Year 13 CIE Statistics Unit Test Mock Paper Solutions | Year 13 CIE 统计：单元测试模拟卷解析

📚 Year 13 CIE Statistics Unit Test Mock Paper Solutions | Year 13 CIE 统计：单元测试模拟卷解析

This article walks you through a full mock unit test designed for the Year 13 CIE Statistics syllabus. Every question is broken down with step‑by‑step reasoning, covering probability generating functions, continuous random variables, hypothesis tests (binomial, normal, chi‑squared), and linear regression. Use these detailed solutions to sharpen your exam technique and deepen your understanding of key statistical concepts.

本文带你完整拆解一套针对 Year 13 CIE 统计课程的单元模拟卷。每一道题都附有分步解析，涵盖概率生成函数、连续随机变量、假设检验（二项、正态、卡方）以及线性回归。通过这份详尽的解答，你可以打磨应试技巧，深化对核心统计概念的理解。

1. Mock Paper Overview | 模拟卷概览

This mock paper contains seven compulsory questions, totalling 75 marks. It is designed to mirror the style and demand of CIE A‑Level Statistics Unit Test components. The coverage includes discrete and continuous distributions, parametric and non‑parametric hypothesis testing, and bivariate data analysis. Each question targets a specific cluster of learning outcomes from the syllabus, ensuring you practise both routine calculations and interpretive reasoning.

本模拟卷包含 7 道必答题，满分 75 分。试卷在风格与难度上贴近 CIE A‑Level 统计单元测试题型。内容覆盖离散与连续分布、参数及非参数假设检验，以及双变量数据分析。每一题都紧扣大纲中特定的学习成果群，帮助你在常规计算与解释性推理两方面同步提升。

You should attempt the paper under timed conditions (90 minutes) before reading the solutions. All required statistical tables are assumed to be available, and exact binomial probabilities or critical values from the normal and chi‑squared tables are used where appropriate.

建议你在阅读解析前，先在限时 90 分钟的条件下独立完成试卷。假定所有必要的统计表均可查阅，解析中将适时使用精确二项概率、正态分布临界值以及卡方分布临界值。

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2. Q1: Discrete Random Variable & PGF | 第1题：离散随机变量与概率生成函数

The first question provides the probability distribution of a discrete random variable X: P(X=0)=0.1, P(X=1)=0.3, P(X=2)=0.4, P(X=3)=0.2. You are asked to write down the probability generating function (PGF) and use it to find E(X) and Var(X).

第一题给出了离散随机变量 X 的分布：P(X=0)=0.1, P(X=1)=0.3, P(X=2)=0.4, P(X=3)=0.2。要求写出概率生成函数 (PGF)，并利用它求 E(X) 与 Var(X)。

The PGF is defined as G(t) = E(t^X) = Σ t^x P(X=x). Substituting the given probabilities yields:

PGF 定义为 G(t) = E(t^X) = Σ t^x P(X=x)。代入给定概率得：

G(t) = 0.1 + 0.3 t + 0.4 t² + 0.2 t³

To obtain E(X), differentiate G(t) and set t = 1. G'(t) = 0.3 + 0.8 t + 0.6 t², so G'(1) = 0.3 + 0.8 + 0.6 = 1.7. Therefore E(X) = 1.7.

为求 E(X)，对 G(t) 求导并令 t=1。G'(t) = 0.3 + 0.8 t + 0.6 t²，因此 G'(1) = 1.7，即 E(X) = 1.7。

For the variance, first find E(X²) using the second derivative. G”(t) = 0.8 + 1.2 t, so G”(1) = 2.0. Then E(X²) = G”(1) + G'(1) = 2.0 + 1.7 = 3.7. Hence Var(X) = E(X²) − [E(X)]² = 3.7 − 1.7² = 3.7 − 2.89 = 0.81.

方差则需先利用二阶导数求 E(X²)。G”(t) = 0.8 + 1.2 t，故 G”(1) = 2.0。于是 E(X²) = G”(1) + G'(1) = 3.7。最终 Var(X) = 3.7 − 2.89 = 0.81。

A common mistake is forgetting to add the first derivative term when computing E(X²). Always recall the identity E(X²) = G”(1) + G'(1).

常见错误是在计算 E(X²) 时忘记加上一阶导数项。务必牢记恒等式 E(X²) = G”(1) + G'(1)。

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3. Q2: Continuous Random Variable | 第2题：连续随机变量

A continuous random variable Y has probability density function f(y) = k y (4 − y) for 0 ≤ y ≤ 4, and zero elsewhere. First, determine the constant k by integrating f(y) over its support and setting the result equal to 1.

连续随机变量 Y 的概率密度函数为 f(y) = k y (4 − y)，0 ≤ y ≤ 4，其他区间为 0。首先通过在整个支撑集上积分并令结果为 1 来求常数 k。

∫₀⁴ k y (4 − y) dy = k ∫₀⁴ (4y − y²) dy = k [2y² − y³/3]₀⁴ = k (32 − 64/3) = k (32/3). Set this equal to 1 → k = 3/32.

∫₀⁴ k y (4 − y) dy = k ∫₀⁴ (4y − y²) dy = k [2y² − y³/3]₀⁴ = k (32 − 64/3) = k (32/3)。令其等于 1，得 k = 3/32。

To find the cumulative distribution function F(y), integrate f(t) from 0 to y for 0 ≤ y ≤ 4. F(y) = (3/32) ∫₀ʸ (4t − t²) dt = (3/32)(2y² − y³/3) = (3y²/16) − (y³/32). For y < 0, F(y)=0; for y > 4, F(y)=1.

累积分布函数 F(y) 在 0 ≤ y ≤ 4 上由对 f(t) 积分得到。F(y) = (3/32) ∫₀ʸ (4t − t²) dt = (3/32)(2y² − y³/3) = (3y²/16) − (y³/32)。当 y < 0 时 F(y)=0，y > 4 时 F(y)=1。

The median m satisfies F(m) = 0.5. Solve (3m²/16) − (m³/32) = 0.5. Multiplying through by 32 gives 6m² − m³ = 16 → m³ − 6m² + 16 = 0. By inspection m=2 is a root (8−24+16=0). Factoring yields (m−2)(m² − 4m − 8)=0. The only root in [0,4] is m=2. Thus the median is 2.

中位数 m 满足 F(m) = 0.5。方程 (3m²/16) − (m³/32) = 0.5 两边乘 32 得 6m² − m³ = 16 → m³ − 6m² + 16 = 0。试根 m=2 满足，因式分解得 (m−2)(m² − 4m − 8)=0，在 [0,4] 内唯一根为 m=2，故中位数为 2。

Finally, P(Y > 1) = 1 − F(1) = 1 − [3/16 − 1/32] = 1 − (6/32 − 1/32) = 1 − 5/32 = 27/32 ≈ 0.84375.

最后 P(Y > 1) = 1 − F(1) = 1 − [3/16 − 1/32] = 1 − 5/32 = 27/32。

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4. Q3: Binomial Hypothesis Test | 第3题：二项假设检验

A factory claims that its defective rate is at most 5%. A random sample of 50 items is inspected, and 6 defectives are found. Conduct a hypothesis test at the 5% significance level to determine whether the claim should be rejected. Let p be the true proportion of defectives.

工厂声称次品率不超过 5%。随机抽取 50 件产品，发现 6 件次品。在 5% 显著性水平下检验该声称是否应被拒绝。记 p 为真实次品率。

Set H₀: p = 0.05 and H₁: p > 0.05 (one‑tailed test). Under H₀, X ~ B(50, 0.05). Since np = 2.5 < 5, the normal approximation is not appropriate; we use exact binomial probabilities.

设 H₀: p = 0.05，H₁: p > 0.05（单尾检验）。在 H₀ 下，X ~ B(50, 0.05)。因 np = 2.5 < 5，不适合用正态近似，故采用精确二项概率。

The p‑value = P(X ≥ 6 | p=0.05) = 1 − P(X ≤ 5). Using binomial tables or computation, P(X ≤ 5) = 0.9622 (check cumulative probability). Hence p‑value ≈ 1 − 0.9622 = 0.0378.

P 值 = P(X ≥ 6 | p=0.05) = 1 − P(X ≤ 5)。查表或计算得 P(X ≤ 5) = 0.9622，因此 p‑值 ≈ 0.0378。

Since p‑value = 0.0378 < 0.05, there is sufficient evidence to reject H₀. The sample data suggest that the true defective rate exceeds 5%. The factory's claim is not supported at the 5% significance level.

由于 p‑值 0.0378 < 0.05，有充分证据拒绝 H₀。样本数据表明真实次品率超过 5%。在 5% 显著性水平下，该厂的声称不被支持。

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5. Q4: Normal Mean Hypothesis Test | 第4题：正态均值假设检验

A bottling process fills bottles with a mean volume claimed to be 500 ml. The standard deviation is known to be 4 ml. A quality inspector takes a random sample of 36 bottles and finds a sample mean of 498 ml. Test at the 1% significance level whether the mean volume is less than 500 ml.

某灌装工序声称平均容量为 500 ml，已知标准差为 4 ml。质检员随机抽取 36 瓶，测得样本均值为 498 ml。在 1% 显著性水平下检验平均容量是否低于 500 ml。

Hypotheses: H₀: μ = 500 vs H₁: μ < 500 (left‑tailed test). The test statistic is Z = (x̄ − μ₀) / (σ/√n) = (498 − 500) / (4/√36) = −2 / (4/6) = −3.

假设：H₀: μ = 500，H₁: μ < 500（左尾检验）。检验统计量 Z = (x̄ − μ₀) / (σ/√n) = (498 − 500) / (4/√36) = −2 / (4/6) = −3。

At the 1% level, the critical value for a one‑tailed test from the standard normal table is z = −2.326 (or −2.33). Since −3 < −2.326, the test statistic falls in the rejection region.

在 1% 水平下，标准正态表单尾临界值为 z = −2.326。因 −3 < −2.326，检验统计量落入拒绝域。

Therefore we reject H₀. There is sufficient evidence at the 1% significance level to conclude that the mean filling volume is less than 500 ml. The process likely needs recalibration.

因此拒绝 H₀。在 1% 显著性水平下有足够证据表明平均灌装量低于 500 ml，工序可能需要重新校准。

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6. Q5: Chi‑squared Goodness‑of‑fit Test | 第5题：卡方拟合优度检验

A six‑sided die is rolled 60 times, yielding the observed frequencies: 1: 8, 2: 9, 3: 12, 4: 10, 5: 11, 6: 10. Perform a goodness‑of‑fit test at the 5% significance level to check if the die is fair.

一颗六面骰子掷 60 次，各面观察频数为：1点8次，2点9次，3点12次，4点10次，5点11次，6点10次。在 5% 显著性水平下进行拟合优度检验，判断骰子是否均匀。

H₀: the die is fair (all pᵢ = 1/6). The expected frequency for each face is 60 × 1/6 = 10. Compute the test statistic χ² = Σ (O − E)² / E.

H₀：骰子均匀（各面概率为 1/6）。每面的期望频数为 10。计算 χ² = Σ (O − E)² / E。

χ² = (8−10)²/10 + (9−10)²/10 + (12−10)²/10 + (10−10)²/10 + (11−10)²/10 + (10−10)²/10 = 0.4 + 0.1 + 0.4 + 0 + 0.1 + 0 = 1.0.

χ² = (8−10)²/10 + (9−10)²/10 + (12−10)²/10 + (10−10)²/10 + (11−10)²/10 + (10−10)²/10 = 0.4 + 0.1 + 0.4 + 0 + 0.1 + 0 = 1.0。

Degrees of freedom = number of categories − 1 = 5. The 5% critical value from the χ²(5) table is 11.07. Since 1.0 < 11.07, we do not reject H₀. There is no evidence to suggest the die is biased.

自由度 = 类别数 − 1 = 5。χ²(5) 的 5% 临界值为 11.07。因 1.0 < 11.07，不拒绝 H₀。没有证据表明骰子不均匀。

It is essential that all expected frequencies are at least 5; here the requirement is satisfied, so the chi‑squared approximation is valid.

务必确认所有期望频数均 ≥ 5；本例满足要求，卡方近似有效。

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7. Q6: Chi‑squared Test for Independence | 第6题：卡方独立性检验

A survey examines the relationship between gender (Male, Female) and favourite sport (Football, Tennis, Basketball). The observed frequencies are summarised in the 2×3 contingency table below. Use a 5% significance level to test for independence.

一项调查考察性别（男、女）与最喜爱运动（足球、网球、篮球）之间的关系。观察频数汇总于以下 2×3 列联表。在 5% 显著性水平下检验独立性。

	Football	Tennis	Basketball	Total
Male	25	10	15	50
Female	20	18	12	50
Total	45	28	27	100

Expected frequencies are calculated using (row total × column total) / grand total. For example, Male–Football: 50 × 45 / 100 = 22.5. The complete set of expected values is shown next.

期望频数按 (行合计 × 列合计) / 总合计 计算。如 男–足球：50 × 45 / 100 = 22.5。完整期望值如下。

Expected: Male: 22.5, 14.0, 13.5; Female: 22.5, 14.0, 13.5

Then χ² = Σ (O − E)² / E = (25−22.5)²/22.5 + (10−14)²/14 + (15−13.5)²/13.5 + (20−22.5)²/22.5 + (18−14)²/14 + (12−13.5)²/13.5 = 0.2778 + 1.1429 + 0.1667 + 0.2778 + 1.1429 + 0.1667 = 3.1748.

计算 χ² = Σ (O − E)² / E = 0.2778 + 1.1429 + 0.1667 + 0.2778 + 1.1429 + 0.1667 ≈ 3.17。

Degrees of freedom = (2−1) × (3−1) = 2. The 5% critical value for χ²(2) is 5.991. Since 3.17 < 5.991, we do not reject H₀. There is no significant association between gender and favourite sport in this sample.

自由度 = (2−1) × (3−1) = 2。χ²(2) 的 5% 临界值为 5.991。因 3.17 < 5.991，不拒绝 H₀。该样本中未发现性别与最喜爱运动之间的显著关联。

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8. Q7: Linear Regression and Correlation | 第7题：线性回归与相关

The final question provides five bivariate observations: (1, 2.1), (2, 3.9), (3, 5.8), (4, 7.8), (5, 9.7). You are asked to fit the regression line of y on x, compute the product‑moment correlation coefficient r, and test whether there is a significant linear relationship.

最后一题给出五对观测值：(1,2.1), (2,3.9), (

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