (15th) ≈ 18 minutes; upper quartile (45th) ≈ 37 minutes. Hence the interquartile range = 37 – 18 = 19 minutes.

累积频率通过依次相加得到：6, 20, 38, 54, 60。在对应组上界（10, 20, 30, 50, 80）处描点并连成光滑曲线。从曲线上可读取：中位数（第 30 个值）约为 27 分钟；下四分位数（第 15 个）≈ 18 分钟；上四分位数（第 45 个）≈ 37 分钟。因此四分位距 IQR = 37 – 18 = 19 分钟。

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4. Probability and Venn Diagrams | 概率与韦恩图

Question 3 gave probabilities for events A, B, and A ∩ B. Specifically, P(A) = 0.4, P(B) = 0.3, and P(A ∩ B) = 0.1. Part (a) required a Venn diagram; part (b) asked for P(A ∪ B) and interpretation of the complement; part (c) required the conditional probability P(A|B).

第 3 题给出事件 A、B 及其交集的概率：P(A) = 0.4，P(B) = 0.3，P(A ∩ B) = 0.1。(a) 小题要求画出韦恩图；(b) 小题求 P(A ∪ B) 并解释补集意义；(c) 小题要求计算条件概率 P(A|B)。

In the Venn diagram, the intersection contains 0.1. The remaining part of A is 0.4 – 0.1 = 0.3, and the remaining part of B is 0.3 – 0.1 = 0.2. The outside region has probability 1 – (0.3 + 0.1 + 0.2) = 0.4.

在韦恩图中，交集部分填 0.1。A 的剩余部分为 0.4 – 0.1 = 0.3，B 的剩余部分为 0.3 – 0.1 = 0.2。外部区域概率为 1 – (0.3 + 0.1 + 0.2) = 0.4。

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.3 − 0.1 = 0.6

The complement of (A ∪ B) is neither A nor B, with probability 0.4. For conditional probability, we use:

(A ∪ B) 的补集表示既不发生 A 也不发生 B，其概率为 0.4。对于条件概率，使用公式：

P(A|B) = P(A ∩ B) / P(B) = 0.1 / 0.3 = 1/3 ≈ 0.333

This demonstrates that given B has occurred, the probability of A is 1/3. Always check that the conditioning event has non-zero probability.

这表明在 B 已发生的前提下，A 发生的概率为 1/3。务必检查条件事件概率非零。

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5. Discrete Random Variables and Expectation | 离散随机变量与期望

Question 4 presented the probability distribution of a discrete random variable X: values 1, 2, 3, 4 with probabilities 0.2, 0.3, 0.4, and 0.1 respectively. Part (a) required verification that the probabilities sum to 1; part (b) asked for E(X) and Var(X); part (c) asked for E(3X + 2) and Var(3X + 2).

第 4 题给出离散随机变量 X 的分布：取值 1, 2, 3, 4，对应概率 0.2, 0.3, 0.4, 0.1。(a) 小题要求验证概率和为 1；(b) 小题求 E(X) 和 Var(X)；(c) 小题求 E(3X + 2) 和 Var(3X + 2)。

Sum of probabilities = 0.2 + 0.3 + 0.4 + 0.1 = 1.0, confirming a valid distribution. The expected value is computed as:

概率总和 = 0.2 + 0.3 + 0.4 + 0.1 = 1.0，确认是有效分布。期望值计算如下：

E(X) = (1×0.2) + (2×0.3) + (3×0.4) + (4×0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4

For variance, first find E(X²):

为求方差，先计算 E(X²)：

E(X²) = (1²×0.2) + (2²×0.3) + (3²×0.4) + (4²×0.1) = 0.2 + 1.2 + 3.6 + 1.6 = 6.6

Var(X) = E(X²) − [E(X)]² = 6.6 − 2.4² = 6.6 − 5.76 = 0.84

Using the linear transformation rules: E(3X + 2) = 3E(X) + 2 = 3(2.4) + 2 = 9.2.

应用线性变换规则：E(3X + 2) = 3E(X) + 2 = 3(2.4) + 2 = 9.2。

Var(3X + 2) = 3² × Var(X) = 9 × 0.84 = 7.56

Notice the addition of a constant does not affect variance. Many candidates incorrectly add the constant inside the square or forget to square the coefficient.

注意加上常数不影响方差。很多考生误把常数放在平方内，或者忘记将系数平方。

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6. Binomial Distribution Calculations | 二项分布计算

Question 5 introduced a binomial variable X ~ B(10, 0.25). Part (a) asked for P(X = 3); part (b) asked for P(X ≥ 2). The paper expected candidates to use the formula or a calculator and to show working.

第 5 题引入二项变量 X ~ B(10, 0.25)。(a) 小题求 P(X = 3)；(b) 小题求 P(X ≥ 2)。考卷期望考生使用公式或计算器，并展示步骤。

The probability mass function is: P(X = r) = ⁿCᵣ pʳ (1−p)ⁿ⁻ʳ. For exactly 3 successes:

概率质量函数为：P(X = r) = ⁿCᵣ pʳ (1−p)ⁿ⁻ʳ。恰有 3 次成功的概率：

P(X = 3) = ¹⁰C₃ × (0.25)³ × (0.75)⁷

¹⁰C₃ = 120, (0.25)³ = 0.015625, (0.75)⁷ ≈ 0.13348. Multiplying gives approximately 120 × 0.015625 × 0.13348 ≈ 0.2503. Therefore P(X = 3) ≈ 0.250.

¹⁰C₃ = 120，(0.25)³ = 0.015625，(0.75)⁷ ≈ 0.13348。相乘得约 120 × 0.015625 × 0.13348 ≈ 0.2503。所以 P(X = 3) ≈ 0.250。

For “at least 2”, it is often easier to use the complement: P(X ≥ 2) = 1 − [P(X = 0) + P(X = 1)].

对于“至少 2 次”，用补集往往更简便：P(X ≥ 2) = 1 − [P(X = 0) + P(X = 1)]。

P(X = 0) = (0.75)¹⁰ ≈ 0.0563; P(X = 1) = 10 × 0.25 × (0.75)⁹ ≈ 0.1877. Sum = 0.2440, so P(X ≥ 2) ≈ 1 − 0.244 = 0.756.

P(X = 0) = (0.75)¹⁰ ≈ 0.0563；P(X = 1) = 10 × 0.25 × (0.75)⁹ ≈ 0.1877。总和为 0.2440，故 P(X ≥ 2) ≈ 1 − 0.244 = 0.756。

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7. Hypothesis Testing for a Binomial Proportion | 二项比例假设检验

Question 6 set up a formal hypothesis test: a manufacturer claims that the proportion of defective items is at most 15%. An inspector tests 20 items and finds 6 defectives. Conduct a hypothesis test at the 5% significance level, stating your hypotheses, the critical region, and your conclusion.

第 6 题构建了一个正式假设检验：某制造商声称次品率不超过 15%。检验员检查 20 件产品，发现 6 件次品。在 5% 显著性水平下进行假设检验，写出假设、拒绝域和结论。

Let p be the true proportion of defective items. Null hypothesis H₀: p = 0.15; alternative H₁: p > 0.15 (one‑tailed test). Under H₀, the test statistic X ~ B(20, 0.15).

设 p 为真实次品率。零假设 H₀：p = 0.15；备择假设 H₁：p > 0.15（单尾检验）。在 H₀ 成立下，检验统计量 X ~ B(20, 0.15)。

We require the smallest x such that P(X ≥ x) ≤ 0.05. Evaluating cumulative probabilities:

我们要找到最小的 x，使得 P(X ≥ x) ≤ 0.05。计算累积概率：

P(X ≥ 5) = 1 − P(X ≤ 4)	≈ 1 − 0.8298 = 0.1702
P(X ≥ 6) = 1 − P(X ≤ 5)	≈ 1 − 0.9327 = 0.0673
P(X ≥ 7) = 1 − P(X ≤ 6)	≈ 1 − 0.9781 = 0.0219

Since P(X ≥ 7) = 0.0219 < 0.05 and P(X ≥ 6) > 0.05, the critical region for the test is X ≥ 7. The observed value is 6, which does not lie in the critical region.

因为 P(X ≥ 7) = 0.0219 < 0.05 而 P(X ≥ 6) > 0.05，检验的拒绝域为 X ≥ 7。观测值为 6，没有落在拒绝域内。

Conclusion: There is insufficient evidence at the 5% significance level to reject the manufacturer’s claim that the proportion of defective items is at most 15%. Always phrase the conclusion in the context of the problem and never say “accept H₀” — we simply do not reject it.

结论：在 5% 显著性水平下，没有充分证据拒绝制造商关于次品率不超过 15% 的说法。务必在问题情境中陈述结论，永远不要说“接受 H₀”——我们只是不拒绝它。

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8. Common Mistakes and Examiner Tips | 常见错误与考官建议

Sampling: A common error is to suggest quota sampling when proportional representation is required; quota sampling does not guarantee randomness and is not covered in AS OCR. Always link your reasons to the elimination of bias.

抽样：常见错误是在需要比例代表时建议配额抽样；配额抽样不保证随机性，且不在 AS OCR 考纲内。解释理由时一定要联系到消除偏差。

Histograms: Candidates often plot frequency instead of frequency density when class widths differ. Remember that the area gives frequency, and the vertical axis must be labelled frequency density. Also confirm that the horizontal scale is continuous.

直方图：当组距不等时，考生常纵轴标为频数而非频率密度。记住面积表示频数，纵轴必须标注“频率密度”。并确保横轴连续。

Venn diagrams: Forgetting to place the intersection probability correctly or muddling union and intersection can lead to lost marks. Practise writing clear expressions before plugging in numbers.

韦恩图：忘记正确放置交集概率，或混淆并集与交集，都会导致失分。先写出清楚表达式再代数值多练习。

Discrete random variables: When computing variance, students often forget to square the expectations of X or misapply the Var(aX + b) formula. Always compute E(X²) carefully and use the identity Var(X) = E(X²) – [E(X)]².

离散随机变量：计算方差时，学生常忘记对 X 的期望平方，或错误使用 Var(aX + b) 公式。务必仔细计算 E(X²)，并应用恒等式 Var(X) = E(X²) – [E(X)]²。

Binomial distribution: Mistakes include using nCr instead of the correct binomial coefficient, forgetting to raise p and (1–p) to the correct powers, or misinterpreting “at least” statements. Practise with the complement rule to save time.

二项分布：错误包括使用错误的二项系数、忘记对 p 和 (1–p) 求正确次方，或误读“至少”含义。多练习运用补集规则可节约时间。

Hypothesis tests: The three most frequent errors are: writing the hypotheses incorrectly (e.g., H₁: p < 0.15 when it should be greater), confusing the critical region with the test statistic, and drawing a conclusion that does not reference the significance level or context. Always state "there is evidence to suggest…" or "there is insufficient evidence…"

假设检验：最常见的三种错误是：假设写错（如本该是 p > 0.15 却写成 p < 0.15）、把拒绝域与检验统计量混淆，以及结论没有提及显著性水平或问题背景。务必使用“有证据表明……”或“没有充分证据……”的表述。

Each section of this mock represents a building block for the AS qualification. Regular practice with such papers, combined with careful review of mark schemes, will sharpen your statistical reasoning and boost exam confidence.

这份模拟卷的每个部分都是 AS 资格的基础模块。通过这类试卷的定期练习，再结合评分方案的仔细研读，你的统计推理能力将得到磨炼，考试信心也会增强。

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