AQA A-level Further Maths Paper 4: FS2 Example Responses | AQA A-level进阶数学试卷4：FS2示例答题

📚 AQA A-level Further Maths Paper 4: FS2 Example Responses | AQA A-level进阶数学试卷4：FS2示例答题

This article presents worked example responses for AQA A-level Further Maths Paper 4 (Further Statistics 2). Each example demonstrates how to structure a clear, mark-worthy solution using correct notation and statistical reasoning.

本文展示AQA A-level进阶数学试卷4（进阶统计2）的示例答题。每个示例都展示如何用清晰的步骤、正确的记号和统计推理组织值得给满分的解答。

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1. Poisson Distribution: Probability Calculations | 泊松分布：概率计算

th mean 4.5. State the probability generating formula and calculate P(X = 3) and P(X ≤ 2).

示例：设X服从均值为4.5的泊松分布。写出概率公式并计算P(X = 3)和P(X ≤ 2)。

For X ~ Po(λ), the probability mass function is P(X = r) = e⁻ˡ · λᵣ / r! for r = 0, 1, 2, ...

对X ~ Po(λ)，概率质量函数为P(X = r) = e⁻ˡ · λᵣ / r!，其中r = 0, 1, 2, ...。

Using λ = 4.5, P(X = 3) = e⁻⁴·⁵ · 4.5³ / 3! = 0.1687 (3 d.p.).

取λ = 4.5，P(X = 3) = e⁻⁴·⁵ · 4.5³ / 3! = 0.1687（保留三位小数）。

P(X ≤ 2) = e⁻⁴·⁵ (1 + 4.5 + 4.5²/2) = e⁻⁴·⁵ × 15.625 = 0.1736 (3 d.p.).

P(X ≤ 2) = e⁻⁴·⁵ (1 + 4.5 + 4.5²/2) = e⁻⁴·⁵ × 15.625 = 0.1736（保留三位小数）。

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2. Poisson Hypothesis Testing | 泊松假设检验

ample: A random variable X is modelled by Po(λ). Test H₀: λ = 2 against H₁: λ > 2 at the 5% significance level. Find the critical region.

示例：设随机变量X服从Po(λ)。在5%显著性水平下检验H₀: λ = 2，备择假设H₁: λ > 2。求临界区域。

Under H₀, X ~ Po(2). We need the smallest value c such that P(X ≥ c) ≤ 0.05.

在H₀下，X ~ Po(2)。我们需要最小的c，使P(X ≥ c) ≤ 0.05。

From tables: P(X ≥ 5) = 0.0527 > 0.05, while P(X ≥ 6) = 0.0166 < 0.05.

查表得：P(X ≥ 5) = 0.0527 > 0.05，而P(X ≥ 6) = 0.0166 < 0.05。

e critical region is x ≥ 6. If the observed value is 6 or more, reject H₀ in favour of H₁.

因此临界区域为x ≥ 6。若观测值不小于6，则拒绝H₀，接受H₁。

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3. Exponential Distribution: pdf and cdf | 指数分布：概率密度函数与累积分布函数

al distribution with rate λ = 0.2. State the pdf, cdf, and find P(X < 3), E(X) and Var(X).

示例：设X服从速率λ = 0.2的指数分布。写出pdf、cdf，并求P(X < 3)、E(X)和Var(X)。

For X ~ Exp(λ), f(x) = λe⁻ˡˣ for x > 0, and F(x) = 1 − e⁻ˡˣ.

对X ~ Exp(λ)，f(x) = λe⁻ˡˣ（x > 0），且F(x) = 1 − e⁻ˡˣ。

P(X < 3) = 1 − e⁻⁰·⁶ = 0.4512 (3 d.p.).

P(X < 3) = 1 − e⁻⁰·⁶ = 0.4512（保留三位小数）。

E(X) = 1/λ = 5 and Var(X) = 1/λ² = 25.

E(X) = 1/λ = 5，Var(X) = 1/λ² = 25。

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4. Memoryless Property of Exponential | 指数分布的无记忆性

Example: Using X ~ Exp(0.2), find P(X > 5 | X > 2).

示例：利用X ~ Exp(0.2)，求P(X > 5 | X > 2)。

The memoryless property states that P(X > s + t | X > s) = P(X > t) = e⁻ˡᵗ.

无记忆性表明P(X > s + t | X > s) = P(X > t) = e⁻ˡᵗ。

Here s = 2 and t = 3, so P(X > 5 | X > 2) = P(X > 3) = e⁻⁰·⁶ = 0.5488 (3 d.p.).

这里s = 2，t = 3，所以P(X > 5 | X > 2) = P(X > 3) = e⁻⁰·⁶ = 0.5488（保留三位小数）。

This result shows that the residual waiting time does not depend on how long we have already waited.

该结果表明剩余等待时间与已经等待的时间长短无关。

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5. Continuous Uniform Distribution | 连续均匀分布

Example: Let X ~ U(2, 8). Find E(X), Var(X), and P(3 ≤ X ≤ 5).

示例：设X ~ U(2, 8)。求E(X)、Var(X)和P(3 ≤ X ≤ 5)。

For X ~ U(a, b), E(X) = (a + b) / 2 and Var(X) = (b − a)² / 12.

对X ~ U(a, b)，E(X) = (a + b) / 2，Var(X) = (b − a)² / 12。

Here a = 2, b = 8, so E(X) = 5 and Var(X) = 6² / 12 = 3.

这里a = 2，b = 8，所以E(X) = 5，Var(X) = 6² / 12 = 3。

P(3 ≤ X ≤ 5) = (5 − 3) / (8 − 2) = 2/6 = 1/3 ≈ 0.3333.

P(3 ≤ X ≤ 5) = (5 − 3) / (8 − 2) = 2/6 = 1/3 ≈ 0.3333。

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6. Chi-squared Goodness of Fit | 卡方拟合优度检验

Example: A die is rolled 60 times. The observed frequencies are (10, 14, 11, 12, 15, 10). Test at the 5% level whether the die is fair.

示例：一枚骰子掷60次，观察频数为(10, 14, 11, 12, 15, 10)。在5%显著性水平下检验骰子是否公平。

If fair, each face has expected frequency 60 × 1/6 = 10.

若公平，每面期望频数为60 × 1/6 = 10。

tic is χ² = Σ (O − E)² / E = [(0)² + (4)² + (1)² + (2)² + (5)² + (0)²] / 10 = 46/10 = 4.6.

检验统计量为χ² = Σ (O − E)² / E = [(0)² + (4)² + (1)² + (2)² + (5)² + (0)²] / 10 = 46/10 = 4.6。

The degrees of freedom are k − 1 = 5. The critical value at 5% is χ²₅(0.05) = 11.07.

自由度为k − 1 = 5。5%水平下临界值χ²₅(0.05) = 11.07。

Since 4.6 < 11.07, we do not reject H₀. There is insufficient evidence that the die is biased.

由于4.6 < 11.07，不拒绝H₀。没有充分证据表明骰子有偏。

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7. Chi-squared Contingency Table | 卡方列联表独立性检验

Example: A 2 × 2 table shows exam results by gender. Test at the 5% level whether gender and pass are independent.

示例：一个2 × 2列联表展示按性别划分的考试成绩。在5%水平下检验性别与通过是否独立。

	Pass | 通过	Fail | 未通过	Total | 合计
Male | 男	30	20	50
Female | 女	40	10	50
Total | 合计	70	30	100

The expected frequency for each cell is (row total × column total) / grand total.

每个单元格的期望频数为（行合计 × 列合计）/ 总合计。

pass expected = 50 × 70 / 100 = 35. The complete expected matrix is (35, 15) and (35, 15).

例如，男-通过期望 = 50 × 70 / 100 = 35。完整期望矩阵为(35, 15)和(35, 15)。

χ² = Σ(O − E)² / E = (30−35)²/35 + (20−15)²/15 + (40−35)²/35 + (10−15)²/15 = 4.762.

χ² = Σ(O − E)² / E = (30−35)²/35 + (20−15)²/15 + (40−35)²/35 + (10−15)²/15 = 4.762。

Degrees of freedom = (2 − 1)(2 − 1) = 1. The critical value is χ²₁(0.05) = 3.841.

自由度 = (2 − 1)(2 − 1) = 1。临界值χ²₁(0.05) = 3.841。

Since 4.762 > 3.841, we reject H₀. There is evidence of an association between gender and pass.

由于4.762 > 3.841，拒绝H₀。有证据表明性别与通过存在关联。

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8. Normal Approximation to Poisson | 泊松分布的正态近似

Example: Let X ~ Po(20). Use a normal approximation with continuity correction to estimate P(X ≤ 15).

示例：设X ~ Po(20)。使用带连续性校正的正态近似估计P(X ≤ 15)。

Since λ = 20 > 10, X is approximately N(λ, λ) = N(20, 20).

由于λ = 20 > 10，X近似服从N(λ, λ) = N(20, 20)。

Apply the continuity correction: P(X ≤ 15) ≈ P(Y ≤ 15.5), where Y ~ N(20, 20).

应用连续性校正：P(X ≤ 15) ≈ P(Y ≤ 15.5)，其中Y ~ N(20, 20)。

Z = (15.5 − 20) / √20 = −1.006. Hence P(Z < −1.006) = 0.157 (3 d.p.).

Z = (15.5 − 20) / √20 = −1.006。因此P(Z < −1.006) = 0.157（保留三位小数）。

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9. Additive Property of Poisson | 泊松分布的可加性

Example: Let X ~ Po(3) and Y ~ Po(4) be independent. Find P(X + Y = 5).

示例：设X ~ Po(3)和Y ~ Po(4)独立。求P(X + Y = 5)。

If independent Poisson variables are added, the sum is also Poisson: X + Y ~ Po(3 + 4) = Po(7).

若独立泊松变量相加，和仍为泊松分布：X + Y ~ Po(3 + 4) = Po(7)。

P(X + Y = 5) = e⁻⁷ · 7⁵ / 5! = 0.1277 (3 d.p.).

P(X + Y = 5) = e⁻⁷ · 7⁵ / 5! = 0.1277（保留三位小数）。

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10. Exam Technique: How to Structure Answers | 考试技巧：如何组织答案

Always write down the null and alternative hypotheses before any calculation.

在任何计算之前，务必写出原假设和备择假设。

Define the distribution of the test statistic under H₀, including all parameters and degrees of freedom.

在H₀下定义检验统计量的分布，包括所有参数和自由度。

Use the exact formula, substitute numbers, then state the inequality or probability used for the decision.

使用准确公式，代入数值，然后说明用于决策的不等式或概率。

End with a conclusion worded in the context of the original problem, not just 'reject H₀'.

结尾要联系原问题语境下结论，而不只是写'拒绝H₀'。

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11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Forgetting the continuity correction when using a normal approximation to a discrete distribution.

用正态近似逼近离散分布时忘记连续性校正。

Using the wrong degrees of freedom: goodness of fit uses k − 1 − p, while contingency tables use (r−1)(c−1).

自由度错误：拟合优度使用k − 1 − p，而列联表使用(r−1)(c−1)。

Confusing one-tailed and two-tailed tests in chi-squared tests; chi-squared tests are always one-tailed.

混淆卡方检验中的单尾和双尾；卡方检验始终是单尾的。

Writing P(X = x) for continuous distributions; use P(a < X < b) or the cdf instead.

对连续分布写P(X = x)；应改用P(a < X < b)或累积分布函数。

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12. Summary of Key FS2 Formulas | FS2核心公式小结

Poisson: P(X = r) = e⁻ˡ λʳ / r!, E(X) = Var(X) = λ.

泊松：P(X = r) = e⁻ˡ λʳ / r!，E(X) = Var(X) = λ。

Exponential: f(x) = λe⁻ˡˣ, E(X) = 1/λ, Var(X) = 1/λ², P(X > x) = e⁻ˡˣ.

指数：f(x) = λe⁻ˡˣ，E(X) = 1/λ，Var(X) = 1/λ²，P(X > x) = e⁻ˡˣ。

Uniform: f(x) = 1/(b−a), E(X) = (a+b)/2, Var(X) = (b−a)²/12.

均匀：f(x) = 1/(b−a)，E(X) = (a+b)/2，Var(X) = (b−a)²/12。

Chi-squared: χ² = Σ(O−E)²/E, with appropriate degrees of freedom.

卡方：χ² = Σ(O−E)²/E，使用相应的自由度。

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