Year 12 OCR Statistics: Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析

📚 Year 12 OCR Statistics: Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for Year 12 OCR Statistics, covering core topics such as probability, discrete random variables, the binomial distribution, and hypothesis testing. Each section breaks down a typical exam‑style question, offering step‑by‑step solutions and key revision points to help you master the techniques required for success.

本文详细解析一份为 Year 12 OCR 统计课程设计的单元测试模拟卷，涵盖概率、离散随机变量、二项分布和假设检验等核心专题。每一节围绕一个典型试题展开，提供分步解答和关键复习要点，帮助你掌握考试所需的方法。

1. Basic Probability and Venn Diagrams | 基础概率与维恩图

A survey of 50 students finds that 28 like football (F), 18 like basketball (B), and 10 like both. One student is chosen at random. We illustrate the information using a Venn diagram and calculate probabilities.

对50名学生调查显示，28人喜欢足球(F)，18人喜欢篮球(B)，10人两者都喜欢。我们使用维恩图表示信息并计算概率。

Draw two overlapping circles. Place 10 in the intersection. The number who like only football is 28 − 10 = 18, only basketball is 18 − 10 = 8, and neither is 50 − (18 + 10 + 8) = 14.

画出两个相交的圆。交集填10。只喜欢足球的人数为28−10=18，只喜欢篮球为18−10=8，都不喜欢为50−(18+10+8)=14。

(i) P(F ∪ B) = (18 + 10 + 8) / 50 = 36/50 = 0.72. This is the probability a student likes at least one sport.

(i) P(F ∪ B) = (18+10+8)/50 = 36/50 = 0.72。 这是学生喜欢至少一种运动的概率。

(ii) P(F’ ∩ B) = 8/50 = 0.16. These students like only basketball.

(ii) P(F’ ∩ B) = 8/50 = 0.16。 这些学生只喜欢篮球。

(iii) P(B | F) = P(B ∩ F) / P(F) = 10/28 ≈ 0.357. Given a student likes football, there is a roughly 35.7% chance they also like basketball.

(iii) P(B | F) = P(B ∩ F) / P(F) = 10/28 ≈ 0.357。 已知一名学生喜欢足球，他们同时也喜欢篮球的概率约为35.7%。

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

The probability distribution of a discrete random variable X is given in the table below. Show that it is a valid distribution and find E(X) and Var(X).

离散随机变量X的概率分布如下表。证明它是合法的分布并求E(X)和Var(X)。

x	0	1	2	3
P(X=x)	0.2	0.35	0.3	0.15

Check Σ P(X=x) = 0.2 + 0.35 + 0.3 + 0.15 = 1.00, and all probabilities are between 0 and 1. The distribution is valid.

验证 Σ P(X=x) = 0.2+0.35+0.3+0.15 = 1.00，且所有概率介于0和1之间，分布合法。

E(X) = Σ x·P(X=x) = (0×0.2) + (1×0.35) + (2×0.3) + (3×0.15) = 0 + 0.35 + 0.6 + 0.45 = 1.4.

E(X²) = (0²×0.2) + (1²×0.35) + (2²×0.3) + (3²×0.15) = 0 + 0.35 + 1.2 + 1.35 = 2.9.

Var(X) = E(X²) − [E(X)]² = 2.9 − 1.4² = 2.9 − 1.96 = 0.94.

The expected value is 1.4 and the variance is 0.94.

期望值为1.4，方差为0.94。

3. Identifying the Binomial Setting | 识别二项分布的条件

A factory produces components with a known defect rate of 2%. A quality inspector randomly selects 30 components and counts the number of defectives, X. Explain why X can be modelled by a binomial distribution.

某工厂生产零件，已知次品率为2%。质检员随机抽取30个零件并记录次品数X。解释为什么X可以用二项分布建模。

A binomial distribution B(n, p) requires: a fixed number of trials (n = 30); each trial is independent; only two outcomes – defective or not defective; and a constant probability of success (defect) p = 0.02. All conditions are met, so X ~ B(30, 0.02).

二项分布B(n, p)要求：试验次数固定(n=30)；每次试验独立；只有两种结果——次品或非次品；每次成功的概率恒定p=0.02。所有条件均满足，因此X ~ B(30, 0.02)。

Always check for independence: the outcome of one component must not affect another. Random sampling from a large production run ensures this holds.

务必检查独立性：一个零件的检测结果不得影响另一个。从大批量生产中随机抽样可保证该条件成立。

4. Calculating Binomial Probabilities | 计算二项概率

Using the model X ~ B(30, 0.02), find the probability that exactly two components are defective, P(X = 2).

利用模型X ~ B(30, 0.02)，计算恰好有两个次品的概率P(X = 2)。

P(X = 2) = ³⁰C₂ × (0.02)² × (0.98)²⁸

³⁰C₂ = 435, so P(X = 2) ≈ 435 × 0.0004 × 0.5688 ≈ 0.0988.

³⁰C₂ = 435，因此P(X = 2) ≈ 435 × 0.0004 × 0.5688 ≈ 0.0988。

To find P(X ≤ 2), sum P(X = 0) + P(X = 1) + P(X = 2). Using respective terms:
P(X = 0) = ⁰.⁹⁸³⁰ ≈ 0.5455, P(X = 1) = ³⁰C₁ × 0.02 × 0.98²⁹ ≈ 0.3340.
P(X ≤ 2) ≈ 0.5455 + 0.3340 + 0.0988 = 0.9783.

求P(X ≤ 2)，将P(X=0)+P(X=1)+P(X=2)相加。各项分别为：P(X=0)≈0.5455，P(X=1)≈0.3340，P(X≤2)≈0.9783。

5. Mean and Variance of a Binomial Distribution | 二项分布的均值与方差

For X ~ B(30, 0.02), calculate the expected number of defectives and the standard deviation.

对于X ~ B(30, 0.02)，求次品的期望个数和标准差。

E(X) = np = 30 × 0.02 = 0.6.
Var(X) = np(1−p) = 30 × 0.02 × 0.98 = 0.588.
Standard deviation = √Var(X) = √0.588 ≈ 0.767.

E(X) = np = 30×0.02 = 0.6；Var(X) = np(1−p) = 30×0.02×0.98 = 0.588；标准差≈0.767。

On average we expect 0.6 defective components in a sample of 30, with a spread of about 0.77.

在30个样本中平均预期有0.6个次品，标准差的波动约0.77。

6. Introduction to Hypothesis Testing | 假设检验入门

A restaurant claims that 90% of customers are satisfied. In a random sample of 50 customers, 42 report being satisfied. Test, at the 5% significance level, whether there is evidence that the satisfaction rate is lower than claimed.

一家餐馆声称顾客满意度为90%。随机抽取50名顾客，其中42人表示满意。在5%显著性水平下检验是否有证据表明满意度低于声称值。

Let p be the true proportion of satisfied customers. Set up hypotheses:
H₀: p = 0.9, H₁: p < 0.9 (one‑tailed test).
Under H₀, the number of satisfied customers in the sample X ~ B(50, 0.9). The observed value is 42.

令p为真实满意比例。建立假设：H₀: p = 0.9，H₁: p < 0.9（单尾检验）。在H₀下，样本中满意人数X ~ B(50, 0.9)。观测值为42。

Calculate p‑value = P(X ≤ 42 | p=0.9). Using binomial tables or calculator, P(X ≤ 42) ≈ 0.043. Since 0.043 < 0.05, we reject H₀.

计算p值 = P(X ≤ 42 | p=0.9)。利用二项分布表或计算器得P(X ≤ 42)≈0.043。由于0.043 < 0.05，拒绝H₀。

Conclusion: There is sufficient evidence at the 5% level to suggest that the satisfaction rate is lower than 90%.

结论：在5%显著性水平下，有充分证据表明满意度低于90%。

7. Setting Up Hypotheses Correctly | 正确建立假设

Many errors occur when students write hypotheses. The null hypothesis H₀ must include an equality (=). The alternative H₁ reflects the suspicion being tested.

许多错误出现在建立假设时。原假设H₀必须包含等号(=)。备择假设H₁反映要检验的疑虑。

Two‑tailed test: H₁: p ≠ value. Use when the claim is “different from” or “changed”.
双尾检验：H₁: p ≠ 数值。当宣称“不同于”或“改变”时使用。
One‑tailed: H₁: p < value (lower tail) or H₁: p > value (upper tail), depending on wording such as “reduced” or “increased”.
单尾：H₁: p < 数值（左侧）或 H₁: p > 数值（右侧），视“降低”或“提高”等措辞而定。

For example, “test whether the proportion of faulty items is greater than 5%” gives H₁: p > 0.05.

例如，“检验次品率是否大于5%”给出H₁: p > 0.05。

Always define p clearly: “p = probability that …” or “p = true proportion of …”.

始终明确定义p：“p = …的概率”或“p = …的真实比例”。

8. Finding Critical Regions and p‑values | 求临界区域与p值

For the coin‑bias example: X ~ B(20, 0.5), H₁: p > 0.5, α = 0.05. Find the critical region.

对于硬币偏差举例：X ~ B(20, 0.5)，H₁: p > 0.5，α=0.05。求临界区域。

We need the smallest c such that P(X ≥ c) ≤ 0.05.
P(X ≥ 15) = 0.0207 ≤ 0.05, P(X ≥ 14) = 0.0577 > 0.05. Hence the critical region is X ≥ 15.

需要最小的c使得P(X ≥ c) ≤ 0.05。P(X ≥ 15)=0.0207 ≤ 0.05，P(X ≥ 14)=0.0577 > 0.05。因此临界区域为X ≥ 15。

The p‑value for an observed x = 15 is P(X ≥ 15) = 0.0207. If the test statistic falls in the critical region, or p‑value < α, reject H₀.

当观测值x=15时，p值 = P(X ≥ 15) = 0.0207。若检验统计量落入临界区域，或p值 < α，则拒绝H₀。

CR (critical region) method is precise for discrete distributions; p‑value method is more common with technology. Both lead to the same conclusion.

临界区域法在离散分布中很精确；p值法使用技术时更常见。两者结论一致。

9. Drawing a Conclusion in Context | 在上下文中得出结论

A conclusion must be written in the context of the problem, not as a generic statistical statement.

Published by TutorHao | Year 12 统计 Revision Series | aleveler.com

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Year 12 OCR Statistics: Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析