📚 Pre-U CAIE Statistics: Unit Test Mock Paper Analysis | Pre-U CAIE 统计：单元测试模拟卷解析
This mock unit test analysis is designed to reinforce key topics in the Pre-U CAIE Statistics syllabus. The paper contains eight structured questions that span discrete random variables, Poisson approximation, the normal distribution, sampling distributions, confidence intervals, hypothesis testing, chi-squared goodness-of-fit, and linear regression. Every question is fully worked through, with detailed steps and commentary in both English and Chinese. Use these solutions to identify common pitfalls and strengthen your exam technique.
本模拟单元测试解析旨在巩固 Pre-U CAIE 统计课程的核心考点。试卷包含八道结构化题目，涵盖离散随机变量、泊松近似、正态分布、抽样分布、置信区间、假设检验、卡方拟合优度以及线性回归。每道题均提供完整演算步骤和中英双语解析，帮助你发现易错点，提升应试技巧。
1. Exam Overview | 考试概览
This mock paper simulates the style of Pre-U Statistics unit tests, blending straightforward calculations with interpretative questions. The eight questions progress from probability foundations to inference and modelling. Candidates should pay close attention to the phrasing of hypotheses, the choice of distributional approximations, and the proper use of statistical tables. Marks are awarded not only for correct numerical answers but also for clear reasoning and accurate notation.
本模拟卷贴近 Pre-U 统计单元测试风格，将基础计算与分析性问题相结合。八道题目由概率基础逐步推进到推断与建模。考生应特别注意假设的表述方式、分布近似的适用条件以及统计表格的规范使用。得分不仅依赖数值正确，更看重清晰的推理过程和准确的符号表达。
The table below summarises how the eight questions map onto the syllabus content and the approximate mark weighting you should expect in a real unit test.
下表汇总了八道题与考纲内容的对应关系，以及在真实单元测试中大致的分值分布。
| Question | Topic 主题 | Marks 分值 | Skill Tested 考查能力 |
|---|---|---|---|
| Q1 | Discrete random variables 离散随机变量 | 6 | E(X), Var(X), linear transformation |
| Q2 | Poisson approximation 泊松近似 | 7 | Conditions, λ = np, tail probability |
| Q3 | Normal distribution 正态分布 | 7 | Standardisation, inverse normal |
| Q4 | Sampling distributions 抽样分布 | 7 | Standard error, CLT |
| Q5 | Confidence intervals 置信区间 | 7 | Construction and interpretation |
| Q6 | Hypothesis testing 假设检验 | 8 | One- and two-tailed tests |
| Q7 | Chi-squared goodness-of-fit 卡方拟合优度 | 8 | Expected frequencies, degrees of freedom |
| Q8 | Linear regression 线性回归 | 8 | Least squares, prediction |
2. Q1: Discrete Random Variables | 第 1 题：离散随机变量
Question: and P(X = 3) = 0.3. (a) Find E(X) and Var(X). (b) Given Y = 3X − 1, find E(Y) and Var(Y).
题目： 离散随机变量 X 满足 P(X = 1) = 0.2，P(X = 2) = 0.5，P(X = 3) = 0.3。(a) 求 E(X) 与 Var(X)。(b) 已知 Y = 3X − 1，求 E(Y) 与 Var(Y)。
Part (a) is a routine expectation calculation, but the marks are concentrated in the layout. Write the probability distribution as a table first, then compute E(X) and E(X²) column by column. This makes the arithmetic easy to check and guarantees method marks even if a single multiplication slips.
第 (a) 问属于常规期望计算，但得分点集中在书写规范上。先把概率分布写成表格，再逐列计算 E(X) 与 E(X²)。这样既便于复核运算，也能确保即使某一步乘法出错仍拿到过程分。
E(X) = Σ x·P(X = x) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1
E(X²) = 1²(0.2) + 2²(0.5) + 3²(0.3) = 0.2 + 2.0 + 2.7 = 4.9
Var(X) = E(X²) − [E(X)]² = 4.9 − 2.1² = 4.9 − 4.41 = 0.49
hifts the mean but never the variance, a point that examiners deliberately probe with distractors such as "add 1 to the variance".
第 (b) 问考查线性变换公式 E(aX + b) = aE(X) + b 与 Var(aX + b) = a²Var(X)。常数项只平移均值、绝不影响方差，这正是考官用“方差加上 1”这类干扰项刻意设置的陷阱。
E(Y) = 3(2.1) − 1 = 5.3, Var(Y) = 3²(0.49) = 4.41
- Always show the E(X²) line explicitly; skipping it loses a method mark even when Var(X) is correct. 务必单独写出 E(X²) 这一行；即使 Var(X) 算对，省略该行也会丢过程分。
- Variance is never negative. If your answer is negative, you have subtracted in the wrong order. 方差不可能为负；若结果为负，说明减号方向写反了。
- Give answers to at least three significant figures unless the question specifies otherwise. 除题目另有要求，答案至少保留三位有效数字。
3. Q2: Poisson Approximation to the Binomial | 第 2 题：二项分布的泊松近似
Question: Defects occur independently at a rate of 0.02 per item. A batch of 200 items is inspected. Using a Poisson approximation, find (a) the probability of no defects and (b) the probability of at most two defects.
题目： 每件产品独立地以 0.02 的概率存在缺陷。现检验 200 件的一批产品。使用泊松近似求 (a) 无缺陷的概率；(b) 至多两件缺陷的概率。
Before touching the calculator, state the two conditions that licence the approximation: n is large and p is small, with np moderate. A useful rule of thumb in the CAIE context is n > 50 and np < 5, though the question itself usually supplies the instruction to approximate.
动笔计算前，先写清使用近似的两个条件：n 很大、p 很小，且 np 适中。CAIE 语境下的经验法则是 n > 50 且 np < 5，不过题目通常会直接给出“使用近似”的指令。
λ = np = 200 × 0.02 = 4, so X ≈ Po(4)
P(X = 0) = e⁻⁴ ≈ 0.0183
P(X ≤ 2) = e⁻⁴(1 + 4 + 8) = 13e⁻⁴ ≈ 0.2381
compute only P(X = 2) = 0.1465 and lose the final mark. Reading the inequality carefully - "at most", "fewer than", "at least" - is worth as much as the arithmetic.
注意 P(X ≤ 2) 是累积概率，必须求和三项。最常见的错误是只算了 P(X = 2) = 0.1465，从而丢掉最后的分。仔细辨析“至多”“少于”“至少”这类不等式，其价值不亚于运算本身。
- Write the Poisson formula P(X = r) = e−λ λ^r ÷ r! before substituting, so the method is visible. 代入前先写出泊松公式 P(X = r) = e−λ λ^r ÷ r!，让方法可见。
- Never use the binomial directly when the question says "using an approximation" - no marks for the exact answer. 题目要求“使用近似”时不要直接算二项分布，算对了也无分。
- Check the tail: P(X ≥ 1) = 1 − P(X = 0) is far quicker than summing from r = 1 upwards. 处理尾部概率用 P(X ≥ 1) = 1 − P(X = 0) 比逐项相加快得多。
4. Q3: The Normal Distribution | 第 3 题：正态分布
Question: The mass of a component is normally distributed with mean 50 g and standard deviation 8 g. (a) Find the probability that a randomly chosen component has mass greater than 62 g. (b) Find the mass m such that 10% of components exceed m.
题目： 某零件质量服从正态分布，均值 50 g、标准差 8 g。(a) 求随机抽取的零件质量大于 62 g 的概率。(b) 求质量 m，使 10% 的零件质量超过 m。
Both parts rely on standardisation. In part (a) you move from a value to a probability; in part (b) you move from a probability back to a value. Recognising the direction of the transformation is the single most useful habit in this topic.
两问都依赖标准化。第 (a) 问由数值求概率，第 (b) 问由概率反推数值。识别转换方向是这一考点中最实用的习惯。
Z = (X − μ) ÷ σ = (62 − 50) ÷ 8 = 1.5
P(X > 62) = P(Z > 1.5) = 1 − Φ(1.5) ≈ 0.0668
ing the correct tail prevents the classic mistake of using z = 1.2816 with the wrong sign.
第 (b) 问上尾面积为 0.10，故 Φ(z) = 0.90，z ≈ 1.2816，于是 m = μ + zσ = 50 + 1.2816(8) ≈ 60.25 g。务必画出草图：标对尾部区域可以避免把 z = 1.2816 的符号写反这一经典错误。
- State the standardisation explicitly as Z = (X − μ) ÷ σ; this is a standard method mark. 明确写出标准化式 Z = (X − μ) ÷ σ，这是固定的方法分。
- Use the continuity correction only when the normal is approximating a discrete distribution. 只有在用正态近似离散分布时才需要使用连续性修正。
- Quote the z-value to at least four decimal places to protect the final accuracy mark. z 值至少保留四位小数，以保住最后的精度分。
5. Q4: Sampling Distributions | 第 4 题：抽样分布
Question: A random sample of 25 components is taken from the population in Q3. Find the probability that the sample mean exceeds 53 g, stating any assumption you make.
题目： 从第 3 题的总体中抽取容量为 25 的随机样本。求样本均值超过 53 g 的概率，并说明你所作的假设。
The key result is that the sample mean has its own distribution: E(X̄) = μ and Var(X̄ square root of the sample size, which is why quadrupling n only halves the standard error.
核心结论是样本均值自身服从一个分布：E(X̄) = μ，Var(X̄) = σ² ÷ n。其平方根即标准误，它随样本量的平方根缩小，这就是为什么 n 变为四倍时标准误只减半。
SE = σ ÷ √n = 8 ÷ √25 = 8 ÷ 5 = 1.6
Z = (53 − 50) ÷ 1.6 = 1.875, P(X̄ > 53) ≈ 0.0304
Because the population itself is normal, X̄ ou would have had to state that n = 25 is large enough for the CLT to apply approximately.
由于总体本身服从正态分布，无论样本量大小，X̄ 都精确服从正态分布，无需借助中心极限定理。若总体非正态，则必须说明 n = 25 足够大、可以近似使用中心极限定理。
- Never confuse σ with σ ÷ √n - the most heavily penalised error in this topic. 切勿把 σ 与 σ ÷ √n 混淆，这是本考点扣分最重的错误。
- When the sample is drawn without replacement from a small population, a finite population correction may be required. 若从小总体中不放回抽样，可能需要使用有限总体修正因子。
- Read the word "assumption" in the question as a direct instruction to justify normality. 题目中出现“假设”一词，即是在要求你说明正态性的依据。
6. Q5: Confidence Intervals | 第 5 题：置信区间
Question: For the sample in Q4, the sample mean is 52 g. Construct a 95% confidence interval for the population mean and interpret it in context.
题目： 第 4 题的样本均值为 52 g。构造总体均值的 95% 置信区间，并结合情境加以解释。
and the population normal, the critical value comes from the standard normal distribution.
置信区间的结构永远是“估计值 ± 临界值 × 标准误”。由于 σ 已知且总体服从正态分布，临界值取自标准正态分布。
x̄ ± z₀.₀₂₅ × (σ ÷ √n) = 52 ± 1.96 × 1.6 = 52 ± 3.136
95% CI ≈ (48.86, 55.14)
The interpretation is where most marks are lost. The correct statement is: we are 95% confident that the true population mean lies between 48.86 g and 55.14 g. It is wrong to say that 95% of components have masses in this range - the interval is about the mean, not about individual observations.
解释部分是失分重灾区。正确说法是：我们有 95% 的把握认为总体均值落在 48.86 g 与 55.14 g 之间。若说“95% 的零件质量落在此区间内”则是错的——该区间针对的是均值，而非单个观测值。
- Interpretation must mention the population mean, the confidence level and the context. 解释中必须包含总体均值、置信水平与具体情境。
- Increasing the confidence level widens the interval; increasing n narrows it. 提高置信水平会加宽区间；增大 n 则会收窄区间。
- If σ is unknown and n is large, replace z with t and use the sample standard deviation. 若 σ 未知且 n 较大，则用 t 替代 z，并改用样本标准差。
7. Q6: Hypothesis Testing | 第 6 题：假设检验
Question: A machine is supposed to produce components with mean mass 50 g. A sample of 25 gives x̄ = 52 g. Test at the 5% level whether the mean has changed, assuming σ = 8 g.
题目： 某机器生产的零件质量均值应为 50 g。抽取 25 件样本得 x̄ = 52 g。在 5% 显著性水平下检验均值是否发生变化，设 σ = 8 g。
The phrase "has changed" signals a two-tailed test. Writing both hypotheses before any calculation secures the first mark and keeps the later comparison honest.
“是否发生变化”提示这是双侧检验。在做任何计算之前先写出两个假设，既能拿下首分，也能保证后续比较的一致性。
H₀: μ = 50 H₁: μ ≠ 50
Z = (52 − 50) ÷ (8 ÷ √25) = 2 ÷ 1.6 = 1.25
Critical values: ±1.96. Since |1.25| < 1.96, do not reject H₀.
The conclusion must be expressed in context and must not claim that H₀ is true. The correct wording is: there is insufficient evidence at the 5% level to conclude that the mean mass has changed. The sample result is consistent with random variation.
结论必须结合情境表述，且不能宣称 H₀ 为真。正确说法是：在 5% 水平下，没有足够证据表明均值发生了改变，样本结果与随机波动一致。
- Compare a test statistic with a critical value, or a p-value with the significance level - never mix the two approaches. 比较检验统计量与临界值，或比较 p 值与显著性水平——两种方法不可混用。
- Use "insufficient evidence" rather than "prove" or "accept". 使用“证据不足”而非“证明”或“接受”。
- For a one-tailed test at 5%, the critical value becomes 1.645, not 1.96. 单侧 5% 检验的临界值为 1.645，而非 1.96。
8. Q7: Chi-Squared Goodness-of-Fit | 第 7 题：卡方拟合优度
Question: A die is rolled 120 times and the observed frequencies of the faces 1 to 6 are 18, 22, 16, 24, 19 and 21. Test at the 5% level whether the die is fair.
题目： 掷骰子 120 次，1 至 6 点的观测频数依次为 18、22、16、24、19、21。在 5% 水平下检验该骰子是否均匀。
ncy for each face is 120 ÷ 6 = 20, provided no parameter has been estimated from the data.
卡方检验比较观测频数与 H₀ 下的期望频数。此处每个点数的期望频数为 120 ÷ 6 = 20，前提是没有从数据中估计任何参数。
H₀: the die is fair H₁: the die is not fair
X² = Σ (O − E)² ÷ E
X² = (4 + 4 + 16 + 16 + 1 + 1) ÷ 20 = 42 ÷ 20 = 2.10
us one, minus the number of parameters estimated from the sample: v = 6 − 1 − 0 = 5. The 5% .07, and since 2.10 < 11.07 we do not reject H₀ - the data are consistent with a fair die.
自由度等于类别数减 1，再减去从样本中估计的参数个数：v = 6 − 1 − 0 = 5。5% 临界值为 11.07，由于 2.10 < 11.07，不拒绝 H₀——数据与均匀骰子相符。
- Every expected frequency must be at least 5; combine adjacent categories if not. 每个期望频数至少为 5，否则需合并相邻类别。
- If a parameter is estimated (for example a Poisson mean), subtract one further degree of freedom. 若估计了参数（例如泊松均值），自由度需再减 1。
- Do not halve the significance level - chi-squared goodness-of-fit is always one-tailed. 不要把显著性水平减半——卡方拟合优度检验始终是单侧的。
9. Q8: Linear Regression | 第 8 题：线性回归
Question: ares regression line of y on x. (b) Predict y when x = 8, commenting on the reliability of the prediction.
题目： 已知成对数据 (x, y) 的样本汇总量为：Σx = 60，Σy = 110，Σx² = 500，Σxy = 900，n = 10。(a) 求 y 关于 x 的最小二乘回归直线方程。(b) 预测 x = 8 时的 y 值，并评论预测的可靠性。
Two quantities drive the whole calculation: Sₓₓ, the corrected sum of squares of x, and Sₓᵧ, the corrected sum of cross-products. Computing them first turns the slope into a single Division.
整个计算由两个量驱动：x 的校正平方和 Sₓₓ，以及校正交叉乘积和 Sₓᵧ。先算出它们，斜率就只需做一次除法。
Sₓₓ = Σx² − (Σx)² ÷ n = 500 − 3600 ÷ 10 = 140
Sₓᵧ = Σxy − (Σx)(Σy) ÷ n = 900 − 6600 ÷ 10 = 240
b = Sₓᵧ ÷ Sₓₓ = 240 ÷ 140 ≈ 1.
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