📚 AS Mathematics Unit 5 June 2019 Question Paper Breakdown | AS数学单元5 2019年6月试卷题型解析
The June 2019 AS Mathematics Unit 5 paper - commonly Statistics 1 (WST01) for the Edexcel International A Level specification - is still one of the most efficient papers to practise, because it samples almost every statistical skill in the specification inside a single 90-minute sitting. In this breakdown we work through the question types in and provide model approaches to help you master the order they normally appear, rebuild the standard methods, show the algebra in full, and flag the exact places where candidates lose marks. All notation is kept in plain, exam-ready form so you can copy the working straight into your own answers.
2019年6月AS数学单元5试卷（通常为爱德思IAL统计学1，代码WST01）至今仍是效率最高的练习卷之一，因为它在短短90分钟内几乎覆盖了考纲中的全部统计技能。本文将按题目通常出现的顺序，逐一梳理题型、重建标准解法、完整展示运算过程，并指出考生最容易失分的具体位置。全文采用与考场一致的规范记号，方便你直接照搬到自己的答卷中。
1. Paper Structure and Specification Mapping | 试卷结构与考纲对应
Unit 5 is a written paper of 1 hour 30 minutes carrying 75 marks, split roughly into a short opening question on probability, two or three mid-length questions on data representation and correlation, and one or two longer questions on discrete random variables and the normal distribution. The June 2019 paper follows that pattern closely, which makes it an ideal diagnostic: if you score below 60% on it, the weakness is almost always in the distribution questions rather than in the opening probability question.
单元5为笔试，时长1小时30分钟，满分75分。结构大致为：开篇一道简短的概率题，中间两到三道关于数据表示与相关性的題目，最后是一到两道关于离散随机变量与正态分布的长题。2019年6月的试卷高度贴合这一模式，因此非常适合做诊断测试：如果你的得分低于60%，薄弱点几乎总是分布在"分布"类题目上，而不是开头那道概率题。
A useful revision habit is to tag every question in the paper against the specification bullet points before you attempt it. That way you learn the paper's architecture, not just its content.
一个有效的复习习惯是：在做题之前，先把试卷中每一题对应到考纲的具体条目上。这样你熟悉的是整份试卷的"架构"，而不仅仅是题目内容。
| Topic area | 主题领域 | Typical marks | 常见分值 | Skill tested | 考查能力 |
| Probability and Venn diagrams | 概率与维恩图 | 5-8 | Addition rule, conditional probability | 加法公式、条件概率 |
| Tree diagrams | 树状图 | 6-9 | Multiplication rule, complement | 乘法法则、补事件 |
| Discrete random variables | 离散随机变量 | 7-10 | E(X), Var(X), linear coding | 期望、方差、线性变换 |
| Normal distribution | 正态分布 | 10-14 | Standardising, inverse normal | 标准化、逆正态 |
| Correlation and regression | 相关与回归 | 8-12 | PMCC, least squares line | 相关系数、最小二乘直线 |
| Representation of data | 数据表示 | 6-10 | Histograms, box plots, outliers | 直方图、箱线图、异常值 |
2. Probability and Venn Diagrams | 概率与维恩图
ou to complete a Venn diagram. A typical set of values is P(A) = 0.35, P(B) = 0.45 and P(A ∪ on those three numbers, so getting this first step right is worth several marks in one go.
2019年6月试卷的开门题通常给出事件A与B的概率，并要求完成维恩图。一组典型数值为 P(A) = 0.35，P(B) = 0.45，P(A ∪ B) = 0.6。后续所有小问都建立在这三个数字之上，因此把第一步做对，一次性就能锁定好几分。
Using the addition rule, you find P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.35 + 0.45 − 0.6 = 0.2.
利用加法公式，可得 P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.35 + 0.45 − 0.6 = 0.2。
45 − 0.2 = 0.25, and the outside region as 1 − 0.6 = 0.4. A typical follow-up asks for P(A ∪ B′) or a conditional probability such as P(A | B).
随后在交集区填 0.2，A 的剩余部分为 0.35 − 0.2 = 0.15，B 的剩余部分为 0.45 − 0.2 = 0.25，外部区域为 1 − 0.6 = 0.4。常见的后续问题是求 P(A ∪ B′) 或条件概率如 P(A | B)。
Remember the conditional probability formula. It is the single most examined relationship in this section.
请记住条件概率公式——它是本节中被考查频率最高的一个关系式。
P(A | B) = P(A ∩ B) ÷ P(B)
In this case, P(A | B) = 0.2 ÷ 0.45 = 4/9 ≈ 0.444.
在此例中，P(A | B) = 0.2 ÷ 0.45 = 4/9 ≈ 0.444。
For P(A ∪ B′), take the whole of A (0.35) plus the part of B′ that lies outside A (0.4), giving 0.75. Drawing the diagram before you calculate keeps this kind of question almost error-proof.
求 P(A ∪ B′) 时，取 A 的全部（0.35）再加上 B′ 中位于 A 之外的部分（0.4），得 0.75。先画图再计算，这类题目几乎不会出错。
Always check whether events are independent by verifying P(A ∩ 75, which is not equal to 0.2, so A and B are not independent. If the values had matched, the events would be independent and the conditional probability would simply collapse to P(A).
务必检验事件是否独立，验证 P(A ∩ B) 是否等于 P(A) × P(B)。此处 0.35 × 0.45 = 0.1575，不等于 0.2，因此 A 与 B 不独立。若两者相等，则事件独立，条件概率直接退化为 P(A)。
Two further variants are worth rehearsing: mutually exclusive events, where P(A ∩ B) = 0 and therefore P(A ∪ B) = P(A) + P(B); and the use of the complement to find P(A′ ∩ B′), which equals 1 − P(A ∪ B).
还有两种变体值得演练：互斥事件，此时 P(A ∩ B) = 0，故 P(A ∪ B) = P(A) + P(B)；以及用补事件求 P(A′ ∩ B′)，它等于 1 − P(A ∪ B)。
- Draw the Venn diagram first and label the four regions. | 先画维恩图并标出四个区域。
- Write down the addition rule before substituting numbers. | 代入数值前先写出加法公式。
- State the independence test explicitly, not just the conclusion. | 明确写出独立性检验过程，而不只给结论。
3. Tree Diagrams and Conditional Probability | 树状图与条件概率
Whether it appears as Question 2 or as part of a longer question, the tree diagram is a permanent fixture of Unit 5. The June 2019 paper is likely to set a two-stage experiment, for example drawing counters from a bag without replacement, or selecting two students from a group and asking about their preferences.
无论是作为第2题还是大题的一部分，树状图都是单元5的常客。2019年6月试卷很可能设置两阶段试验，例如不放回地从袋中抽取筹码，或从一组学生中抽取两人并询问其偏好。
Label every branch with its probability, and make sure the probabilities on the second set of branches sum to 1 for each starting branch. Use the multiplication rule along a branch and the addition rule between branches.
在每条分支上标注概率，并确保对每一个起始分支而言，第二层分支的概率之和为1。沿分支使用乘法法则，不同分支之间使用加法法则。
e P(exactly one red) requires both orders: (5/8) × (3/7) + (3/8) × (5/7) = 15/56 + 15/56 = 30/56 = 15/28 ≈ 0.536.
假设袋中有5个红筹码和3个蓝筹码，不放回地抽取两个。则 P(红后红) = (5/8) × (4/7) = 20/56 = 5/14 ≈ 0.357；而"恰好一个红"需要两种顺序：(5/8) × (3/7) + (3/8) × (5/7) = 15/56 + 15/56 = 30/56 = 15/28 ≈ 0.536。
To find 'at least one' type probabilities, it is nearly always easier to use the complement:
求"至少一个"类型的概率时，使用补事件几乎总是更快：
P(at least one) = 1 − P(none)
ding the three cases RR, RB, BR - the complement route saves time and reduces arithmetic slips.
因此 P(至少一个红) = 1 − P(蓝后蓝) = 1 − (3/8) × (2/7) = 1 − 6/56 = 50/56 = 25/28 ≈ 0.893。与把 RR、RB、BR 三种情形相加相比，补事件路线更省时，也更能减少算术失误。
Conditional probability appears here too, usually as P(second draw is red | was blue) = 5/7, because one blue counter has left the bag and the total has dropped to 7.
条件概率同样会在这里出现，通常写作 P(第二次抽到红 | 第一次抽到蓝) = 5/7，因为一个蓝筹码已被取走，总数降为7。
If the counters were replaced each time, the second-layer fractions would stay at 5/8 and 3/8, and the two draws would be independent. Spotting the words 'without replacement' versus 'with replacement' is worth an easy mark.
若每次抽取后放回，第二层的分数将保持 5/8 和 3/8，两次抽取相互独立。能否识别"不放回"与"有放回"的措辞，往往就是一个白送的分。
4. Discrete Random Variables and Expectation | 离散随机变量与期望
Discrete random variable questions usually present a probability distribution in table form and then ask for the value of a constant, the mean, and the variance. The three essential properties are that every probability lies between 0 and 1, that the probabilities sum to 1, and that expectations are computed with the definition rather than by intuition.
离散随机变量题通常以表格形式给出概率分布，然后要求求某个常数、均值与方差。三个基本性质是：每个概率介于0与1之间；所有概率之和为1；期望必须按定义计算，而不能凭直觉。
E(X) = Σ x P(X = x)
Var(X) = E(X²) − [E(X)]²
9 × 0.3 = 4.9, so Var(X) = 4.9 − 2.1² = 4.9 − 4.41 = 0.49, giving a standard deviation of √0.49 = 0.7.
例如，若 X 取值为 1、2、3，对应概率 0.2、0.5、0.3，则 E(X) = 1 × 0.2 + 2 × 0.5 + 3 × 0.3 = 2.1，E(X²) = 1 × 0.2 + 4 × 0.5 + 9 × 0.3 = 4.9，故 Var(X) = 4.9 − 2.1² = 4.9 − 4.41 = 0.49，标准差为 √0.49 = 0.7。
Linear transformations of a random variable are examined almost every session. The rules are worth memorising in their general form, because they also save time in the normal distribution question.
随机变量的线性变换几乎每场考试都会考到。这两个一般形式的规则值得背熟，因为它们在正态分布题中同样能节省时间。
E(aX + b) = a E(X) + b Var(aX + b) = a² Var(X)
A very common error is to write Var(aX + b) = a Var(X) + b, which is wrong on both counts.
请特别注意：加上常数 b 会平移均值，但不改变方差；系数 a 在方差中要平方。最常见的错误是写成 Var(aX + b) = a Var(X) + b，这在两处都是错的。
n as P(X = x) = k x for x = 1, 2, 3, 4. In that case 1 + 2 + 3 + 4 = 10, so k = 1/10 = 0.1, and the rest of the question follows mechanically.
有时题目要求利用 Σ P(X = x) = 1 求未知概率，或求形如 P(X = x) = k x（x = 1, 2, 3, 4）中的常数 k。此时 1 + 2 + 3 + 4 = 10，故 k = 1/10 = 0.1，后续小问便可机械推进。
- Check that the probabilities sum to 1 before doing anything else. | 动手之前先检查概率之和是否为1。
- Lay out E(X) and E(X²) in two separate columns to avoid mixing terms. | 把 E(X) 与 E(X²) 分两列计算，避免混淆。
- Write the variance formula before substituting, so method marks are secured. | 代入前先写出方差公式，以锁定方法分。
5. Binomial Distribution and Repeated Trials | 二项分布与重复试验
The binomial distribution belongs to the same specification family as Unit 5 and shares its probability techniques, so it is worth keeping fresh. A binomial model applies when there is a fixed number of trials n, each trial has exactly two outcomes, the probability of success p is constant, and the trials are independent.
二项分布与单元5同属一个考纲体系，且共用概率技巧，因此值得保持熟练。当试验次数 n 固定、每次试验只有两种结果、成功概率 p 恒定、且各次试验相互独立时，就可以使用二项模型。
P(X = r) = C(n, r) × pʳ × (1 − p)ⁿ⁻ʳ
The binomial coefficients C(n, r) count the number of orders in which r successes can occur, which is exactly why the factor appears in front of the probability of any single ordering.
二项系数 C(n, r) 统计的是 r 次成功可能出现的位置数，这正是它出现在"任一特定顺序概率"之前的原因。
The mean and variance of a binomial variable are given directly, so no tables of values are needed:
二项分布的均值与方差可直接写出，无需逐项列表计算：
E(X) = np Var(X) = np(1 − p)
For example, if X ~ B(20, 0.3), then E(X) = 20 × 0.3 = 6 and Var(X) = 20 × 0.3 × 0.7 = 4.2, so σ = √4.2 ≈ 2.049. Examine a typical phrasing such as 'find the probability that at least 8 succeed' X ≤ 7) and is best evaluated with the cumulative distribution function on your calculator.
例如，若 X ~ B(20, 0.3)，则 E(X) = 20 × 0.3 = 6，Var(X) = 20 × 0.3 × 0.7 = 4.2，标准差 σ = √4.2 ≈ 2.049。遇到"至少8次成功"这类表述，应写作 P(X ≥ 8) = 1 − P(X ≤ 7)，用计算器的累积分布函数求解最佳。
Watch the inequality direction carefully. 'At least' means ≥, 'at most' means ≤, 'more than' means >, and 'fewer than' means <. Mixing these up costs marks even when the method is perfect.
请仔细辨别不等号方向："至少"是 ≥，"至多"是 ≤，"多于"是 >，"少于"是 <。即使方法完全正确，写错方向也会丢分。
6. The Normal Distribution | 正态分布
The normal distribution carries the heaviest mark weighting on the paper and rewards precision more than speed. Its density curve is symmetric about the mean μ, has total area 1, and is described by the standard deviation σ, which controls the spread.
正态分布在整份试卷中分值权重最高，它奖励精准而非速度。其密度曲线关于均值 μ 对称，总面积为1，由标准差 σ 描述，σ 控制分布的分散程度。
Because the normal distribution cannot be integrated in closed form at this level, every calculation is done by standardising to the standard normal variable Z, for which tables and calculator functions are available.
由于在本阶段正态分布无法用闭式积分处理，所有计算都通过标准化转换为标准正态变量 Z 来完成，而 Z 有现成的表和计算器功能。
Z = (X − μ) ÷ σ
For example, if X ~ ÷ 6 = 11/6 ≈ 1.833, so P(X > 61) = P(Z > 1.833) = 1 − Φ(1.833) ≈ 0.0334. Roughly 3.3% of the distribution lies above this value.
例如，若 X ~ N(50, 6²)，要求 P(X > 61)，则 Z = (61 − 50) ÷ 6 = 11/6 ≈ 1.833，故 P(X > 61) = P(Z > 1.833) = 1 − Φ(1.833) ≈ 0.0334，约有3.3%的分布位于该值之上。
Three standard readings appear repeatedly and should be memorised, because they let you sanity-check any answer in seconds.
三个标准结论反复出现，值得记牢，因为它们能让你在几秒内对答案做合理性检验。
| Interval | 区间 | Probability | 概率 |
| μ ± 1σ | ≈ 0.683 |
| μ ± 2σ | style="border:1px
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