A-Level AQA Statistics: In-Depth Analysis of Past Paper Questions | A-Level AQA 统计：历年真题深度解析

📚 A-Level AQA Statistics: In-Depth Analysis of Past Paper Questions | A-Level AQA 统计：历年真题深度解析

Past paper questions are the most valuable resource for mastering AQA Statistics. They reveal recurring question styles, hidden pitfalls, and the precise depth of reasoning expected by examiners. This article dissects authentic exam-style problems across all major topics, offering step‑by‑step solutions, common mistakes, and strategies that will sharpen your performance.

历年真题是攻克 AQA 统计的最宝贵资源。它们揭示了反复出现的问题类型、隐藏的陷阱以及考官期望的精确推理深度。本文深度剖析横跨各个主要主题的真实考题风格问题，提供逐步求解、常见错误和提升表现的策略。

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1. Normal Distribution: Probability & Inverse | 正态分布：概率与反查表

Question: The mass of grapefruits from an orchard is normally distributed with mean 310 g and standard deviation 8 g. A grapefruit is classified as 'large' if its mass exceeds 320 g. Find the probability that a randomly chosen grapefruit is large. Also, determine the mass exceeded by only 2% of grapefruits.

问题：某果园的葡萄柚质量服从正态分布，均值为 310 g，标准差为 8 g。若质量超过 320 g 则被归类为'大果'。求随机抽取一个葡萄柚为大果的概率。同时，求只有 2% 的葡萄柚能超过的质量值。

Solution & Analysis: = 0.1056. For the inverse calculation, let the required mass be m. Then P(X > m) = 0.02 ⇒ P(Z > (m − 310)/8) = 0.02. From tables, the z‑ upper tail of 0.02 is z = 2.0537 (interpolation may be needed). Hence (m − 310)/8 = 2.0537 ⇒ m = 310 + 8 × 2.0537 ≈ 326.43 g. A common error is reading the wrong tail area—always draw a diagram and check whether you need Φ(z) or 1 − Φ(z).

解析与分析：首先标准化边界：z = (320 − 310) / 8 = 1.25。查正态分布表，P(Z < 1.25) = 0.8944，因此 P(X > 320) = 1 − 0.8944 = 0.1056。对于反查，设所需质量为 m，则 P(X > m) = 0.02 ⇒ P(Z > (m − 310)/8) = 0.02。查表得上尾概率 0.02 对应的 z 值为 2.0537（可能需要插值）。于是 (m − 310)/8 = 2.0537 ⇒ m = 310 + 8 × 2.0537 ≈ 326.43 g。常见错误是看错尾部面积——始终画出图示，检查需要 Φ(z) 还是 1 − Φ(z)。

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2. Binomial Hypothesis Test | 二项分布假设检验

Question: A company claims that 40% of its seeds germinate. A gardener suspects the germination rate is lower. She plants 20 seeds and observes 5 germinate. Test, at the 5% significance level, whether the true proportion p is less than 0.4. State the hypotheses and the critical region.

问题：一家公司声称其种子发芽率为 40%。一位园丁怀疑发芽率更低。她种植了 20 粒种子，观察到 5 粒发芽。检验在 5% 显著性水平下，真实比例 p 是否小于 0.4。陈述假设和拒绝域。

Solutions: Let X ~ r calculator: P(X ≤ 4) = 0.0510 (not ≤ 0.05), P(X ≤ 3) = 0.0160. Hence the critical region is X ≤ 3. Since the observed value 5 is not in the critical region, we do not reject H₀. The evidence is insufficient to conclude that the germination rate is below 40%. Many students mistakenly use P(X ≤ 5) as the p‑ , still leading to the same decision but missing the formal definition of critical region.

解析：设 X ~ B(20, 0.4)。H₀: p = 0.4, H₁: p < 0.4。检验统计量为发芽种子数。我们需要找到最大的 c，使得 P(X ≤ c) ≤ 0.05。查表或使用计算器：P(X ≤ 4) = 0.0510（不 ≤ 0.05），P(X ≤ 3) = 0.0160。因此拒绝域为 X ≤ 3。由于观测值 5 不在拒绝域内，我们不拒绝 H₀。证据不足以得出种子发芽率低于 40% 的结论。许多学生误将 P(X ≤ 5) 当作 p 值（0.1256），这仍然大于 0.05，得出相同决策，但忽略了拒绝域的形式定义。

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3. Conditional Probability & Tree Diagrams | 条件概率与树状图

Question: In a factory, machines A and B produce 60% and 40% of items respectively. Machine A has a defect rate of 2%, machine B has 3%. An item is randomly selected and found defective. What is the probability it came from machine B?

问题：在一家工厂中，机器 A 和 B 分别生产 60% 和 40% 的产品。机器 A 的次品率为 2%，机器 B 为 3%。随机抽取一件产品，发现是次品。求它来自机器 B 的概率。

Solution: robability of defective: P(D) = 0.6×0.02 + 0.4×0.03 = 0.012 + 0.012 = 0.024. Then by Bayes' theorem, P(B|D) = P(B∩D) / P(D) = 0.4×0.03 / 0.024 = 0.012 / 0.024 = 0.5. So there is a 50% fective item is from machine B. A typical error is swapping the conditional in the formula—always identify which is the event already known (defective).

解析：画树状图：P(A) = 0.6, P(B) = 0.4。次品条件概率：A 下为 0.02，B 下为 0.03。次品总概率：P(D) = 0.6×0.02 + 0.4×0.03 = 0.012 + 0.012 = 0.024。由贝叶斯定理，P(B|D) = P(B∩D) / P(D) = 0.4×0.03 / 0.024 = 0.012 / 0.024 = 0.5。因此次品来自机器 B 的概率为 50%。典型错误是交换条件公式中的分子分母——始终确定哪个是已知事件（次品）。

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4. Poisson Approximation to Binomial | 泊松分布近似二项

Question: A rare disease affects 0.1% of a population. In a random sample of 4000 people, find the exact binomial probability that exactly 3 have the disease, and compare with the Poisson approximation.

问题：一种罕见病影响人口的 0.1%。在 4000 人的随机样本中，求恰好有 3 人患病的精确二项概率，并与泊松近似进行比较。

Analysis: Exact: X ~ ard but possible. Poisson approximation: λ = np = 4000×0.001 = 4. Then P(Y=3) = (e⁻⁴×4³)/3! n exams, acceptance criteria: n > 50 and np < 5 are typically given. Remember to apply a continuity correction when needed, though not for point probabilities.

解析：精确计算：X ~ B(4000, 0.001)。P(X=3) = C(4000,3)×0.001³×0.999³⁹⁹⁷，计算不便但可行。泊松近似：λ = np = 4000×0.001 = 4。则 P(Y=3) = (e⁻⁴×4³)/3! = (0.018316×64)/6 ≈ 0.1954。精确二项概率约为 0.1952，匹配极好，因为 n 很大而 p 很小。在考试中常用条件：n > 50 且 np < 5。注意需要时做连续性校正，尽管点概率不需要。

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5. Continuous Random Variables: PDF & CDF | 连续型随机变量：概率密度函数与分布函数

Question: ind the constant k, the cumulative distribution function F(x), and P(0.5 < X < 1.5).

问题：连续型随机变量 X 具有概率密度函数 f(x) = kx² (0 ≤ x ≤ 2)，其他处为 0。求 k，累积分布函数 F(x) 以及 P(0.5 < X < 1.5)。

Solution: The total area under the pdf must be 1: ∫₀² kx² dx = 1. k × [x³/3]₀² = k×(8/3) = 1 ⇒ k = 3/8. For F(x): for 0 ≤ x ≤ 2, F(x) = ∫₀ˣ (3/8)t² dt = (3/8)×[t³/3]₀ˣ = 3.25/8 = 0.40625. Common oversight: forgetting to check the support when defining F(x).

解析：概率密度函数下方面积必须为 1：∫₀² kx² dx = 1。k × [x³/3]₀² = k×(8/3) = 1 ⇒ k = 3/8。当 0 ≤ x ≤ 2，F(x) = ∫₀ˣ (3/8)t² dt = (3/8)×[t³/3]₀ˣ = x³/8。因此 F(x) 分段为：x < 0 时为 0，0 ≤ x ≤ 2 时为 x³/8，x > 2 时为 1。则 lt; X < 1.5) = F(1.5) − F(0.5) = (1.5³ − 0.5³)/8 = (3.375 − 0.125)/8 = 3.25/8 = 0.40625。常见疏忽：定义 F(x) 时忘记检查定义域。

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6. Chi-Squared Test for Independence | 卡方独立性检验

Question: A survey of 200 students records study method (visual, auditory, kinesthetic) and exam result (pass, fail). The observed frequencies are given. Perform a χ² test at the 5% level to determine whether study method and result are independent.

问题：一项针对 200 名学生的调查记录了学习方式（视觉型、听觉型、动觉型）和考试成绩（及格、不及格）。给定观测频数。在 5% 水平下进行 χ² 检验，判断学习方式与成绩是否独立。

	Visual	Auditory	Kinesthetic	Total
Pass	45	50	25	120
Fail	15	10	55	80
Total	60	60	80	200

Analysis: = 48. Visual-Fail: 24, Auditory-Fail: 24, Kinesthetic-Fail: 32. Compute χ² = Σ (O−E)²/E. Degrees of freedom = (2−1)(3−1) = 2. Critical value χ²(0.05,2) = 5.991. The computed statistic is very large, so we reject H₀: study method and result are associated. Be careful: categories must be exclusive and expected frequencies > 5.

解析：期望频数：(行合计×列合计)/总计。视觉-及格：60×120/200 = 36，听觉-及格：36，动觉-及格：80×120/200 = 48。视觉-不及格：24，听觉-不及格：24，动觉-不及格：32。计算 χ² = Σ (O−E)²/E。自由度 df = (2−1)(3−1) = 2。临界值 χ²(0.05,2) = 5.991。检验统计量远大于临界值，故拒绝 H₀：学习方式与成绩存在关联。须注意类别互斥且期望频数 > 5。

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7. Correlation and Regression | 相关与回归

Question: The following data shows hours of revision (x) and test score (y): (5, 60), (8, 72), (3, 45), (10, 88), (6, 65). Calculate the product moment correlation coefficient r, and find the equation of the regression line of y on x. Interpret the slope.

问题：以下数据展示复习小时数 (x) 与测试分数 (y)：(5, 60), (8, 72), (3, 45), (10, 88), (6, 65)。计算积矩相关系数 r，并求 y 对 x 的回归线方程。解释斜率。

Solution: Summarise: Σx = 32, Σy = 330, Σx² = 234, Σy² = 22538, Σxy = 2271, n = 5. S_xx = 234 − 32²/5 = 234 − 204.8 = 29.2; S_yy = 22538 − 330²/5 = 22538 − 21780 = 758; S_xy = 2271 − 32×330/5 = 2271 − 2112 = 159. Then r = S_xy / √S_xxS_yy = 159 / √29.2×758 ≈ 159 / 148.75 ≈ 1.069? Wait, that' mpossible: r must be ≤1. Recalculate: Σy² = 3600+5184+2025+7744+4225 = 22778, not 22538. I'll correct: Σy² should be 22778. Then S_yy = 22778 − 21780 = 998. r = 159 / √29.2×998 = 159 / √29141.6 ≈ 159 / 170.71 ≈ 0.931. So strong positive correlation. Regression line y = a + bx: b = S_xy/S_xx = 159 / 29.2 ≈ 5.445; a = ȳ − b x̄ ch extra hour of revision adds about 5.45 marks. Do not extrapolate beyond the data range.

解析：求和得 Σx = 32, Σy = 330, Σx² = 234, Σy² = 22778, Σxy = 2271, n = 5。S_xx = 234 − 32²/5 = 29.2；S_yy = 22778 − 330²/5 = 998；S_xy = 2271 − 32×330/5 = 159。r = 159 / √29.2×998 ≈ 0.931，强正相关。回归线 y = a + bx：b = 159 / 29.2 ≈ 5.445；a = 66 − 5.445×6.4 ≈ 31.15，得 y = 31.15 + 5.445x。斜率：每多一小时复习约增加 5.45 分。不要超出数据范围外推。

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8. Sampling Distribution & Central Limit Theorem | 抽样分布与中心极限定理

Question: The daily sales in a store have mean £250 and standard deviation £40. A random sample of 50 days is taken. Find the probability that the sample mean exceeds £258. Explain why you do not need to know the shape of the population.

问题：某商店的日销售额均值为 £250，标准差 £40。随机选取 50 天为样本。求样本均值超过 £258 的概率。解释为何无需知道总体分布形状。

Solution: By the Central Limit Theorem, for n = 50 (large), the sample mean X̄ is approximately normal with μ_x̄ = 250 and standard error = 40/√50 − 250) / 5.657 ≈ 1.414. P(Z > 1.414) = 1 − Φ(1.41) ≈ 1 − 0.9207 = 0.0793. So about 7.93% ntly large. Common mistake: using population standard deviation instead of standard error.

解析：由中心极限定理，n = 50 很大，样本均值 X̄ 近似服从正态分布，均值为 250，标准误为 40/√50 ≈ 5.657。z = (258 − 250) / 5.657 ≈ 1.414。P(Z > 1.414) = 1 − 0.9207 ≈ 0.0793。故约 7.93% 的概率。中心极限定理确保无论总体分布如何，样本均值都近似正态，因为样本量足够大。常见错误：使用总体标准差而非标准误。

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9. Confidence Intervals for the Mean | 均值的置信区间

Question: A sample of 25 bolts has mean diameter 10.2 mm with standard deviation 0.4 mm. Construct a 95% confidence interval for the population mean diameter, assuming normality.

问题：一个 25 根螺栓的样本，平均直径为 10.2 mm，标准差为 0.4 mm。假设正态分布，构建总体均值直径的 95% 置信区间。

Analysis: Since σ is unknown, use the t‑distribution with ν = 24 degrees of freedom. The critical value t_0.025,24 = 2.064. Standard error = s/√n tion σ, we would use z = 1.96. Common pitfall: wrongly using z when σ is estimated by s. Check whether 'standard deviation' refers to sample or population.

解析：因 σ 未知，使用自由度为 ν = 24 的 t 分布。临界值 t_0.025,24 = 2.064。标准误 = 0.4/5 = 0.08。误差界限 = 2.064 × 0.08 = 0.16512。置信区间：(10.2 − 0.165, 10.2 + 0.165) = (10.035, 10.365) mm。若已知总体标准差，应使用 z = 1.96。常见误区：当 σ 用 s 估计时错误使用 z。务必检查'标准差'指样本还是总体。

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10. Type I & Type II Errors and Power | 第一类与第二类错误及功效

Question: esis test for the population mean, H₀: μ = 100 vs H₁: μ > 100, the critical region is X̄ le size 36, and σ = 10. Find the probability of a Type II error and the power of the test.

问题：在总体均值的假设检验中，H₀: μ = 100 vs H₁: μ > 100，拒绝域为 X̄ > 104。真实均值为 106，样本量 n = 36，σ = 10。求发生第二类错误的概率以及检验的功效。

Solution: Type II error occurs when we fail to reject H₀ while H₁ is true. Here β = P(X̄ ≤ 104 | μ = 106). Under the true μ = 106, the sampling distribution of X̄ = −1.2. Then β = P(Z ≤ −1.2) = 0.1151. Power = 1 − β = 0.8849. So the test has about 88.5% power. Students often confuse the conditions for α and β—alpha is set under H₀, beta under H₁.

解析：第二类错误发生在 H₁ 为真时却未拒绝 H₀。此处 β = P(X̄ ≤ 104 | μ = 106)。在真实 μ = 106 下，X̄ 的抽样分布为 N(106, 1.667²)。标准化：z = (104 − 106) / 1.667 = −1.2。β = P(Z ≤ −1.2) = 0.1151。功效 = 1 − β = 0.8849，检验约有 88.5% 的功效。学生常混淆 α 和 β 的条件——α 是在 H₀ 下设定，β 是在 H₁ 下计算。

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Published by TutorHao | Statistics Revision Series | aleveler.com

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