📚 Mastering Pre-U OCR Statistics: In-depth Analysis of Past Papers | 掌握 Pre-U OCR 统计：历年真题深度解析
Past papers are the most powerful revision tool for Pre-U OCR Statistics (9794). By dissecting real exam questions, you develop the ability to recognise underlying structures, apply the correct distributional theory, and avoid subtle traps that separate A* candidates from the rest. This article explores key topics through the lens of genuine past-paper challenges, providing step-by-step reasoning and bilingual commentary.
历年真题是复习 Pre-U OCR 统计（9794）最强大的武器。通过剖析真实的考试题目，你可以培养识别底层结构、应用恰当分布理论的能力，并避开那些区分 A* 考生与常人的陷阱。本文透过一道道具代表性的真题，提供逐步解析和中英双语点评。
1. Discrete Random Variables & Probability Distributions | 离散随机变量与概率分布
A typical Pre-U question gives a small probability mass function with an unknown constant and asks for the full distribution, E(X) and Var(X). Often candidates misread the condition that two probabilities are equal or that a cumulative probability is provided. Always verify that the sum of probabilities equals one before proceeding.
典型的 Pre-U 题目会给出一个含有未知常数的小概率质量函数，并要求写出完整分布、E(X) 和 Var(X)。考生经常误读两个概率相等的条件，或者忽略给出的累积概率。务必在继续之前验证概率之和为 1。
Example: lues 0, 1, 2, 3. It is given that P(X=0) = 0.1, P(X=1) = 2k, P(X=2) = k, and P(X=3) = 0.3.
例题：随机变量 X 取值 0, 1, 2, 3。已知 P(X=0) = 0.1, P(X=1) = 2k, P(X=2) = k, P(X=3) = 0.3。
- Step 1: Solve 0.1 + 2k + k + 0.3 = 1 ⇒ 3k = 0.6 ⇒ k = 0.2.
- 步骤1：解方程 0.1 + 2k + k + 0.3 = 1 ⇒ 3k = 0.6 ⇒ k = 0.2。
- Step 2: =0.4, P(X=2)=0.2. Use these to find E(X) = Σ x·p(x) = 0×0.1 + 1×0.4 + 2×0.2 + 3×0.3 = 1.7.
- 步骤2：因此 P(X=1)=0.4, P(X=2)=0.2。计算 E(X) = Σ x·p(x) = 0×0.1 + 1×0.4 + 2×0.2 + 3×0.3 = 1.7。
- Step 3: (X²) = 0²×0.1 + 1²×0.4 + 2²×0.2 + 3²×0.3 = 0 + 0.4 + 0.8 + 2.7 = 3.9. Hence Var(X) = E(X²) – [E(X)]² = 3.9 – 1.7² = 3.9 – 2.89 = 1.01.
- 步骤3：然后 E(X²) = 0²×0.1 + 1²×0.4 + 2²×0.2 + 3²×0.3 = 0 + 0.4 + 0.8 + 2.7 = 3.9。因此 Var(X) = E(X²) – [E(X)]² = 3.9 – 1.7² = 3.9 – 2.89 = 1.01。
This foundation is crucial because it reappears in PGFs and linear combinations, where errors in basic expectation algebra cascade.
这个基础至关重要，因为它在 PGF 和线性组合中反复出现，基础期望代数中的错误会引发连锁反应。
2. PGFs: Finding Distributions and Moments | 概率生成函数：求分布与矩
s G(t) and asks to recover the probability distribution, then find E(X) and Var(X) using G'(1) and G''(1). Remember that the coefficient of tˣ in the power series expansion gives P(X=x).
概率生成函数是 Pre-U 的高频考点。常见题型给出 G(t)，要求还原概率分布，然后利用 G'(1) 和 G''(1) 求 E(X) 与 Var(X)。记住，幂级数展开式中 tˣ 的系数就是 P(X=x)。
Example from a past paper: The PGF of X is G(t) = (0.2 + 0.8t)⁴.
真题示例：X 的 PGF 为 G(t) = (0.2 + 0.8t)⁴。
By inspection, this is a binomial expansion. Expanding gives P(X=x) = ⁴Cₓ (0.8)ˣ (0.2)⁴⁻ˣ for x = 0,1,2,3,4. Differentiation is quicker: G'(t) = 4×0.8 (0.2+0.8t)³ = 3.2 (0.2+0.8t)³, so E(X) = G'(1) = 3.2×1³ = 3.2, as expected since np = 4×0.8 = 3.2.
通过观察可知这是二项展开式。展开后得到 P(X=x) = ⁴Cₓ (0.8)ˣ (0.2)⁴⁻ˣ，x=0至4。用求导更快：G'(t) = 4×0.8 (0.2+0.8t)³ = 3.2 (0.2+0.8t)³，所以 E(X) = G'(1) = 3.2×1³ = 3.2，与二项分布的 np = 4×0.8 = 3.2 一致。
Next, G''(t) = 3.2×3×0.8 (0.2+0.8t)² = 7.68 (0.2+0.8t)², so G''(1) = 7.68. Then Var(X) = G''(1) + G'(1) – [G'(1)]² = 7.68 + 3.2 – 3.2² = 10.88 – 10.24 = 0.64, matching npq = 4×0.8×0.2 = 0.64.
接着，G''(t) = 3.2×3×0.8 (0.2+0.8t)² = 7.68 (0.2+0.8t)²，故 G''(1) = 7.68。于是 Var(X) = G''(1) + G'(1) – [G'(1)]² = 7.68 + 3.2 – 3.2² = 10.88 – 10.24 = 0.64，与 npq = 4×0.8×0.2 = 0.64 相符。
3. Continuous Random Variables: PDFs and CDFs | 连续随机变量：PDF 与 CDF
ys check that the integral of the pdf over its domain equals 1. For the median m, solve F(m) = 0.5.
Pre-U 常考分段概率密度函数 f(x)，要求求出常数、累积分布函数 F(x) 和中位数。务必验证 pdf 在定义域上的积分等于 1。对于中位数 m，解方程 F(m) = 0.5。
Example: f(x) = { kx² 0 ≤ x ≤ 2, { k(4 – x) 2 < x ≤ 4, { 0 otherwise. Find k, F(x), and the median.
例题：f(x) = { kx² 0 ≤ x ≤ 2, { k(4 – x) 2 < x ≤ 4, { 0 其他。求 k, F(x) 和 中位数。
Integrate: ∫₀² kx² dx + ∫₂⁴ k(4 – x) dx = k [x³/3]₀² + k [4x – x²/2]₂⁴ = k(8/3) + k[(16 – 8) – (8 – 2)] = k(8/3 + 8 – 6) = k(8/3 + 2) = (14k/3) = 1 ⇒ k = 3/14.
积分：∫₀² kx² dx + ∫₂⁴ k(4 – x) dx = k [x³/3]₀² + k [4x – x²/2]₂⁴ = k(8/3) + k[(16 – 8) – (8 – 2)] = k(8/3 + 8 – 6) = k(8/3 + 2) = (14k/3) = 1 ⇒ k = 3/14。
For 0 ≤ x ≤ 2: F(x) = ∫₀ˣ (3/14) t² dt = (3/14)(x³/3) = x³/14. For 2 < x ≤ 4: F(x) = F(2) + ∫₂ˣ (3/14)(4 – t) dt = 8/14 + (3/14)[4t – t²/2]₂ˣ = 8/14 + (3/14)[(4x – x²/2) – (8 – 2)] = 4/7 + (3/14)(4x – x²/2 – 6) = 4/7 + (12x/14 – 3x²/28 – 18/14) = (8/14 – 18/14) + (12x/14) – (3x²/28) = –5/7 + (6x/7) – (3x²/28).
0 ≤ x ≤ 2 时: F(x) = x³/14. 2 < x ≤ 4 时: F(x) = 4/7 + (3/14)(4x – x²/2 – 6)，整理得 F(x) = –5/7 + (6x/7) – (3x²/28)。
n lies where F(m)=0.5. Test F(2)=8/14≈0.571, so median < 2. Solve m³/14 = 0.5 → m³=7 → m = ∛7 ≈ 1.913.
中位数满足 F(m)=0.5。检验 F(2)=8/14≈0.571，因此中位数 < 2。解 m³/14 = 0.5 → m³=7 → m = ∛7 ≈ 1.913。
4. Expectation Algebra & Linear Combinations | 期望代数与线性组合
(Y) = aE(X₁) + bE(X₂) + c, and Var(Y) = a²Var(X₁) + b²Var(X₂) if independent. A common trap is forgetting to square the coefficients for variance.
Pre-U 卷二经常要求计算独立随机变量线性组合的期望与方差，如 Y = aX₁ + bX₂ + c。使用 E(Y) = aE(X₁) + bE(X₂) + c，以及独立时 Var(Y) = a²Var(X₁) + b²Var(X₂)。常见的陷阱是对方差忘记系数平方。
Example: X₁ ~ N(μ₁, σ₁²) and X₂ ~ N(μ₂, σ₂²) independently. Let T = 3X₁ – 2X₂. Find E(T) and Var(T).
例题：X₁ ~ N(μ₁, σ₁²) 且 X₂ ~ N(μ₂, σ₂²) 相互独立。设 T = 3X₁ – 2X₂。求 E(T) 和 Var(T)。
E(T) = 3μ₁ – 2μ₂. Var(T) = 3²σ₁² + (– n confidence intervals for the difference of means or in hypothesis tests about contrasts.
E(T) = 3μ₁ – 2μ₂。Var(T) = 3²σ₁² + (–2)²σ₂² = 9σ₁² + 4σ₂²。这一结果后续会用于均值差值的置信区间或关于对照的假设检验。
5. Sampling & The Central Limit Theorem | 抽样与中心极限定理
Questions on the Central Limit Theorem (CLT) often give a non-normal population with known mean μ and variance σ². For a sample of size n, the sample mean X̄ is approximately N(μ, σ²/n) if n is large. The Pre-U exam expects you to state the CLT clearly and justify the approximation before calculating probabilities.
中心极限定理（CLT）的题目常常给定一个非正态总体，已知均值 μ 和方差 σ²。对于样本量 n 足够大的样本，样本均值 X̄ 近似服从 N(μ, σ²/n)。Pre-U 考试要求你清晰陈述 CLT 并进行近似论证，然后计算概率。
Example: A shop’s daily sales X are skewed with mean £820 and standard deviation £150. Find the probability that the mean sales over 50 randomly chosen days exceed £850.
例题：某店日销售额 X 呈偏态，均值 £820，标准差 £150。求随机50天的日均销售额超过 £850 的概率。
By the CLT, X̄ ≈ N(820, 150²/50 = 450) for n=50. Thus Z = (850 – 820)/√450 ≈ 30 / 21.21 = 1.414. P(X̄ lways check that n is large enough; the exam may ask why n=50 justifies the approximation.
由 CLT，n=50 时 X̄ ≈ N(820, 150²/50 = 450)。因此 Z = (850 – 820)/√450 ≈ 30 / 21.21 = 1.414。P(X̄ > 850) = P(Z > 1.414) ≈ 0.0786。务必检查 n 是否足够大；试题可能要求解释 n=50 为何可以合理近似。
6. Confidence Intervals for the Mean | 均值的置信区间
Constructing a 95% confidence interval when σ is known uses the formula X̄ ± z₀.₀₂₅ × σ/√n. Past papers love to then ask for the interpretation: “We are 95% confident that the true population mean lies within this interval.” The interval does not make a probability statement about μ once it is computed.
当 σ 已知时，构建 95% 置信区间的公式为 X̄ ± z₀.₀₂₅ × σ/√n。真题常要求解释：“我们有 95% 的信心认为真实总体均值位于该区间内。”区间一旦计算出来，就不再是关于 μ 的概率陈述。
Example: A random sample of 40 items gives X̄ = 52.0, population σ = 5.0. Find the 90% confidence interval.
例题：一个 40 个产品的随机样本给出 X̄ = 52.0，总体 σ = 5.0。求 90% 置信区间。
For 90% confidence, z₀.₀₅ = 1.645. The interval is 52.0 ± 1.645 × (5.0/√40 e question asked for the minimum sample size to achieve a width of 2, solve 2 × 1.645 × 5/√n = 2 ⇒ √n = 8.225 ⇒ n ≈ 67.7, so need 68.
90% 置信水平下，z₀.₀₅ = 1.645。区间为 52.0 ± 1.645 × (5.0/√40) = 52.0 ± 1.645 × 0.7906 = (50.70, 53.30)。若题目要求达到宽度为 2 的最小样本量，解 2 × 1.645 × 5/√n = 2 ⇒ √n = 8.225 ⇒ n ≈ 67.7，故需 68 个样本。
7. Hypothesis Testing: Type I/II Errors & Power | 假设检验：两类错误与功效
A well-crafted Pre-U question asks for the probability of a Type I error (α) and the power of a test (1 – β). You need to identify the critical region from the null distribution, then compute the probability under the alternative distribution.
精心设计的 Pre-U 题目会要求计算 I 型错误的概率（α）和检验功效（1 – β）。你需要从原假设分布确定拒绝域，然后在备择假设分布下计算概率。
Example: H₀: μ = 100, σ = 15, n = 25. Test at the 5% significance level against H₁: μ > 100. Determine the power when μ = 105.
例题：H₀: μ = 100, σ = 15, n = 25。在 5% 显著性水平下检验 H₁: μ > 100。求 μ = 105 时检验的功效。
Under H₀, X̄ ~ N(100, 15²/25 = 9). Critical value c = 100 + 1.645×3 = 104.935. Reject if X̄ > 104.935. Type I error = P(reject | H₀ true) = 0.05. Under H₁ with μ = 105, X̄ ~ N(105, 9). Power = P(X̄ > 104.935 | μ=105) = P( Z > (104.935 – 105)/3 ) = P(Z > –0.0217) ≈ 0.5087. So the test has about 51% power to detect a shift of 5 units.
在原假设下 X̄ ~ N(100, 9)，临界值 c = 100 + 1.645×3 = 104.935。当 X̄ > 104.935 时拒绝。I 型错误 = 0.05。在备择假设 μ = 105 时，X̄ ~ N(105, 9)。功效 = P(X̄ > 104.935) = P( Z > –0.0217 ) ≈ 0.5087。因此该检验检测 5 个单位偏移的能力约为 51%。
8. Chi-Squared Goodness of Fit | 卡方拟合优度检验
For goodness-of-fit tests, the test statistic is Σ (Oᵢ – Eᵢ)² / Eᵢ, which under the null follows a χ² distribution with (k – 1 – p) degrees of freedom, where p parameters are estimated from the data. A common mistake is using the wrong degrees of freedom or forgetting to combine categories with small expected frequencies.
拟合优度检验的检验统计量为 Σ (Oᵢ – Eᵢ)² / Eᵢ，在原假设下服从自由度为 (k – 1 – p) 的 χ² 分布，其中 p 是从数据中估计的参数个数。常见错误是使用错误的自由度，或忘记合并期望频数较小的类别。
Example: Observed counts for 5 categories: 18, 25, 22, 20, 15. Test the hypothesis that they follow a uniform distribution.
例题：5 个类别的观测频数：18, 25, 22, 20, 15。检验它们是否服从均匀分布。
Total n = 100. Under uniform, each Eᵢ = 20. χ² = (18–20)²/20 + (25–20)²/20 + (22–20)²/20 + (20–20)²/20 + (15–20)²/20 = 4/20 + 25/20 + 4/20 + 0 + 25/20 = 58/20 = 2.9. Degrees of freedom = 5 – 1 = 4. Critical value at 5% is 9.488, so we do not reject H₀. The fit to a uniform distribution is adequate.
总样本量 n = 100。均匀分布下每个 Eᵢ = 20。χ² = (18–20)²/20 + (25–20)²/20 + (22–20)²/20 + (20–20)²/20 + (15–20)²/20 = 2.9。自由度 df = 5 – 1 = 4。5% 临界值为 9.488，不拒绝 H₀，数据与均匀分布拟合尚可。
9. Correlation & Regression Inference | 相关与回归推断
Pre-U often includes a full analysis of bivariate data: compute Pearson’s r, find the least-squares regression line, and perform a t-test for the correlation coefficient or the slope. Remind yourself of the formula Sxy / √Sxx Syy for r and b = Sxy / Sxx for the slope.
Pre-U 经常要求对双变量数据进行完整分析：计算 Pearson 相关系数 r，求最小二乘回归直线，并对相关系数或斜率进行 t 检验。请熟记公式：r = Sxy / √Sxx Syy，斜率 b = Sxy / Sxx。
Example: Sxx = 100, Syy = 80, Sxy = 70. Find the regression line y = a + bx and test H₀: β = 0 at 5% level.
例题：n = 12, Σx = 60, Σy = 48, Sxx = 100, Syy = 80, Sxy = 70。求回归直线 y = a + bx，并在 5% 水平检验 H₀: β = 0。
Mean x̄ = 5, ȳ = 4. Slope b = Sxy/Sxx = 70/100 = 0.7. Intercept a = ȳ – b x̄ = 4 – 0.7×5 = 0.5. So line: y = 0.5 + 0.7x.
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