📚 A-Level AQA Statistics: Unit Test Mock Paper Walkthrough | A-Level AQA 统计学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for AQA A-Level Statistics, covering key topics and common question types to help you prepare effectively. Each section pairs a worked example with bilingual explanations, highlighting essential techniques and examiner expectations.
本文详细解析了一份 AQA A-Level 统计学单元测试模拟卷，涵盖核心主题和常见题型，帮助你高效备考。每一小节都通过例题与中英双语解析，突出必备技巧和评分要求。
1. Probability and Conditional Probability | 概率与条件概率
Q1. A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random without replacement. Find the probability that both marbles are red.
问题1. 一个袋子装有5颗红色弹珠和3颗蓝色弹珠。随机无放回地抽取两颗弹珠。求两颗都是红色的概率。
We model the situation using a tree diagram or the multiplication rule for dependent events. The probability the first marble is red is 5/8. After removing one red marble, 4 red and 3 blue remain, so the second red probability is 4/7.
我们可以用树状图或相依事件乘法法则分析。第一颗是红色的概率为5/8。抽走一颗红色后，剩下4红3蓝，因此第二颗仍为红色的概率是4/7。
Multiply along the branch: P(both red) = (5/8) × (4/7) = 20/56, which simplifies to 5/14. Always check if the fraction can be reduced and present your final answer in its simplest form.
沿该分支相乘：P(两红) = (5/8) × (4/7) = 20/56，约分后为5/14。务必检查分数是否可以化简，并以最简形式呈现最终答案。
Conditional probability questions often require careful attention to the reduced sample space. Here, without replacement, the outcomes are not independent — a classic AQA scenario.
条件概率题需要特别注意缩小的样本空间。本题中无放回抽取使事件不独立，这是AQA考试中的典型情境。
2. Binomial Distribution | 二项分布
Q2. In a manufacturing process, the probability that an item is defective is 0.2. A random sample of 10 items is selected. Let X be the number of defective items. Find: (a) P(X = 3); (b) E(X) and Var(X).
问题2. 某生产过程中，任一产品为次品的概率是0.2。随机抽取10件产品，设X为次品数量。求：(a) P(X = 3)；(b) E(X) 与 Var(X)。
Here X ~ B(10, 0.2). For (a), use the probability mass function: P(X = k) = ⁿCₖ pᵏ (1-p)ⁿ⁻ᵏ. With n = 10, p = 0.2, k = 3, the combination term is ¹⁰C₃ = 120.
这里 X ~ B(10, 0.2)。(a) 使用概率质量函数：P(X = k) = ⁿCₖ pᵏ (1-p)ⁿ⁻ᵏ。代入 n=10, p=0.2, k=3，组合数 ¹⁰C₃ = 120。
Then P(X = 3) = 120 × (0.2)³ × (0.8)⁷. Compute (0.2)³ = 0.008, (0.8)⁷ ≈ 0.2097, giving approximately 120 × 0.008 × 0.2097 = 0.2013 (4 d.p.).
因此 P(X=3) = 120 × (0.2)³ × (0.8)⁷。计算得 (0.2)³ = 0.008，(0.8)⁷ ≈ 0.2097，乘积约 120 × 0.008 × 0.2097 = 0.2013（保留四位小数）。
For (b), theoretical moments: E(X) = np = 10 × 0.2 = 2; Var(X) = np(1-p) = 10 × 0.2 × 0.8 = 1.6. These provide quick checks for consistency.
(b) 理论矩：E(X) = np = 10×0.2 = 2；Var(X) = np(1-p) = 10×0.2×0.8 = 1.6。这些可作为一致性检验的快速参考。
3. Normal Distribution | 正态分布
Q3. The weights of bags of flour are Normally distributed with mean 1500 g and standard deviation 15 g. A bag is rejected if its weight is less than 1475 g. Find the proportion of bags rejected.
问题3. 袋装面粉的重量服从均值为1500 g、标准差为15 g的正态分布。重量低于1475 g的袋子将被拒收。求被拒收的袋子所占比例。
Standardise: Z = (X – μ) / σ = (1475 – 1500) / 15 = -1.667 (approx.). We need Φ(-1.667), which equals 1 – Φ(1.667) by symmetry.
标准化：Z = (X – μ)/σ = (1475 – 1500)/15 ≈ -1.667。要求 Φ(-1.667)，由对称性等于 1 – Φ(1.667)。
From tables, Φ(1.67) = 0.9525, so Φ(1.667) is interpolated or taken as 0.9523. Hence proportion rejected ≈ 1 – 0.9523 = 0.0477, i.e. about 4.77%.
查表得 Φ(1.67) = 0.9525，插值后 Φ(1.667) 约为0.9523。因此被拒收比例 ≈ 1 – 0.9523 = 0.0477，即约 4.77%。
Always sketch the bell curve and mark the rejection region to avoid tail errors. AQA expects clear standardisation steps and correct table reading.
始终画出正态曲线并标记拒绝域，以避免尾部错误。AQA考试要求清晰的标准化步骤和准确的表格读取。
4. Poisson Distribution | 泊松分布
Q4. The number of accidents occurring in a factory per week follows a Poisson distribution with mean 2.5. Find the probability that in a randomly chosen week there are exactly 2 accidents.
问题4. 某工厂每周发生的事故数服从均值为2.5的泊松分布。求随机选取的一周内恰好发生2起事故的概率。
For Poisson(λ), P(X = k) = e⁻λ λᵏ / k!. Here λ = 2.5, k = 2. So P(X = 2) = e⁻²·⁵ × 2.5² / 2!.
对于 Poisson(λ)，P(X=k) = e⁻λ λᵏ / k!。此处 λ=2.5, k=2，故 P(X=2) = e⁻²·⁵ × 2.5² / 2!。
Compute 2.5² = 6.25, 2! = 2, and e⁻²·⁵ ≈ 0.0821 (from calculator, or e⁻²·⁵ = 1/e²·⁵). Multiply: 0.0821 × 6.25 / 2 ≈ 0.2566. So the required probability is about 0.257 (3 d.p.).
计算得 2.5² = 6.25, 2! = 2, e⁻²·⁵ ≈ 0.0821。相乘：0.0821 × 6.25 / 2 ≈ 0.2566。因此所求概率约为 0.257（三位小数）。
Poisson calculations often involve the exponential constant e. AQA allows calculator use, but you must show the substituted formula before the final answer.
泊松计算常涉及自然常数 e。AQA 允许使用计算器，但必须先展示代入后的公式，再写出最终结果。
5. Sampling and the Central Limit Theorem | 抽样与中心极限定理
Q5. A population has a mean μ = 60 and standard deviation σ = 8. A random sample of size n = 36 is drawn. Describe the approximate distribution of the sample mean X̄, and find P(X̄ > 62).
问题5. 总体均值为60，标准差为8。随机抽取大小为36的样本。描述样本均值X̄的近似分布，并求 P(X̄ > 62)。
By the Central Limit Theorem, for large n (n ≥ 30), X̄ is approximately Normal with mean μ = 60 and standard error σ/√n = 8/√36 = 8/6 ≈ 1.333.
根据中心极限定理，当n较大（n ≥ 30）时，X̄ 近似服从正态分布，均值为μ=60，标准误 σ/√n = 8/√36 = 8/6 ≈ 1.333。
Then standardise: Z = (62 – 60) / 1.333 = 2 / 1.333 = 1.5 exactly (since 2 ÷ (8/6) = 12/8 = 1.5). So P(X̄ > 62) = P(Z > 1.5) = 1 – Φ(1.5).
标准化：Z = (62 – 60)/1.333 = 2÷1.333 = 1.5。因此 P(X̄ > 62) = P(Z > 1.5) = 1 – Φ(1.5)。
From tables, Φ(1.5) = 0.9332, hence P(X̄ > 62) = 0.0668. Even if the population is not Normal, the CLT justifies the Normal approximation here.
查表，Φ(1.5) = 0.9332，因此 P(X̄ > 62) = 0.0668。即使总体不服从正态分布，中心极限定理仍保证了这里的正态近似。
6. Confidence Intervals | 置信区间
Q6. A sample of 50 rods gives a mean length x̄ = 25.4 cm. The population standard deviation is known to be 0.8 cm. Construct a 95% confidence interval for the population mean μ.
问题6. 某样本50根金属棒的平均长度 x̄ = 25.4 cm。已知总体标准差为0.8 cm。求总体均值μ的95%置信区间。
When σ is known, the 95% confidence interval is x̄ ± z × σ/√n, with z = 1.96 for two-tailed 5% significance. Here standard error = 0.8/√50 ≈ 0.1131.
当σ已知时，95% 置信区间为 x̄ ± z × σ/√n，其中双尾5%显著性对应的z值为1.96。此处标准误 = 0.8/√50 ≈ 0.1131。
Margin of error = 1.96 × 0.1131 ≈ 0.2217. Lower bound = 25.4 – 0.2217 = 25.1783; upper bound = 25.4 + 0.2217 = 25.6217. We can round to 25.18 to 25.62 (2 d.p.).
误差幅度 = 1.96 × 0.1131 ≈ 0.2217。下限 = 25.4 – 0.2217 = 25.1783；上限 = 25.4 + 0.2217 = 25.6217。可四舍五入为 25.18 至 25.62（两位小数）。
Interpretation: we are 95% confident that the true mean length lies between 25.18 cm and 25.62 cm. The interval narrows with larger sample size.
解读：我们有95%的把握认为真实平均长度介于25.18 cm与25.62 cm之间。样本量越大，区间越窄。
7. Hypothesis Testing for a Mean | 均值的假设检验
Q7. A machine fills bottles with a mean volume of 500 ml. A sample of 40 bottles has a mean of 498.5 ml. The filling volume is known to have a standard deviation of 4 ml. Test at the 5% significance level whether the machine is underfilling.
问题7. 一台灌装机设定的平均灌装量为500 ml。从40瓶中测得样本均值为498.5 ml。已知灌装量的标准差为4 ml。在5%显著性水平下检验该机器是否存在灌装不足。
Hypotheses: H₀: μ = 500; H₁: μ < 500 (one-tailed, as we suspect underfilling). Test statistic: Z = (x̄ - μ₀) / (σ/√n) = (498.5 - 500) / (4/√40) = -1.5 / 0.6325 ≈ -2.37.
假设：H₀: μ = 500；H₁: μ < 500（单尾检验，因为怀疑灌装不足）。检验统计量：Z = (x̄ - μ₀)/(σ/√n) = (498.5 - 500)/(4/√40) ≈ -1.5/0.6325 ≈ -2.37。
Critical value at 5% left-tail: z = -1.645. Since -2.37 < -1.645, the test statistic falls in the rejection region. We reject H₀.
5%左侧临界值 z = -1.645。因为 -2.37 < -1.645，检验统计量落入拒绝域。拒绝 H₀。
Conclusion: there is sufficient evidence at the 5% level to suggest the machine is underfilling. Always state the conclusion in context.
结论：在5%显著性水平下，有充分证据表明该机器存在灌装不足。一定要结合背景陈述结论。
8. Product Moment Correlation | 积矩相关系数
Q8. For a sample of 10 paired (x, y) values, Sₓₓ = 140, Sᵧᵧ = 90, Sₓᵧ = 84. Calculate the product moment correlation coefficient r and test whether the population correlation ρ is positive, using a 1% significance level.
问题8. 一组10对数据的样本中，Sₓₓ = 140, Sᵧᵧ = 90, Sₓᵧ = 84。计算积矩相关系数 r，并在1%显著性水平下检验总体相关系数 ρ 是否为正。
r = Sₓᵧ / √(Sₓₓ Sᵧᵧ) = 84 / √(140 × 90) = 84 / √12600 ≈ 84 / 112.25 = 0.748 (3 d.p.). This suggests a positive linear relationship.
r = Sₓᵧ / √(Sₓₓ Sᵧᵧ) = 84 / √(140×90) = 84 / √12600 ≈ 84/112.25 = 0.748（三位小数）。这表明存在正向线性关系。
For test: H₀: ρ = 0, H₁: ρ > 0. Test statistic t = r √(n-2) / √(1 – r²) with n-2 = 8 df. t = 0.748 × √8 / √(1 – 0.748²) ≈ 2.116 / 0.713 ≈ 2.97.
检验：H₀: ρ = 0，H₁: ρ > 0。检验统计量 t = r √(n-2)/√(1-r²)，自由度=8。t = 0.748×√8 / √(1-0.748²) ≈ 2.116/0.713 ≈ 2.97。
Critical value from t-tables for 8 df and one-tail 1% is 2.896. Since 2.97 > 2.896, reject H₀. There is evidence of positive correlation.
查t分布表，单尾1%显著性水平下自由度为8的临界值为2.896。因为2.97 > 2.896，拒绝H₀。有证据表明存在正相关。
9. Chi-Squared Test for Independence | 卡方独立性检验
Q9. A survey of 200 people examined the relationship between age group and preference for tea or coffee. The observed frequencies are given in the table below. Test at the 5% level whether age and drink preference are independent.
| Tea | Coffee | Total | |
| Under 30 | 40 | 60 | 100 |
| 30 and over | 70 | 30 | 100 |
| Total | 110 | 90 | 200 |
问题9. 一项200人的调查考察了年龄组与茶或咖啡偏好的关系。观测频数见下表。在5%显著性水平下检验年龄与饮品偏好是否独立。
Compute expected frequencies under independence: Expected = (row total × column total) / grand total. For Under 30 & Tea: (100 × 110)/200 = 55. Under 30 & Coffee: (100×90)/200 = 45. 30+ & Tea: (100×110)/200 = 55. 30+ & Coffee: 45.
计算独立假设下的期望频数：期望值 = (行合计 × 列合计) / 总计。30岁以下 & 茶：(100×110)/200 = 55。30岁以下 & 咖啡：45。30岁及以上 & 茶：55；咖啡：45。
χ² = Σ (O – E)² / E. For Tea/Under30: (40-55)²/55 = 4.091. For Coffee/Under30: (60-45)²/45 = 5.0. For Tea/30+: (70-55)²/55 = 4.091. For Coffee/30+: (30-45)²/45 = 5.0. Sum = 18.182.
χ² = Σ (O-E)²/E。茶/30以下：(40-55)²/55=4.091；咖啡/30以下：5.0；茶/30+：4.091；咖啡/30+：5.0。总和 = 18.182。
Degrees of freedom = (2-1)×(2-1) = 1. 5% critical value for χ²(1) = 3.841. Since 18.182 > 3.841, reject H₀. There is a significant association between age and preference.
自由度 = (2-1)×(2-1)=1。χ²(1) 在5%显著性下的临界值为3.841。因18.182 > 3.841，拒绝H₀。年龄与偏好之间存在显著关联。
10. Common Pitfalls and Revision Tips | 常见易错点与备考建议
Pitfall: Confusing P(A|B) and P(B|A). Always re-read the question to identify which event is given. A tree diagram helps visualise conditional structure.
易错点：混淆 P(A|B) 与 P(B|A)。务必重读题干以确定哪个事件为条件。树状图有助于将条件结构可视化。
Pitfall: Forgetting to apply continuity correction when using the Normal approximation to the Binomial. Check if the question explicitly states that an approximation is suitable and adjust boundaries by 0.5.
易错点：用正态分布近似二项分布时忘记连续性校正。若题目明确说明可使用近似，记得将边界值调整 ±0.5。
Pitfall: Using population standard deviation in confidence intervals when the sample standard deviation should be used with the t-distribution. Distinguish between ‘σ known’ and ‘σ unknown’ scenarios.
易错点：在应使用样本标准差和t分布时误用了总体标准差。注意区分“σ已知”与“σ未知”的情境。
Pitfall: Stating ‘accept H₀’ instead of ‘do not reject H₀’. AQA insists on correct phrasing:
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