AQA Year 13 Statistics Mock Paper Walkthrough | AQA 高三统计模拟卷解析

📚 AQA Year 13 Statistics Mock Paper Walkthrough | AQA 高三统计模拟卷解析

This article presents a complete walkthrough of a mock paper designed for the Year 13 AQA Statistics specification. The paper tests key topics including the Poisson distribution, hypothesis testing for proportions, the Normal approximation to the binomial, linear regression, product moment correlation, and continuous random variables. Each question is broken down with a full worked solution, examiner-style commentary, and common pitfalls to avoid. This resource is ideal for self-assessment, revision lessons, and building confidence ahead of the final exam.

本文为AQA高三统计课程设计了一套完整的模拟试卷并逐题解析。试卷涵盖泊松分布、比例假设检验、二项分布的正态近似、线性回归、积矩相关系数以及连续随机变量等核心主题。每道题都配有详细解答、阅卷人风格的点评以及常见错误的避坑指南。本文特别适合自我评估、复习课教学及考前建立信心。

1. Question 1 – Poisson Distribution: Probabilities and Cumulative Functions | 第1题 – 泊松分布：概率与累积函数

The number of cars arriving at a remote petrol station in a ten-minute interval is modelled by a Poisson distribution with mean 2.4.
Find (a) the probability that exactly 3 cars arrive in a given ten-minute interval, (b) the probability that at most 1 car arrives in a given ten-minute interval, and (c) the smallest value of k such that P(X ≤ k) > 0.95.

某偏远加油站在每十分钟内到达的车辆数服从均值为2.4的泊松分布。计算 (a) 在某个十分钟内恰好有3辆车到达的概率，(b) 在某个十分钟内最多有1辆车到达的概率，(c) 满足 P(X ≤ k) > 0.95 的最小 k 值。

Solution (a): Let X ~ Po(2.4). Using the Poisson probability formula, P(X = 3) = e⁻²·⁴ × 2.4³ / 3! = e⁻²·⁴ × 13.824 / 6 ≈ 0.0907 × 2.304 = 0.2090 (4 s.f.).

解析 (a)： 设 X ~ Po(2.4)。代入泊松概率公式，P(X = 3) = e⁻²·⁴ × 2.4³ / 3! = e⁻²·⁴ × 13.824 / 6 ≈ 0.0907 × 2.304 = 0.2090（保留四位有效数字）。

Solution (b): P(X ≤ 1) = P(X = 0) + P(X = 1) = e⁻²·⁴ + e⁻²·⁴ × 2.4 = e⁻²·⁴ (1 + 2.4) = 0.0907 × 3.4 ≈ 0.3084.

解析 (b)： P(X ≤ 1) = P(X = 0) + P(X = 1) = e⁻²·⁴ + e⁻²·⁴ × 2.4 = e⁻²·⁴ (1 + 2.4) = 0.0907 × 3.4 ≈ 0.3084。

Solution (c): From Poisson cumulative tables with λ = 2.4, P(X ≤ 4) = 0.9041, P(X ≤ 5) = 0.9643. The smallest k for which P(X ≤ k) > 0.95 is 5.

解析 (c)： 查 λ = 2.4 的泊松累积分布表，P(X ≤ 4) = 0.9041，P(X ≤ 5) = 0.9643。满足 P(X ≤ k) > 0.95 的最小 k 值为5。

Examiner tip: Always state the distribution clearly and check whether the question asks for the smallest k or the value of k that gives a probability just exceeding the threshold.

考官提示：务必先明确写出分布名称，同时仔细审题，看清题目要求的是最小 k 还是概率刚刚超过阈值所对应的 k。

2. Question 2 – Hypothesis Test for a Binomial Proportion | 第2题 – 二项分布比例的假设检验

A seed supplier claims that the germination rate of a new batch of tomato seeds is at least 85%. A gardener plants 40 seeds and 29 germinate. Test, at the 5% significance level, whether there is sufficient evidence to contradict the supplier’s claim. State your hypotheses clearly and interpret the result in context.

某种子供应商声称一批新番茄种子的发芽率不低于85%。一位园丁播种了40粒种子，其中29粒发芽。在5%的显著性水平下检验是否有充分证据反驳供应商的说法。请清晰陈述假设并联系实际解读结果。

Solution: Let p be the true germination proportion. H₀: p = 0.85, H₁: p < 0.85 (one-tailed lower-tail test). Under H₀, X ~ B(40, 0.85) where X is the number of seeds that germinate. Observed x = 29.

解析： 设 p 为真实发芽率。H₀: p = 0.85，H₁: p < 0.85（单尾左侧检验）。在 H₀ 下，发芽种子数 X ~ B(40, 0.85)。实际观测值 x = 29。

P(X ≤ 29) = 0.0198 (from cumulative binomial tables or calculator). Since 0.0198 < 0.05, the result is significant. We reject H₀.

P(X ≤ 29) = 0.0198（查二项分布累积表或使用计算器）。由于 0.0198 < 0.05，结果为显著。拒绝零假设 H₀。

There is sufficient evidence at the 5% level to contradict the supplier’s claim. The data suggest the true germination rate is below 85%.

在5%显著性水平下有充分证据反驳供应商的说法。数据显示真实发芽率低于85%。

Common mistake: Students often use a two-tailed test when the wording “at least 85%” points clearly to a lower-tail test. Also, ensure the p-value is compared to the significance level and the conclusion is written in context, not just “reject H₀”.

常见错误：题目中“不低于85%”的表述明确指向单尾左侧检验，但常有学生误用双尾检验。此外，务必把P值与显著性水平进行比较，并用实际背景写出结论，不要只写“拒绝H₀”。

3. Question 3 – Normal Approximation to the Binomial | 第3题 – 二项分布的正态近似

A multiple-choice quiz consists of 80 questions, each with 4 options. A student guesses every answer completely at random. Use a Normal approximation to estimate the probability that the student scores more than 28 correct answers. Apply a continuity correction.

一次选择题测验共80题，每题4个选项。某学生完全随机猜测所有答案。使用正态近似估计该生答对28题以上的概率，并应用连续性校正。

Solution: Let Y ~ B(80, 0.25). μ = np = 80 × 0.25 = 20, σ² = np(1-p) = 80 × 0.25 × 0.75 = 15, so σ = √15 ≈ 3.873. To approximate P(Y > 28), use continuity correction: P(Y > 28.5).

解析： 设 Y ~ B(80, 0.25)。μ = np = 80 × 0.25 = 20，σ² = np(1-p) = 80 × 0.25 × 0.75 = 15，故 σ = √15 ≈ 3.873。为了近似 P(Y > 28)，采用连续性校正：P(Y > 28.5)。

Z = (28.5 – 20) / 3.873 ≈ 8.5 / 3.873 ≈ 2.195. Then P(Z > 2.195) = 1 – Φ(2.195) ≈ 1 – 0.9859 = 0.0141.

Z = (28.5 – 20) / 3.873 ≈ 8.5 / 3.873 ≈ 2.195。则 P(Z > 2.195) = 1 – Φ(2.195) ≈ 1 – 0.9859 = 0.0141。

Check conditions: np = 20 > 5 and n(1-p) = 60 > 5, so the Normal approximation is appropriate. The continuity correction is essential because a discrete distribution is being approximated by a continuous one.

检查条件：np = 20 > 5 且 n(1-p) = 60 > 5，因此正态近似适用。由于是将离散分布近似为连续分布，连续性校正是必不可少的。

4. Question 4 – Product Moment Correlation Coefficient and Hypothesis Testing | 第4题 – 积矩相关系数与假设检验

A biologist records the wing length (x mm) and body mass (y g) of 12 adult dragonflies. The summary statistics are: Σx = 672, Σy = 9.72, Σx² = 38820, Σy² = 8.34, Σxy = 556.4. (a) Calculate the product moment correlation coefficient r. (b) Test, at the 1% significance level, whether there is evidence of positive correlation in the population. Use the appropriate critical value from tables.

一位生物学家记录了12只成年蜻蜓的翅长（x mm）与体重（y g）。汇总统计量为：Σx = 672，Σy = 9.72，Σx² = 38820，Σy² = 8.34，Σxy = 556.4。(a) 计算积矩相关系数 r。(b) 在1%的显著性水平下，检验是否存在总体正相关的证据。请查表得出合适的临界值。

Solution (a): n = 12. Sxx = Σx² – (Σx)²/n = 38820 – 672²/12 = 38820 – 451584/12 = 38820 – 37632 = 1188. Syy = Σy² – (Σy)²/n = 8.34 – 9.72²/12 = 8.34 – 94.4784/12 = 8.34 – 7.8732 = 0.4668. Sxy = Σxy – (Σx)(Σy)/n = 556.4 – 672×9.72/12 = 556.4 – 6531.84/12 = 556.4 – 544.32 = 12.08.

解析 (a)： n = 12。Sxx = Σx² – (Σx)²/n = 38820 – 672²/12 = 38820 – 451584/12 = 38820 – 37632 = 1188。Syy = Σy² – (Σy)²/n = 8.34 – 9.72²/12 = 8.34 – 94.4784/12 = 8.34 – 7.8732 = 0.4668。Sxy = Σxy – (Σx)(Σy)/n = 556.4 – 672×9.72/12 = 556.4 – 6531.84/12 = 556.4 – 544.32 = 12.08。

r = Sxy / √(Sxx × Syy) = 12.08 / √(1188 × 0.4668) = 12.08 / √554.5584 = 12.08 / 23.54 ≈ 0.513 (3 s.f.).

r = Sxy / √(Sxx × Syy) = 12.08 / √(1188 × 0.4668) = 12.08 / √554.5584 = 12.08 / 23.54 ≈ 0.513（保留三位有效数字）。

Solution (b): H₀: ρ = 0, H₁: ρ > 0 (one-tailed test). n = 12, degrees of freedom = 10. At the 1% level for a one-tailed test, the critical value from the PMCC table is 0.6580. Since 0.513 < 0.6580, we do not reject H₀. There is insufficient evidence at the 1% level to conclude a positive correlation exists in the population.

解析 (b)： H₀: ρ = 0，H₁: ρ > 0（单尾检验）。n = 12，自由度 = 10。在1%显著性水平下单尾检验的临界值查PMCC表为0.6580。由于0.513 < 0.6580，不能拒绝H₀。在1%水平下，没有充分证据表明总体中存在正相关。

5. Question 5 – Linear Regression and Prediction | 第5题 – 线性回归与预测

Using the data from Question 4, the equation of the regression line of y on x is given by y = a + bx where b = Sxy / Sxx and a = (Σy/n) – b(Σx/n). (a) Calculate the values of a and b. Give the equation in the form y = a + bx. (b) Interpret the gradient in context. (c) Predict the body mass of a dragonfly with a wing length of 58 mm. Comment on the reliability of this prediction.

使用第4题的数据，y 对 x 的回归直线方程为 y = a + bx，其中 b = Sxy / Sxx，a = (Σy/n) – b(Σx/n)。(a) 计算 a 和 b 的值，并写出形如 y = a + bx 的方程。(b) 结合背景解释斜率的含义。(c) 预测翅长为58 mm的蜻蜓的体重，并评述该预测的可靠性。

Solution (a): b = Sxy / Sxx = 12.08 / 1188 ≈ 0.01017. Mean of x = 672/12 = 56, mean of y = 9.72/12 = 0.81. a = 0.81 – 0.01017 × 56 = 0.81 – 0.56952 = 0.2405. Equation: y = 0.241 + 0.0102x (3 s.f.).

解析 (a)： b = Sxy / Sxx = 12.08 / 1188 ≈ 0.01017。x 的均值 = 672/12 = 56，y 的均值 = 9.72/12 = 0.81。a = 0.81 – 0.01017 × 56 = 0.81 – 0.56952 = 0.2405。方程：y = 0.241 + 0.0102x（保留三位有效数字）。

Solution (b): The gradient 0.0102 indicates that for every additional 1 mm in wing length, the predicted body mass increases by approximately 0.0102 g, on average.

解析 (b)： 斜率0.0102表示翅长每增加1 mm，预计体重平均增加约0.0102 g。

Solution (c): When x = 58, predicted y = 0.241 + 0.0102 × 58 = 0.241 + 0.5916 = 0.8326 g. However, 58 mm is within the range of the original data (interpolation), so the prediction is reasonably reliable in terms of data range. Yet, the correlation coefficient was only 0.513, which suggests a moderate linear relationship; individual predictions may have substantial variability.

解析 (c)： 当 x = 58，预计 y = 0.241 + 0.0102 × 58 = 0.241 + 0.5916 = 0.8326 g。58 mm在原数据范围之内（内插），因此就数据范围而言该预测相对可靠。但相关系数仅为0.513，表明线性关系中等，个别预测值可能存在较大波动。

6. Question 6 – Continuous Random Variables and PDFs | 第6题 – 连续随机变量与概率密度函数

A continuous random variable X has probability density function f(x) = kx(5 – x) for 0 ≤ x ≤ 5, and f(x) = 0 otherwise. (a) Show that k = 6/125. (b) Find E(X). (c) Calculate P(1 < X < 3).

连续随机变量 X 的概率密度函数为 f(x) = kx(5 – x)，定义域 0 ≤ x ≤ 5，其余为0。(a) 证明 k = 6/125。(b) 求 E(X)。(c) 计算 P(1 < X < 3)。

Solution (a): ∫₀⁵ kx(5 – x) dx = k ∫₀⁵ (5x – x²) dx = k [ (5x²/2) – (x³/3) ]₀⁵ = k [ (125/2) – (125/3) ] = k [ 375/6 – 250/6 ] = k × 125/6. Set = 1, so k = 6/125.

解析 (a)： ∫₀⁵ kx(5 – x) dx = k ∫₀⁵ (5x – x²) dx = k [ (5x²/2) – (x³/3) ]₀⁵ = k [ (125/2) – (125/3) ] = k [ 375/6 – 250/6 ] = k × 125/6。令其等于1，得 k = 6/125。

Solution (b): E(X) = ∫₀⁵ x f(x) dx = (6/125) ∫₀⁵ (5x² – x³) dx = (6/125) [ (5x³/3) – (x⁴/4) ]₀⁵ = (6/125) [ (625/3) – (625/4) ] = (6/125) × 625 × (1/3 – 1/4) = (6/125) × 625 × (1/12) = (6 × 625) / (125 × 12) = 3750 / 1500 = 2.5.

解析 (b)： E(X) = ∫₀⁵ x f(x) dx = (6/125) ∫₀⁵ (5x² – x³) dx = (6/125) [ (5x³/3) – (x⁴/4) ]₀⁵ = (6/125) [ (625/3) – (625/4) ] = (6/125) × 625 × (1/3 – 1/4) = (6/125) × 625 × (1/12) = (6 × 625) / (125 × 12) = 3750 / 1500 = 2.5。

Solution (c): P(1 < X < 3) = (6/125) ∫₁³ (5x - x²) dx = (6/125) [ (5x²/2) - (x³/3) ]₁³ = (6/125) { [ (45/2) - (27/3) ] - [ (5/2) - (1/3) ] } = (6/125) { [22.5 - 9] - [2.5 - 0.3333] } = (6/125) {13.5 - 2.1667} = (6/125) × 11.3333 ≈ 0.5440.

解析 (c)： P(1 < X < 3) = (6/125) ∫₁³ (5x - x²) dx = (6/125) [ (5x²/2) - (x³/3) ]₁³ = (6/125) { [ (45/2) - (27/3) ] - [ (5/2) - (1/3) ] } = (6/125) { [22.5 - 9] - [2.5 - 0.3333] } = (6/125) {13.5 - 2.1667} = (6/125) × 11.3333 ≈ 0.5440。

7. Question 7 – Poisson Hypothesis Test (Critical Region Method) | 第7题 – 泊松假设检验（临界域法）

Calls to a helpline are believed to occur at a rate of 3.2 calls per hour during off-peak times. The manager suspects the rate has decreased. In a randomly selected one-hour off-peak period, 1 call is received. Using a 10% significance level, find the critical region and state the conclusion.

某求助热线在非高峰时段来电率据信为每小时3.2通。经理怀疑该比率已经下降。在某随机选取的一小时非高峰时段内，只接到1通来电。使用10%的显著性水平，找出临界域并给出结论。

Solution: H₀: λ = 3.2, H₁: λ < 3.2 (lower-tail test). Test statistic: X ~ Po(3.2). We need the largest k such that P(X ≤ k) ≤ 0.10. From tables, P(X ≤ 1) = P(X=0) + P(X=1) = 0.0408 + 0.1304 = 0.1712 (> 0.10). P(X ≤ 0) = 0.0408 (≤ 0.10). So the critical region at the 10% level is X = 0. The observed value x = 1 is not in the critical region. Do not reject H₀. There is insufficient evidence at the 10% significance level to conclude the rate has decreased.

解析： H₀: λ = 3.2，H₁: λ < 3.2（左侧检验）。检验统计量：X ~ Po(3.2)。需要找到满足 P(X ≤ k) ≤ 0.10 的最大 k 值。查表得，P(X ≤ 1) = P(X=0) + P(X=1) = 0.0408 + 0.1304 = 0.1712 (> 0.10)。P(X ≤ 0) = 0.0408 (≤ 0.10)。因此10%水平下的临界域为 X = 0。观测值 x = 1 不在临界域内。不能拒绝 H₀。在10%显著性水平下，没有充分证据表明来电率已下降。

8. Question 8 – Chi-Squared Test for Association | 第8题 – 卡方独立性检验

A researcher surveys 200 adults on their preferred news source (TV, Online, Print) and age group (Under 30, 30–50, Over 50). The observed frequencies are given in the table below. Test, at the 5% significance level, whether there is an association between age group and news source preference. State the degrees of freedom and the critical value.

一位研究者调查了200名成年人偏好的新闻来源（电视、网络、纸质）及其年龄段（30岁以下、30-50岁、50岁以上）。观测频数如下表所示。在5%的显著性水平下检验年龄段与新闻来源偏好之间是否存在关联。说出自由度与临界值。

Observed / 观测值	TV / 电视	Online / 网络	Print / 纸质	Total
Under 30	20	45	5	70
30–50	35	30	10	75
Over 50	30	10	15	55
Total	85	85	30	200

Solution: H₀: No association between age group and news source. H₁: There is an association. Degrees of freedom = (3-1)(3-1) = 4. Expected frequencies are calculated row total × column total / grand total, e.g., E(TV, Under 30) = 70×85/200 = 29.75.

解析： H₀：年龄段与新闻来源之间无关联。H₁：存在关联。自由度 = (3-1)(3-1) = 4。期望频数按行合计 × 列合计 / 总计计算，例如 E(电视, 30岁以下) = 70×85/200 = 29.75。

Χ² = Σ (O – E)² / E ≈ 5.18 + 7.82 + 4.51 + 1.09 + 4.24 + 1.52 + 2.67 + 9.85 + 4.37 ≈ 39.2. From chi-squared tables, critical value at 5% with 4 df is 9.488. Since 39.2 >> 9.488, reject H₀. There is strong evidence of an association between age group and news source preference.

Χ² = Σ (O – E)² / E ≈ 5.18 + 7.82 + 4.51 + 1.09 + 4.24 + 1.52 + 2.67 + 9.85 + 4.37 ≈ 39.2。查卡方表，5%水平下自由度为4的临界值为9.488。由于39.2远大于9.488，拒绝H₀。有强烈证据表明年龄段与新闻来源偏好之间存在关联。

9. Common Mistakes and Key Takeaways | 第9题 – 常见错误与关键要点

Across all questions, the most frequent errors involve forgetting continuity corrections, misidentifying one- or two-tailed tests, and confusing Sxx with Sxy in correlation and regression formulas. For the Poisson distribution, students sometimes fail to use cumulative tables correctly for finding critical values. In chi-squared tests, always check that expected frequencies are all at least 5, and never use percentages in place of actual frequencies. Showing clear hypotheses, stating significance levels, and writing conclusions in the context of the problem are essential for gaining full marks.

在所有题目中，最常见的错误包括遗漏连续性校正、误判单尾或双尾检验、在相关与回归公式中把 Sxx 和 Sxy 混淆。对于泊松分布，学生有时不能正确使用累积分布表来求临界值。在卡方检验中，务必检查所有期望频数是否都至少为5，且切勿用百分比代替实际频数。清晰写出假设、标明显著性水平、并在实际背景下给出结论，是获得满分的必要条件。

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AQA Year 13 Statistics Mock Paper Walkthrough | AQA 高三统计模拟卷解析