📚 A-Level Maths Unit 5 Mark Scheme Jan22 Question Type Analysis | A-Level数学 Unit 5 2022年1月评分方案题型解析
This article breaks down the question types that appeared in the January 2022 A-Level Mathematics Unit 5 (Probability & Statistics 1) paper and analyses how the official mark scheme allocated marks for method, accuracy, and interpretation. Understanding these patterns helps you focus on what examiners truly reward and avoid losing marks on technicalities.
本文拆解了2022年1月A-Level数学 Unit 5（概率与统计1）试卷中出现的题型，并分析了官方评分方案如何对方法、准确性和解释进行给分。理解这些模式能帮助你聚焦考官真正给分的点，避免因技术细节丢分。
1. Measures of Central Tendency and Spread | 集中趋势与离散程度的度量
In grouped frequency tasks, the mark scheme always awards a method mark (M1) for using class midpoints to estimate the mean, and an accuracy mark (A1) for the correct final value. A common trap is rounding intermediate midpoints too early, which often leads to accuracy loss.
在分组频数任务中，评分方案总是为使用组中值估计均值给予方法分（M1），并为正确的最终值给予准确分（A1）。一个常见陷阱是过早对组中值进行取整，这通常会导致精确度丢失。
For standard deviation, the formula √(Σfx²/Σf − (Σfx/Σf)²) must be clearly substituted. The mark scheme checks whether you have shown the correct substitution before you reach the final answer, so always write the numerical working steps.
对于标准差，必须明确代入公式 √(Σfx²/Σf − (Σfx/Σf)²)。评分方案会检查你是否在得出最终答案前展示了正确代入过程，因此务必写出数值运算步骤。
2. Probability and Venn Diagrams | 概率与韦恩图
Venn diagram questions often carry marks for correctly placing given probabilities into regions and for deducing missing values using the addition rule. The January 2022 mark scheme shows that a clear diagram with all intersections labelled is rewarded as B1, even if the final numerical answer is slightly off due to a copying error.
韦恩图题目通常因正确将给定概率放入区域，并利用加法法则推导出缺失值而得分。2022年1月的评分方案显示，只要图示清晰且所有交集都标出，即使最终数值因誊写错误略有偏差，也能得到B1分。
A typical two‑mark part asks for P(A ∩ B’): one mark is for identifying the correct region, the other for giving its probability. Never just write the decimal; explain which region you are using, e.g., “A only = 0.35”.
典型的两分小题要求 P(A ∩ B’)：一分用于识别正确区域，另一分用于给出其概率。绝不能只写小数；要说明你使用的是哪个区域，例如 “A only = 0.35”。
3. Discrete Random Variables and Probability Distributions | 离散随机变量与概率分布
When constructing a probability distribution table, the mark scheme ensures the sum of probabilities equals 1 before any further marks are awarded. If you are given an unknown k, the equation Σ P(X=x) = 1 is the essential starting point, and the scheme awards M1 for forming this equation correctly.
构建概率分布表时，评分方案确保概率总和为1后再给后续分。如果给出了未知数 k，方程 Σ P(X=x) = 1 是必要起点，方案对正确列出该方程给予M1分。
For E(X) and Var(X), you must show the use of the formulas Σ x·p and Σ x²·p − (E(X))². The mark scheme permits using either the definition formula or the computational shortcut, but a correct substitution must be visible. Units are not required for E(X) unless the variable is defined with a unit.
对于 E(X) 和 Var(X)，你必须展示使用公式 Σ x·p 和 Σ x²·p − (E(X))²。评分方案允许使用定义公式或计算捷径，但正确的代入过程必须可见。除非变量定义带有单位，否则 E(X) 不需要单位。
4. Permutations and Combinations | 排列与组合
This topic regularly appears with constraints such as “particular items must be together” or “not at the ends”. The mark scheme breaks the solution into logical steps: treating a group as a single block, then permuting blocks and internal items separately. Each step earns a method mark.
这一主题经常伴随“特定物品必须相邻”或“不能放在两端”等限制条件出现。评分方案将解法分解为逻辑步骤：先把一组视为一个整体，然后分别排列整体和内部元素。每一步都可获得方法分。
In combination problems, the phrase “at least one” is a signal to use the complement method. The January 2022 mark scheme explicitly awards B1 for stating the total number of selections without restriction, then another B1 for the complement calculation, making it an easy mark if you remember this strategy.
在组合问题中，“至少一个”这个短语是使用补集法的信号。2022年1月的评分方案明确对写出无限制选择总数给一个B1分，再对补集计算给一个B1分，只要记住这一策略，这类分数很容易得到。
5. Binomial Distribution | 二项分布
Binomial questions test both exact probabilities and cumulative probabilities. The mark scheme expects the notation X ~ B(n, p) to be clearly stated, and for P(X = r) the explicit formula (ⁿCᵣ)·pʳ·(1−p)ⁿ⁻ʳ is often awarded an M1 when substituted numerically. Calculators are allowed, but the working must indicate the distribution used.
二项分布题目既考查精确概率也考查累积概率。评分方案期望清楚写出记号 X ~ B(n, p)，对于 P(X = r)，当数值代入明确的公式 (ⁿCᵣ)·pʳ·(1−p)ⁿ⁻ʳ 时，通常会得到M1分。允许使用计算器，但解题过程必须指出所使用的分布。
For inequalities like P(X ≥ 4), the scheme rewards converting to 1 − P(X ≤ 3) and then using tables or a calculator function. If you simply write the answer with no explanation, you risk losing the method mark even if the number is correct.
对于 P(X ≥ 4) 这样的不等式，方案奖励转化为 1 − P(X ≤ 3) 并使用表格或计算器函数。如果只是写出答案而无解释，即使数值正确也可能丢失方法分。
6. Normal Distribution and Inverse Normal | 正态分布与逆正态
The standardisation formula z = (x − μ)/σ must be applied whenever you are working with an arbitrary normal variable. The mark scheme allocates M1 for the substitution and A1 for the resulting z-value. If the question gives the probability and asks for an unknown mean or standard deviation, you must set up an equation with the inverse z-value.
处理任意正态变量时必须使用标准化公式 z = (x − μ)/σ。评分方案为代入过程分配M1，为得出的z值分配A1。如果题目给出概率并要求求未知均值或标准差，你必须利用逆z值建立方程。
Sketching a bell curve and shading the relevant region is recommended even if not directly marked, because it helps avoid the common error of using the wrong tail probability. The January 2022 paper required identifying symmetric regions for a given confidence level, where the mark scheme awards B1 for noting that the interval is μ ± z·σ/√n.
推荐绘制钟形曲线并给相关区域涂阴影，即使不直接给分，因为这有助于避免使用错误尾部概率的常见错误。2022年1月的试卷要求识别给定置信水平的对称区域，评分方案对指出区间为 μ ± z·σ/√n 给予B1分。
7. Data Representation: Histograms and Cumulative Frequency | 数据表示：直方图与累积频数
Histogram tasks demand careful calculation of frequency density = frequency ÷ class width. The mark scheme often gives B1 for correctly scaling the vertical axis and plotting bars, and another B1 for labelling axes. Missing labels can cost you marks even when the bars are perfect.
直方图任务要求仔细计算频率密度 = 频数 ÷ 组距。评分方案通常对正确设置纵轴刻度并绘制条形图给予B1，再对坐标轴标注给予另一个B1分。即使条形图完美，缺少标注也可能丢分。
For cumulative frequency curves, the scheme expects points plotted at upper class boundaries, joined by a smooth curve. When estimating medians or quartiles, draw clear horizontal and vertical lines on the graph—these construction lines are a source of M1 marks and must not be omitted.
对于累积频数曲线，方案期望将点绘制在组上限处，并用光滑曲线连接。在估计中位数或四分位数时，要在图上画出清晰的水平和垂直线——这些构造线是M1得分的依据，绝不能省略。
8. Conditional Probability and Tree Diagrams | 条件概率与树状图
Tree diagrams with two or three stages appear regularly. The mark scheme gives M1 for placing the correct complementary probabilities on the branches and for multiplying along a path. Always write the multiplication step out explicitly, e.g., 0.6 × 0.3 = 0.18, rather than just the final decimal.
两个或三个阶段树状图经常出现。评分方案对在树枝上放置正确的补全概率，以及沿路径相乘给予M1。务必明确写出乘法步骤，例如 0.6 × 0.3 = 0.18，而不仅仅是最终小数。
Conditional probability questions from a two‑way table require the formula P(A|B) = P(A ∩ B) / P(B). The mark scheme awards B1 for identifying the correct intersection and another B1 for the division. If the denominator is taken from a marginal total, make sure it is the total for the conditioned event.
来自双向表的条件概率题需要公式 P(A|B) = P(A ∩ B) / P(B)。评分方案对识别正确的交集给一个B1，对除法给另一个B1。如果分母取自边际总和，要确保它是所条件事件的总和。
9. Linear Combinations of Independent Random Variables | 独立随机变量的线性组合
This topic tests the rules E(aX ± bY) = aE(X) ± bE(Y) and Var(aX ± bY) = a²Var(X) + b²Var(Y) assuming independence. The mark scheme is strict about showing the variance formula with squared coefficients; missing the square is a common blunder that loses the M1 instantly.
这一主题考查 E(aX ± bY) = aE(X) ± bE(Y) 和 Var(aX ± bY) = a²Var(X) + b²Var(Y) 的规则（假设独立）。评分方案严格要求展示带平方系数的方差公式；漏掉平方是常见的失分错误，会立即失去M1。
When the result is a normal distribution, you must also state the distribution of the combination, e.g., W ~ N(μ_W, σ_W²). The mark scheme includes an A1 for this step, so always conclude with a clear distribution statement.
当结果为正态分布时，还必须写出该组合的分布，例如 W ~ N(μ_W, σ_W²)。评分方案为这一步包含A1分，因此务必以清晰的分布陈述作为结论。
10. Hypothesis Testing for a Population Mean | 总体均值的假设检验
A structured approach is essential: state hypotheses H₀ and H₁ using μ, choose the test statistic, calculate the standardised value or the p‑value, and compare with the critical value or significance level. The mark scheme awards B1 for both hypotheses, M1 for the standardisation, and A1 for a correct conclusion in context.
结构化解题至关重要：用 μ 陈述假设 H₀ 和 H₁，选择检验统计量，计算标准化值或 p 值，并与临界值或显著性水平比较。评分方案对两个假设给B1，对标准化过程给M1，对正确的情境结论给A1。
The conclusion must be written in non‑technical language, e.g., “There is sufficient evidence to reject H₀, suggesting that the mean has changed.” A purely numerical statement like “Reject H₀” without context loses the final A1 in many mark schemes, including January 2022.
结论必须用非技术性语言表述，例如 “有足够证据拒绝 H₀，表明均值已发生变化”。诸如 “Reject H₀” 这种纯粹数值陈述而不结合情境，在许多评分方案（包括2022年1月）中会丢掉最后的A1分。
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