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

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

This article provides a detailed walkthrough of a mock unit test paper for A-Level OCR Statistics. It covers a representative set of topics, including probability, discrete random variables, binomial and Poisson distributions, normal distribution, hypothesis testing for a proportion, chi-squared tests for independence, and correlation. Each question is solved step-by-step with clear explanations, formulas, and calculator tips. The aim is to help you practise exam-style questions and deepen your understanding of key statistical methods required by the OCR specification.

本文对一份A-Level OCR统计单元测试模拟卷进行了详细解析。试卷涵盖了概率、离散随机变量、二项分布与泊松分布、正态分布、比例假设检验、独立性卡方检验以及相关分析等代表性主题。每道题目都给出了分步解答，包括清晰的解释、公式和计算器使用技巧，旨在帮助你练习考试风格的题目，加深对OCR考纲要求的关键统计方法的理解。

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1. Mock Paper Overview | 模拟卷概览

This mock paper consists of eight compulsory questions designed to be completed in 90 minutes, mirroring the pace of an actual A-Level unit test. The total mark is 60. Questions progressively assess the ability to recall definitions, perform calculations, apply statistical tests, and interpret results in context. A calculator with statistical functions (such as binomial CD, inverse normal, and correlation) is assumed.

本模拟卷共8道必答题，设计为90分钟完成，贴近实际单元考试的节奏。满分60分。题目逐步考查概念回忆、计算执行、统计检验应用以及在情境中解读结果的能力。考试假设学生可使用具有统计功能（如二项分布累积、逆正态、相关计算）的计算器。

The following sections present each question in the same format: the problem statement, a worked solution in English, followed by the corresponding Chinese explanation. Key formulas are displayed centred and in bold. All calculations are shown with a level of accuracy appropriate for A-Level marking.

后续各小节按统一格式呈现：题目陈述、英文解答步骤以及对应的中文说明。核心公式居中加粗显示，所有计算精度均符合A-Level评分标准。

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2. Q1: Probability with Venn Diagrams | 第1题：概率与韦恩图

Question 1: In a group of 50 students, 30 study Mathematics, 25 study Physics, and 18 study both subjects. Find the probability that a randomly chosen student studies neither Mathematics nor Physics.

第1题：一组50名学生中，30人选修数学，25人选修物理，18人同时选修两门课程。求随机抽取一名学生既不选修数学也不选修物理的概率。

athematics students and P be the set of Physics students. Using the principle of inclusion–exclusion, the number studying at least one subject is n(M ∪ P) = n(M) + n(P) − n(M ∩ studying neither subject is 50 − 37 = 13. The required probability is 13/50 = 0.26 (or 26% e drawn with the intersection 18, M-only 12, and P-only 7, which confirms the calculation.

解答：设M为选修数学的学生集合，P为选修物理的学生集合。利用容斥原理，至少选修一门的人数 n(M ∪ P) = n(M) + n(P) − n(M ∩ P) = 30 + 25 − 18 = 37。因此两门都不选的人数为 50 − 37 = 13。所求概率为 13/50 = 0.26（或26%）。可绘制韦恩图：交集18，仅选数学12，仅选物理7，验证了计算结果。

OCR examiners expect clear notation and, where helpful, a diagram. Always subtract the intersection to avoid double counting, and remember that probabilities are expressed as fractions, decimals, or percentages.

OCR考官期望清晰的符号，并在有帮助时提供图示。务必减去交集以避免重复计数，并记住概率可用分数、小数或百分比表示。

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3. Q2: Discrete Random Variable | 第2题：离散随机变量

Question 2: A discrete random variable X has the following probability distribution.

第2题：离散随机变量X具有如下概率分布。

x	0	1	2	3
P(X = x)	0.2	0.3	0.4	0.1

Calculate E(X) and Var(X).

计算E(X)和Var(X)。

Solution: The expected value is given by

E(X) = Σ x·P(X = x) = 0×0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 0 + 0.3 + 0.8 + 0.3 = 1.4.

解答：期望值公式为 E(X) = Σ x·P(X = x) = 0×0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 1.4。

To find the variance, first compute

E(X²) = Σ x²·P(X = x) = 0²×0.2 + 1²×0.3 + 2²×0.4 + 3²×0.1 = 0 + 0.3 + 1.6 + 0.9 = 2.8.

为求方差，先计算 E(X²) = Σ x²·P(X = x) = 0²×0.2 + 1²×0.3 + 2²×0.4 + 3²×0.1 = 2.8。

Then, using Var(X) = E(X²) − [E(X)]², we obtain Var(X) = 2.8 − (1.4)² = 2.8 − 1.96 = 0.84.

然后利用 Var(X) = E(X²) − [E(X)]²，得 Var(X) = 2.8 − 1.96 = 0.84。

rete distributions, E(X) is the mean of the distribution, and the variance measures spread around that mean.

务必验证概率总和为1 (0.2+0.3+0.4+0.1 = 1)。对于离散分布，E(X)是分布的均值，方差衡量均值周围的离散程度。

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4. Q3: Binomial Distribution | 第3题：二项分布

dom sample of 12 components is selected. Let X be the number of defective components, so X ~ B(12, 0.25). Find (a) P(X ≥ 3), (b) P(X = 5).

第3题：某工厂生产的元件次品率为0.25。随机抽取12个元件，记次品数为X，则X ~ B(12, 0.25)。求 (a) P(X ≥ 3), (b) P(X = 5)。

P(X ≤ 2). Using a calculator with binomial cumulative distribution function or the formula

P(X = k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ,

解答(a)：P(X ≥ 3) = 1 − P(X ≤ 2)。使用计算器的二项概率累积功能或者公式 P(X = k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ，

≥ 3) = 1 − 0.3907 = 0.6093 (to 4 decimal places). A calculator directly gives P(X ≥ 3) = 0.6093.

将k=0,1,2的概率相加。P(X=0)≈0.0317，P(X=1)≈0.1267，P(X=2)≈0.2323，总和P(X ≤ 2) ≈ 0.3907，故 P(X ≥ 3) = 1 − 0.3907 = 0.6093。计算器可直接给出结果。

(b) P(X = 5): using the formula or calculator,

P(X = 5) = C(12,5) × (0.25)⁵ × (0.75)⁷ ≈ 792 × 9.766×10⁻⁴ × 0.1335 ≈ 0.1032 (4 d.p.).

(b) P(X = 5)：使用公式或计算器得 P(X = 5) = C(12,5) × 0.25⁵ × 0.75⁷ ≈ 0.1032。

Remember to specify the number of trials n and success probability p clearly. When using an approximation or table, check the conditions; here the binomial is exact. At A-Level, answers are expected to four decimal places unless stated otherwise.

务必明确给出试验次数n和成功概率p。此处二项分布为精确分布，无需近似。A-Level考试中除非另有说明，答案一般保留四位小数。

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5. Q4: Poisson Distribution | 第4题：泊松分布

Question 4: The number of calls received by a call centre follows a Poisson distribution with a mean of 4 calls per hour. Find the probability that exactly 2 calls are received in a 30-minute period.

第4题：某呼叫中心接听电话的次数服从泊松分布，平均每小时4次。求在30分钟内恰好接到2次电话的概率。

eriod, the mean rate λ becomes 4 × 0.5 = 2 calls. The Poisson probability mass function is

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