Pre-U Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥Pre-U统计：单元测试模拟卷解析

📚 Pre-U Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥Pre-U统计：单元测试模拟卷解析

This article presents a full mock unit test paper for the Cambridge Pre-U Statistics syllabus, followed by detailed step-by-step solutions. Each question targets key assessment objectives such as probability laws, discrete and continuous distributions, sampling theory, hypothesis testing, confidence intervals, chi‑squared tests, and bivariate analysis. Careful exposition of the reasoning helps students consolidate both computational skills and conceptual understanding.

本文提供一份根据剑桥 Pre‑U 统计课程设计的完整单元测试模拟卷，并逐题给出详细解析。试题覆盖概率法则、离散与连续分布、抽样理论、假设检验、置信区间、卡方检验和双变量分析等核心考点。通过逐步拆解解题思路，帮助同学们巩固运算技能并深化概念理解。

1. Probability without Replacement | 不放回概率问题

Problem: A bag contains six red balls and four green balls. Two balls are drawn at random without replacement. Find the probability that (a) both balls are red, (b) at least one ball is green.

题目：一个袋子中有 6 个红球和 4 个绿球，随机不放回地抽取两球。求 (a) 两球均为红色的概率，(b) 至少有一球为绿色的概率。

second draw probability of red is 5/9. Multiplying, P(RR) = (6/10) × (5/9) = 30/90 = 1/3.

在 (a) 部分，第一次抽到红球的概率为 6/10。取出后剩 5 个红球及共 9 个球，因此第二次抽到红球的条件概率为 5/9。相乘得 P(RR) = (6/10) × (5/9) = 30/90 = 1/3。

For part (b), the event "at least one green" is the complement of "both red". Hence P(at least one green) = 1 − P(RR) = 1 − 1/3 = 2/3.

在 (b) 部分，“至少有一绿”是“两球皆红”的补事件，故 P(至少一绿) = 1 − P(RR) = 1 − 1/3 = 2/3。

Always check that the sample space changes under without‑replacement sampling. The multiplication rule for dependent events is essential, and using the complement often simplifies calculations.

务必留意不放回抽样下样本空间的变动。依赖事件的乘法法则是关键，而利用补事件常常能简化计算。

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2. Binomial Distribution Calculations | 二项分布计算

Problem: Let X ~ B(12, 0.25). Evaluate, correct to four decimal places: (a) P(X = 3), (b) P(X ≥ 10).

题目：设 X ~ B(12, 0.25)，求以下概率并保留四位小数：(a) P(X = 3), (b) P(X ≥ 10)。

For a binomial probability mass function, P(X = x) = ⁿCₓ pˣ (1−p)ⁿ⁻ˣ C3 × (0.25)³ × (0.75)⁹. 12C3 = 220, (0.25)³ = 0.015625, (0.75)⁹ = 0.0750847 approximately.

二项概率质量函数为 P(X = x) = ⁿCₓ pˣ (1−p)ⁿ⁻ˣ。这里 n = 12, p = 0.25, q = 0.75。(a) P(X = 3) = 12C3 × (0.25)³ × (0.75)⁹，其中 12C3 = 220，(0.25)³ = 0.015625，(0.75)⁹ ≈ 0.0750847。

P(X = 3) = 220 × 0.015625 × 0.0750847 ≈ 0.2581

Thus P(X = 3) ≈ 0.2581 (to 4 d.p.).

因此 P(X = 3) ≈ 0.2581（保留四位小数）。

0.5625, giving product ≈ 66 × 9.5367×10⁻⁷ × 0.5625 ≈ 3.54×10⁻⁵. P(X = 11) = 12 × (0.25)¹¹ × 0.75 ≈ 1.19×10⁻⁶; P(X = 12) = (0.25)¹² ≈ 5.96×10⁻⁸. Summing yields approximately 3.66×10⁻⁵, which to 4 d.p. is 0.0000.

(b) P(X ≥ 10) = P(10) + P(11) + P(12)。分别计算：P(X = 10) = 12C10 × (0.25)¹⁰ × (0.75)² ≈ 66 × 9.5367×10⁻⁷ × 0.5625 ≈ 3.54×10⁻⁵；P(X = 11) = 12 × (0.25)¹¹ × 0.75 ≈ 1.19×10⁻⁶；P(X = 12) = (0.25)¹² ≈ 5.96×10⁻⁸。累加后约 3.66×10⁻⁵，四舍五入至四位小数得 0.0000。

When p is small and n is moderate, probabilities in the upper tail can become extremely small. Using statistical tables or software is strongly recommended, but showing the underlying formula is essential in the exam.

当 p 较小而 n 中等时，上尾概率可能极其微小。考试中强烈建议使用统计表或计算器，但写出底层公式仍是必需的。

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3. Poisson Distribution Problem | 泊松分布问题

Problem: The number of emails received by a server in one hour follows a Poisson distribution with mean 3. Find the probability that in a randomly chosen hour (a) exactly two emails are received, (b) at most one email is received.

题目：某服务器每小时收到邮件数服从均值为 3 的泊松分布。求随机选取的一个小时内 (a) 恰好收到两封邮件的概率，(b) 至多收到一封邮件的概率。

For Poisson(λ), P(X = x) = (e⁻λ λˣ) / x!. With λ = 3, (a) P(X = 2) = (e⁻³ × 3²) / 2! = (e⁻³ × 9)/2. Using e⁻³ ≈ 0.049787, we obtain 9 × 0.049787 / 2 ≈ 0.2240.

对于泊松分布 Poisson(λ)，P(X = x) = (e⁻λ λˣ) / x!。λ = 3，(a) P(X = 2) = (e⁻³ × 3²) / 2! = (e⁻³ × 9)/2。取 e⁻³ ≈ 0.049787，计算得 9 × 0.049787 / 2 ≈ 0.2240。

≤ 1) = P(X = 0) + P(X = 1) = e⁻³ + (e⁻³ × 3) = e⁻³(1 + 3) = 4 e⁻³ ≈ 4 × 0.049787 = 0.1991.

(b) P(X ≤ 1) = P(0) + P(1) = e⁻³ + (e⁻³ × 3) = e⁻³(1 + 3) = 4 e⁻³ ≈ 4 × 0.049787 = 0.1991。

The Poisson distribution is a one‑parameter model that often describes rare events. Remember that the mean equals the variance, and it can approximate the binomial distribution when n is large and p is small.

泊松分布是描述稀有事件的单参数模型，其均值等于方差。当 n 很大且 p 很小时，它常用来逼近二项分布，这一性质需加以留意。

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4. Normal Distribution and Sampling Distribution | 正态分布与抽样分布

Problem: Adult male heights in a region are normally distributed with mean 175 cm and standard deviation 8 cm. A random sample of 64 men is selected. Find the probability that the sample mean height lies between 174 cm and 176 cm.

题目：某地区成年男性身高服从正态分布，均值为 175 cm，标准差为 8 cm。随机抽取 64 名男性，求样本均值介于 174 cm 与 176 cm 之间的概率。

By the Central Limit Theorem (or exact normality here), the sampling distribution of the sample mean X̄ is N(μ, σ/√n) = N(175, 8/√64 = 1). So the standard error is 1 cm.

根据中心极限定理（在此亦为精确正态），样本均值 X̄ 的抽样分布为 N(μ, σ/√n) = N(175, 8/√64 = 1)，即标准误为 1 cm。

We need P(174 < X̄ < 176) = P((174−175)/1 < Z < (176−175)/1) = P(−1 < Z < 1).

需求 P(174 < X̄ < 176) = P((174−175)/1 < Z < (176−175)/1) = P(−1 < Z < 1)。

P(−1 < Z < 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6826

rd normal tables, Φ(1) = 0.8413, Φ(−1) = 0.1587, giving 0.6826. Thus there is about a 68.3% chance that the sample mean falls within 1 cm of the population mean.

查标准正态表，Φ(1) = 0.8413，Φ(−1) = 0.1587，得 0.6826。故样本均值落在总体均值±1 cm 范围内的概率约为 68.3%。

Always remember to divide the population standard deviation by √n when working with sample means. This question reinforces the 68–95–99.7 rule for normal distributions.

处理样本均值时务必用总体标准差除以 √n。本题也巩固了正态分布的 68–95–99.7 经验法则。

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5. Hypothesis Testing for a Population Mean | 总体均值的假设检验

Problem: A machine fills packets of sugar labelled as containing 500 g. A random sample of 25 packets gives a mean mass of 498 g and a standard deviation of 6 g. Test, at the 5% significance level, whether the machine is underfilling. State your hypotheses, the test statistic, the critical value, and your conclusion.

题目：一台机器包装标注 500 g 的糖包。随机抽取 25 包，得样本均值 498 g，标准差 6 g。在 5% 显著性水平下检验机器是否存在不足量灌装。写出假设、检验统计量、临界值以及结论。

Let μ be the true mean fill mass. Hypotheses: H₀: μ = 500 g vs H₁: μ < 500 g (one‑tailed). Because the population standard deviation is unknown, a one‑sample t‑test is appropriate.

设 μ 为真实的平均灌装量。假设：H₀: μ = 500 g，H₁: μ < 500 g（单尾）。由于总体标准差未知，适合采用单样本 t 检验。

Test statistic: t = (x̄ − μ₀) / (s/√n) = (498 − 500) / (6/√25) = −2 / 1.2 = −1.6667 (to 4 d.p.). Degrees of freedom ν = 24.

检验统计量：t = (498 − 500) / (6/√25) = −2 / 1.2 = −1.6667（保留四位）。自由度 ν = 24。

At the 5% significance level (one‑tailed), the critical value from t‑tables is t_crit 711, the observed t is not more extreme than the critical value. Thus we do not reject H₀.

在 5% 显著性水平（单尾）下，查 t 分布表得临界值 t_crit = −1.711。因 −1.667 > −1.711，观测到的 t 值并未比临界值更极端，故不拒绝 H₀。

Conclusion: There is insufficient evidence at the 5% level to conclude that the machine is underfilling. The shortfall could reasonably be due to sampling variability.

结论：在 5% 水平下没有足够证据表明机器灌装不足，观测到的短缺可能由抽样波动引起。

Always clearly specify whether the test is one‑ or two‑tailed, and check whether the sample supports the direction of the alternative hypothesis. Reporting the p‑value (approximately 0.055) would further illustrate the marginal result.

务必明确说明是单尾或双尾检验，并检查样本是否支持备择假设的方向。额外报告 p 值（约 0.055）能进一步说明结果处于临界边缘。

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6. Confidence Interval for a Proportion | 比例的置信区间

Problem: In a random poll of 400 voters, 220 indicate they will vote for Party A. Construct a 95% confidence interval for the true proportion of voters supporting Party A.

题目：一项随机民调显示，400 名选民中有 220 人表示支持 A 党。构建总体支持比例的 95% 置信区间。

Sample proportion: p̂ = 220/400 = 0.55. The standard error of p̂ is √(p̂(1 − p̂)/n) = √0.55 × 0.45 / 400.

样本比例：p̂ = 220/400 = 0.55。标准误 SE = √0.55×0.45 / 400。

SE = √0.2475 / 400 = √0.00061875 ≈ 0.02487

For a 95% confidence level, the critical z‑value is 1.96. Hence the margin of error is 1.96 × 0.02487 ≈ 0.0487.

对于 95% 置信水平，临界 z 值为 1.96。误差界限 = 1.96 × 0.02487 ≈ 0.0487。

The 95% confidence interval is p̂ ± z* × SE = 0.55 ± 0.0487 = (0.5013, 0.5987).

95% 置信区间为 p̂ ± z* × SE = 0.55 ± 0.0487 = (0.5013, 0.5987)。

Interpretation: We are 95% confident that the true proportion of support for Party A lies between 50.13% and 59.87%. The method relies on the normal approximation, which is valid because np̂ and n(1−p̂) both exceed 10.

解释：我们有 95% 的信心认为 A 党的真实支持率介于 50.13% 到 59.87% 之间。该方法依赖正态近似，因 np̂ 与 n(1−p̂) 均大于 10，正态条件成立。

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7. Chi‑Squared Test of Independence | 卡方独立性检验

Problem: A researcher wishes to test whether gender and drink preference (coffee or tea) are associated. Observed frequencies are: Male – coffee 40, tea 30; Female – coffee 35, tea 45. Carry out a chi‑squared test at the 5% significance level, stating hypotheses, expected frequencies, test statistic, degrees of freedom, critical value, and conclusion.

题目：研究者想检验性别与饮品偏好（咖啡或茶）是否有关联。观测频数为：男性——咖啡 40，茶 30；女性——咖啡 35，茶 45。在 5% 显著性水平下进行卡方检验，写出假设、期望频数、检验统计量、自由度、临界值和结论。

w totals: Male = 70, Female = 80; Column totals: Coffee = 75, Tea = 75; Grand total = 150.

H₀：性别与饮品偏好独立；H₁：两者不独立。行合计：男 70，女 80；列合计：咖啡 75，茶 75；总计 150。

xpected frequencies under independence: E = (row total × column total) / grand total. Male‑coffee: 70 × 75 / 150 = 35; Male‑tea: 35; Female‑coffee: 80 × 75 /

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