📚 A-Level CAIE Statistics: Unit Test Mock Paper Walkthrough | A-Level CAIE 统计：单元测试模拟卷解析
This article walks through a typical CAIE A-Level Statistics unit test, covering data representation, probability, discrete and normal distributions, sampling, regression and hypothesis testing. Each section explains a representative question with the key working, common errors and exam technique.
本文解析一套典型的 CAIE A-Level 统计单元测试，涵盖数据表示、概率、离散与正态分布、抽样、回归和假设检验。每节通过一道代表性题目，说明关键步骤、常见错误与应试技巧。
1. Exam Overview and Paper Structure | 考试概览与试卷结构
The mock paper follows the CAIE Probability & Statistics 1 format: 50 marks in 75 minutes, with all questions compulsory. A typical paper includes one data display question, one probability question, one combinatorics question, one discrete random variable question, one binomial distribution question, one normal distribution question, and one inference or hypothesis testing question.
本模拟卷遵循 CAIE 概率与统计 1 的格式：满分 50 分，时间 75 分钟，全部为必答题。典型试卷包含一道数据展示题、一道概率题、一道排列组合题、一道离散随机变量题、一道二项分布题、一道正态分布题以及一道推断或假设检验题。
- Data handling: stem-and-leaf diagram, median, quartiles | 数据处理：茎叶图、中位数、四分位数
- Probability: Venn diagrams, conditional probability, independence | 概率：韦恩图、条件概率、独立性
- Combinatorics: permutations, combinations, repeated items | 排列组合：排列、组合、重复元素
- Distributions: discrete random variables, binomial, normal | 分布：离散随机变量、二项分布、正态分布
- Inference: confidence intervals, hypothesis tests, correlation, regression | 推断：置信区间、假设检验、相关与回归
2. Question 1: Stem-and-Leaf Diagram and Data Description | 第1题：茎叶图与数据描述
A stem-and-leaf diagram arranges data in ascending order, making the median and quartiles easy to locate. For a data set with n values, the median is at position (n + 1) / 2, while the lower quartile Q₁ and upper quartile Q₃ are found from the median of the lower half and upper half respectively.
茎叶图将数据按升序排列，便于确定中位数和四分位数。对于容量为 n 的数据集，中位数位于第 (n + 1) / 2 个位置，而下四分位数 Q₁ 与上四分位数 Q₃ 分别由下半部分和上半部分的中位数得到。
Consider the ordered stem-and-leaf values: 31, 32, 34, 35, 40, 43, 43, 46, 47, 48, 51, 54. Since n = 12, the median is the average of the 6th and 7th values: (43 + 43) / 2 = 43. The lower quartile is the median of the first six values, (34 + 35) / 2 = 34.5, and the upper quartile is the median of the last six values, (47 + 48) / 2 = 47.5.
考虑有序茎叶数据：31、32、34、35、40、43、43、46、47、48、51、54。因为 n = 12，中位数为第 6 个和第 7 个值的平均值：(43 + 43) / 2 = 43。下四分位数为前六个值的中位数，(34 + 35) / 2 = 34.5，上四分位数为后六个值的中位数，(47 + 48) / 2 = 47.5。
IQR = Q₃ − Q₁ = 47.5 − 34.5 = 13
Always state the interquartile range with the correct upper minus lower order, because IQR measures the spread of the middle 50% of the data and is not affected by extreme outliers.
计算四分位距时必须用上四分位数减去下四分位数，因为 IQR 衡量中间 50% 数据的离散程度，且不受极端异常值影响。
3. Question 2: Probability and Venn Diagrams | 第2题：概率与韦恩图
A Venn diagram question may ask for P(A ∪ B), the conditional probability P(A | B), or a test for independence. Use the addition rule and the conditional probability formula, and always check that probabilities lie between 0 and 1.
韦恩图题可能要求计算 P(A ∪ B)、条件概率 P(A | B) 或判断独立性。使用加法公式和条件概率公式，并始终检查概率是否在 0 到 1 之间。
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A | B) = P(A ∩ B) / P(B)
For example, if P(A) = 0.4, P(B) = 0.3 and P(A ∩ B) = 0.1, then P(A ∪ B) = 0.4 + 0.3 − 0.1 = 0.6. Also P(A | B) = 0.1 / 0.3 = 1/3. Since P(A ∩ B) = 0.1 is not equal to P(A)P(B) = 0.12, the events are not independent.
例如，若 P(A) = 0.4、P(B) = 0.3 且 P(A ∩ B) = 0.1，则 P(A ∪ B) = 0.4 + 0.3 − 0.1 = 0.6。同时 P(A | B) = 0.1 / 0.3 = 1/3。由于 P(A ∩ B) = 0.1 不等于 P(A)P(B) = 0.12，所以事件不独立。
Common mistakes include using P(A ∪ B) = P(A) + P(B) without subtracting the intersection, or swapping the conditional probability numerator and denominator.
常见错误包括计算 P(A ∪ B) 时没有减去交集，或将条件概率的分子和分母颠倒。
4. Question 3: Permutations and Combinations | 第3题：排列与组合
Permutation questions require the number of ordered arrangements, while combination questions require the number of unordered selections. Decide whether order matters before choosing a formula, and remember to divide by factorial terms for repeated items.
排列题求有序安排的数量，组合题求无序选择的数量。选用公式前先判断顺序是否重要，并记住如果有重复元素需要除以相应的阶乘项。
nPr = n! / (n − r)! and nCr = n! / [r!(n − r)!]
For example, the number of ways to arrange 5 people in 3 chairs is 5P3 = 60, while the number of ways to choose a committee of 3 from 10 people is 10C3 = 120. If arranging the letters of the word M A M M A, the number of distinct arrangements is 5! / (3! × 2!) = 10 because the three M letters and two A letters are identical.
例如，将 5 人安排在 3 把椅子上的方法数为 5P3 = 60，而从 10 人中选出 3 人组成委员会的方法数为 10C3 = 120。若排列单词 M A M M A 的字母，不同排列数为 5! / (3! × 2!) = 10，因为三个 M 和两个 A 是相同字母。
In mixed questions, multiply the number of choices for each stage if the stages are independent, and add them if the situations are mutually exclusive.
在混合题型中，若各阶段相互独立，则相乘；若各情况互斥，则相加。
5. Question 4: Discrete Random Variables and Expectation | 第4题：离散随机变量与期望
A discrete random variable question gives a probability distribution table listing each value of X and its corresponding probability. The expectation and variance are calculated from first principles, and the probabilities must sum to 1.
离散随机变量题会给出概率分布表，列出 X 的每个取值及对应概率。期望和方差由基本定义计算，且所有概率之和必须等于 1。
E(X) = Σ x p(x) and Var(X) = E(X²) − [E(X)]²
For the distribution P(X = 1) = 0.2, P(X = 2) = 0.3, P(X = 3) = 0.5, the expectation is E(X) = 1 × 0.2 + 2 × 0.3 + 3 × 0.5 = 2.3. Then E(X²) = 1² × 0.2 + 2² × 0.3 + 3² × 0.5 = 5.9, so Var(X) = 5.9 − 2.3² = 0.61.
对于分布 P(X = 1) = 0.2、P(X = 2) = 0.3、P(X = 3) = 0.5，期望为 E(X) = 1 × 0.2 + 2 × 0.3 + 3 × 0.5 = 2.3。进一步 E(X²) = 1² × 0.2 + 2² × 0.3 + 3² × 0.5 = 5.9，所以 Var(X) = 5.9 − 2.3² = 0.61。
The variance formula is quicker than using Σ(x − μ)² p(x), but it is sensitive to rounding in E(X²), so keep full calculator accuracy throughout the working.
方差公式比使用 Σ(x − μ)² p(x) 更快，但对 E(X²) 的四舍五入敏感，因此计算过程中应保留计算器的完整精度。
6. Question 5: Binomial Distribution | 第5题：二项分布
If a random variable X follows a binomial distribution, written X ~ B(n, p), the probability of exactly r successes is found using the binomial probability formula. For inequalities, sum the relevant probabilities or use cumulative binomial tables where allowed.
若随机变量 X 服从二项分布，记作 X ~ B(n, p)，则恰好 r 次成功的概率由二项概率公式给出。对于不等式，将相关概率相加，或使用允许的累积二项分布表。
P(X = r) = C(n, r) pʳ (1 − p)ⁿ⁻ʳ
For X ~ B(10, 0.3), the probability P(X = 2) = C(10, 2) × 0.3² × 0.7⁸ ≈ 45 × 0.09 × 0.057648 = 0.2335. To find P(X ≤ 2), calculate P(X = 0) + P(X = 1) + P(X = 2) because these events are mutually exclusive.
对于 X ~ B(10, 0.3)，P(X = 2) = C(10, 2) × 0.3² × 0.7⁸ ≈ 45 × 0.09 × 0.057648 = 0.2335。求 P(X ≤ 2) 时，需计算 P(X = 0) + P(X = 1) + P(X = 2)，因为这些事件互斥。
Candidates often confuse P(X = r) with P(X ≤ r) and lose accuracy by rounding intermediate values. Write the full expression before using the calculator to show clear method marks.
考生常将 P(X = r) 与 P(X ≤ r) 混淆，并因中间值四舍五入而失分。应先写出完整表达式，再用计算器求值，以展示清晰的方法分。
7. Question 6: Normal Distribution and Continuity Correction | 第6题：正态分布与连续性校正
When X follows a normal distribution with mean μ and variance σ², convert to the standard normal variable Z to find probabilities. If a binomial distribution is approximated by a normal distribution, apply a continuity correction by adjusting the boundary by 0.5.
当 X 服从均值为 μ、方差为 σ² 的正态分布时，需转换为标准正态变量 Z 来求概率。若用正态分布近似二项分布，需进行连续性校正，将边界值调整 0.5。
Z = (X − μ) / σ
For X ~ N(50, 4²), the probability P(X > 55) is found by Z = (55 − 50) / 4 = 1.25. Then P(Z > 1.25) = 1 − P(Z < 1.25) = 1 − 0.8944 = 0.1056.
对于 X ~ N(50, 4²)，概率 P(X > 55) 通过 Z = (55 − 50) / 4 = 1.25 求得。因此 P(Z > 1.25) = 1 − P(Z < 1.25) = 1 − 0.8944 = 0.1056。
For a binomial approximation, suppose X ~ B(100, 0.5) and we approximate P(X ≤ 48). The normal approximation uses the boundary 48.5, giving Z = (48.5 − 50) / 5 = −0.3. Without the correction, Z = (48 − 50) / 5 = −0.4 and the probability would be understated.
对于二项分布近似，假设 X ~ B(100, 0.5)，要近似计算 P(X ≤ 48)。正态近似使用边界值 48.5，得到 Z = (48.5 − 50) / 5 = −0.3。若不校正，Z = (48 − 50) / 5 = −0.4，概率会被低估。
8. Question 7: Sampling and Estimation | 第7题：抽样与估计
The sample mean x̄ is an unbiased estimator of the population mean μ. If the population standard deviation σ is known, the sampling distribution of the mean has standard error σ / √n, and a confidence interval is constructed around the sample mean.
样本均值 x̄ 是总体均值
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