📚 A-Level Further Maths Example Responses Paper 4 Unit FS2 | A-Level进阶数学例解：第四卷 统计2知识点精讲
In the A-Level Further Mathematics specification, Paper 4 Unit FS2 (Further Statistics 2) deepens your understanding of probability models, statistical inference, and the underpinning theory of errors and estimation. Mastering FS2 requires not only fluency with distributions like Poisson and exponential, but also the ability to construct and interpret chi‑squared tests, work with probability generating functions, and apply the Central Limit Theorem to construct confidence intervals. This article unpacks the key topics through clear example responses, highlighting common pitfalls and examiner expectations.
在A-Level进阶数学课程中，第四卷统计2（FS2）深化了我们对概率模型、统计推断以及误差与估计理论的理解。要掌握FS2，不仅需要熟练运用泊松分布和指数分布，还需要能够构建和解释卡方检验、运用概率生成函数，并应用中心极限定理来构造置信区间。本文通过清晰的例解来剖析核心知识点，突出常见错误及考官预期。
1. Continuous Random Variables – PDF, CDF and Moments | 连续随机变量——概率密度函数、累积分布函数与矩
For a continuous random variable X, the probability density function f(x) satisfies ∫−∞∞ f(x) dx = 1 and probabilities are found by integration: P(a < X < b) = ∫ab f(x) dx. The cumulative distribution function is F(x) = P(X ≤ x) = ∫−∞x f(t) dt. The mean is E(X) = ∫ x f(x) dx and the variance Var(X) = E(X²) − [E(X)]².
对于连续随机变量X，概率密度函数f(x)满足∫−∞∞ f(x) dx = 1，概率通过积分求得：P(a < X < b) = ∫ab f(x) dx。累积分布函数为F(x) = P(X ≤ x) = ∫−∞x f(t) dt。均值为E(X) = ∫ x f(x) dx，方差Var(X) = E(X²) − [E(X)]²。
Example response: The continuous random variable X has f(x) = k(4 − x²) for −2 ≤ x ≤ 2, and 0 otherwise. Find k and the median m.
Solution: Integrate: ∫−22 k(4 − x²) dx = k [4x − x³/3]−22 = k (16 − 16/3) = (32/3)k = 1 ⇒ k = 3/32. For the median, solve ∫−2m (3/32)(4 − x²) dx = 0.5. This yields (3/32)[4m − m³/3 + 16/3] = 0.5. Solving the cubic 4m − m³/3 = 16/3 gives m = 0 (by symmetry), indeed median = 0. A common error is forgetting to check that the median lies within the support.
示例解答：连续随机变量X的概率密度函数为f(x) = k(4 − x²)，−2 ≤ x ≤ 2，其余为0。求k和中位数m。
解：积分：∫−22 k(4 − x²) dx = k [4x − x³/3]−22 = k (16 − 16/3) = (32/3)k = 1 ⇒ k = 3/32。对于中位数，解∫−2m (3/32)(4 − x²) dx = 0.5。得到(3/32)[4m − m³/3 + 16/3] = 0.5。化简三次方程4m − m³/3 = 16/3，解得m = 0（由对称性），因此中位数为0。常见错误是忘记验证中位数是否落在定义域内。
2. Exponential Distribution – Memoryless Property | 指数分布——无记忆性
The exponential distribution with rate parameter λ has pdf f(x) = λ e−λx for x ≥ 0, and cdf F(x) = 1 − e−λx. Its mean is 1/λ and variance 1/λ². A distinctive feature is the memoryless property: P(X > s + t | X > s) = P(X > t). This property often appears in questions modelling waiting times or the lifetime of components.
指数分布具有速率参数λ，其概率密度函数为f(x) = λ e−λx（x ≥ 0），累积分布函数为F(x) = 1 − e−λx。均值为1/λ，方差为1/λ²。其显著特征是无记忆性：P(X > s + t | X > s) = P(X > t)。这一性质常出现在对等待时间或元件寿命建模的题目中。
Example response: The lifetime T of a light bulb (in 1000 hours) follows an exponential distribution with mean 2. Find P(T > 3 | T > 1).
Solution: λ = 1/mean = 0.5. P(T > t) = e−0.5t. Then using memoryless property, P(T > 3 | T > 1) = P(T > 2) = e−0.5×2 = e−1 ≈ 0.368. Alternatively, compute P(T > 3)/P(T > 1) = e−1.5/e−0.5 = e−1. The key is to recognise that the conditional probability reduces to the unconditional probability of surviving an additional 2000 hours.
示例解答：灯泡的寿命T（以千小时计）服从均值为2的指数分布。求P(T > 3 | T > 1)。
解：λ = 1/均值 = 0.5。P(T > t) = e−0.5t。利用无记忆性，P(T > 3 | T > 1) = P(T > 2) = e−0.5×2 = e−1 ≈ 0.368。或者直接计算P(T > 3)/P(T > 1) = e−1.5/e−0.5 = e−1。关键是识别出条件概率简化为再存活2000小时的无条件概率。
3. Poisson Distribution & Hypothesis Testing | 泊松分布与假设检验
The Poisson distribution models the number of events in a fixed interval, with probability mass function P(X = r) = λr e−λ / r! for r = 0,1,2,… . In FS2, you are expected to test hypotheses about the population mean λ. For a one‑tailed test, you calculate the probability of obtaining the observed count or more extreme under H0. If the total probability is less than the significance level, reject H0.
泊松分布用于描述固定区间内事件发生的次数，其概率质量函数为P(X = r) = λr e−λ / r!，r = 0,1,2,…。在FS2中，你需要对总体均值λ进行假设检验。对于单尾检验，计算在H0下得到观测频数或更极端情况的概率。若累计概率小于显著性水平，则拒绝H0。
Example response: A company claims its machine produces 2 defects per roll of fabric. A new roll contains 6 defects. Test at 5% significance whether the defect rate has increased.
Solution: H0: λ = 2, H1: λ > 2. Test statistic: observed count = 6. X ~ Po(2). P(X ≥ 6) = 1 − P(X ≤ 5). Using tables, P(X ≤ 5) = 0.9834 (approx), so P-value = 0.0166 < 0.05. Reject H0; there is evidence of increase. It is crucial to define the test statistic properly and to use the correct tail.
示例解答：一家公司声称其机器每卷布平均产生2个瑕疵。一卷新布有6个瑕疵。在5%显著性水平下检验瑕疵率是否上升。
解：H0: λ = 2，H1: λ > 2。检验统计量：观测次数=6。X ~ Po(2)。P(X ≥ 6) = 1 − P(X ≤ 5)。查表得P(X ≤ 5) ≈ 0.9834，因此P值 = 0.0166 < 0.05。拒绝H0，有证据表明瑕疵率上升。正确设定检验统计量并使用正确的尾部至关重要。
4. Type I and II Errors & Power of a Test | 第一类、第二类错误与检验功效
A Type I error occurs when H0 is rejected but true (probability α, the significance level). A Type II error occurs when H0 is not rejected but false (probability β). The power of a test is 1 − β, the probability of correctly rejecting a false H0. In FS2, you are required to calculate these probabilities for Poisson and binomial models, and to determine the sample size needed to achieve a desired power.
当H0真而被拒绝时发生第一类错误（概率α，即显著性水平）；当H0假但未被拒绝时发生第二类错误（概率β）。检验功效为1 − β，即正确拒绝错误H0的概率。在FS2中，你需要针对泊松和二项模型计算这些概率，并确定达到所需功效所需的样本量。
Example response: Given X ~ Po(μ), test H0: μ = 4 against H1: μ = 6. The critical region is X ≥ 7. Find α and the power.
Solution: α = P(X ≥ 7 | μ = 4). Using Po(4) tables, P(X ≤ 6) = 0.8893, so α = 1 − 0.8893 = 0.1107. For power, β = P(X < 7 | μ = 6) = P(X ≤ 6 | μ = 6) = 0.6063 (from Po(6) table). Hence power = 1 − 0.6063 = 0.3937. This shows that while α is close to 0.1, the power is low, indicating the test is not very effective at detecting the increase to 6.
示例解答：设X ~ Po(μ)，检验H0: μ = 4 vs H1: μ = 6。拒绝域为X ≥ 7。求α和功效。
解：α = P(X ≥ 7 | μ = 4)。查Po(4)表，P(X ≤ 6) = 0.8893，故α = 1 − 0.8893 = 0.1107。对于功效，β = P(X < 7 | μ = 6) = P(X ≤ 6 | μ = 6) = 0.6063，因此功效 = 1 − 0.6063 = 0.3937。这表明虽然α接近0.1，但功效较低，说明该检验对检测μ增加到6的效果不佳。
5. Chi-Squared Goodness of Fit Tests | 卡方拟合优度检验
The chi‑squared goodness-of-fit test assesses whether an observed frequency distribution differs from a theoretical distribution. The test statistic is χ² = Σ (Oi − Ei)² / Ei, where Oi are observed and Ei expected frequencies. Degrees of freedom ν = number of classes − 1 − number of estimated parameters. You must ensure all Ei ≥ 5, combining categories if necessary. A common exam task is to test whether data follow a Poisson distribution.
卡方拟合优度检验用于评估观测频数分布是否与理论分布存在差异。检验统计量为χ² = Σ (Oi − Ei)² / Ei，其中Oi为观测频数，Ei为期望频数。自由度ν = 类别数 − 1 − 估计参数个数。必须确保所有Ei ≥ 5，必要时合并类别。常见的考题是检验数据是否符合泊松分布。
Example response: The number of accidents per day at a factory is recorded for 50 days. The mean is 1.2. Frequencies for 0,1,2,3+ are 18, 20, 8, 4. Test at 5% significance whether a Poisson distribution fits.
Solution: Estimate λ = 1.2. Expected frequencies using Po(1.2): P(0)=0.3012 → 15.06, P(1)=0.3614 → 18.07, P(2)=0.2169 → 10.845, P(≥3)=0.1205 → 6.025. Combine last class with ≥2 to meet the E ≥ 5 requirement? Actually ≥3 expected 6.025 > 5, so no need combine. Classes: 0,1,2,≥3. χ² = (18−15.06)²/15.06 + (20−18.07)²/18.07 + (8−10.845)²/10.845 + (4−6.025)²/6.025 = 0.573 + 0.207 + 0.741 + 0.680 = 2.201. ν = 4 − 1 − 1 = 2. Critical value at 5% is 5.991. Since 2.201 < 5.991, do not reject H0; Poisson model is acceptable.
示例解答：某工厂记录50天的事故次数，均值为1.2。频数分布：0次18天，1次20天，2次8天，3次及以上4天。在5%显著性水平下检验是否服从泊松分布。
解：估计λ = 1.2。用Po(1.2)计算期望频数：P(0)=0.3012→15.06，P(1)=0.3614→18.07，P(2)=0.2169→10.845，P(≥3)=0.1205→6.025。所有期望值均大于5，无需合并。χ² = (18−15.06)²/15.06 + (20−18.07)²/18.07 + (8−10.845)²/10.845 + (4−6.025)²/6.025 = 2.201。ν = 4 − 1 − 1 = 2。5%临界值为5.991。因2.201 < 5.991，不拒绝H0，泊松模型可接受。
6. Contingency Tables – Test for Association | 列联表——独立性检验
Contingency tables test whether two categorical variables are associated. Expected frequencies are computed as (row total × column total) / grand total. The chi‑squared statistic again uses Σ (O − E)²/E. Degrees of freedom = (rows − 1) × (columns − 1). A special case is the 2×2 table with Yates’ correction: χ² = Σ (|O − E| − 0.5)² / E, applied to improve the approximation when frequencies are small.
列联表检验两个类别变量是否关联。期望频数计算为（行合计 × 列合计）/ 总计。卡方统计量同样使用Σ (O − E)²/E。自由度为（行数−1）×（列数−1）。对于2×2表格，可使用Yates校正：χ² = Σ (|O − E| − 0.5)² / E，以在频数较小时改善近似效果。
Example response: A sample of 200 people are asked about their smoking habit (Smoker/Non‑smoker) and exercise frequency (Regular/Irregular). The table shows Smoker&Regular 30, Smoker&Irregular 60, Non‑smoker&Regular 70, Non‑smoker&Irregular 40. Test for association at 1% level.
Solution: Expected values: row1 total 90, row2 110; col1 100, col2 100. E₁₁ = 90×100/200=45, E₁₂=45, E₂₁=55, E₂₂=55. Using Yates’ correction: χ² = (|30−45|−0.5)²/45 + (|60−45|−0.5)²/45 + (|70−55|−0.5)²/55 + (|40−55|−0.5)²/55 = (14.5²/45)+(14.5²/45)+(14.5²/55)+(14.5²/55) ≈ 4.672+4.672+3.823+3.823=16.99. ν=1, critical 1% value 6.635. Reject H0; there is strong evidence of association. Without correction, χ² would be larger, so the correction is conservative.
示例解答：调查200人的吸烟习惯（吸烟/不吸烟）与运动频率（规律/不规律）。数据：吸烟且规律30人，吸烟且不规律60人，不吸烟且规律70人，不吸烟且不规律40人。在1%水平下检验关联性。
解：期望值：行1合计90，行2合计110；列1合计100，列2合计100。E₁₁=90×100/200=45，E₁₂=45，E₂₁=55，E₂₂=55。用Yates校正：χ² = (|30−45|−0.5)²/45 + (|60−45|−0.5)²/45 + (|70−55|−0.5)²/55 + (|40−55|−0.5)²/55 = 16.99。ν=1，1%临界值6.635。拒绝H0，有极强关联证据。不校正时χ²更大，因此校正是保守的。
7. Spearman’s Rank Correlation Coefficient | 斯皮尔曼等级相关系数
Spearman’s rank correlation coefficient rs measures the strength of monotonic association between two variables. It is calculated by rs = 1 − 6 Σ di² / [n(n² − 1)], where di are the differences in ranks. A hypothesis test for zero correlation uses the test statistic t = rs √(n−2) / √(1 − rs²), which follows a t‑distribution with n−2 degrees of freedom.
斯皮尔曼等级相关系数rs衡量两个变量之间单调关系的强度。计算公式为rs = 1 − 6 Σ di² / [n(n² − 1)]，其中di为秩次之差。对零相关进行假设检验时，使用检验统计量t = rs √(n−2) / √(1 − rs²)，服从自由度为n−2的t分布。
Example response: The ranks of 8 students in mathematics and physics are given. The rank differences d: 1, −2, 0, 3, −1, −2, 1, 0. Compute rs and test at 5% whether there is positive correlation.
Solution: Σd² = 1+4+0+9+1+4+1+0 = 20. rs = 1 − 6×20 / [8(64−1)] = 1 − 120 / (8×63)=1 − 120/504 = 1 − 0.2381 = 0.7619. t = 0.7619 √6 / √(1−0.5805) = 0.7619×2.449 / √0.4195 ≈ 1.866 / 0.6477 = 2.88. Critical t at 6 df, one‑tailed 5% is 1.943. Since 2.88 > 1.943, reject H0; significant positive correlation.
示例解答：8名学生的数学与物理排名秩差d为：1, −2, 0, 3, −1, −2, 1, 0。计算rs并在5%水平下检验是否存在正相关。
解：Σd²=20。rs=1−6×20/[8(64−1)]=1−120/504≈0.7619。t=0.7619√6/√(1−0.5805
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