📚 A-Level Maths Unit 5 (Jan 2021) Mark Scheme: Common Mistakes Summary | A-Level 数学 Unit 5 2021年1月评分方案易错点总结
Analysing the mark scheme from the January 2021 A-Level Mathematics Unit 5 paper reveals consistent patterns of errors that prevented many candidates from securing top marks. Whether you are preparing for a resit or aiming to refine your exam technique, understanding these pitfalls will help you anticipate common traps and apply mark-scheme-friendly reasoning. This article breaks down the most frequent mistakes, explains why they occur, and shows you how to avoid them in future assessments.
分析 2021 年 1 月 A-Level 数学第五单元的评分方案可以发现，有许多反复出现的错误模式导致大量考生未能获得高分。无论你是在准备重考还是想优化自己的考试技巧，理解这些易错点都能帮助你预判常见陷阱，并运用符合评分方案要求的推理。本文将逐一剖析最常见的错误，解释其产生原因，并展示如何在今后的考试中避免它们。
1. Misapplying Conditional Probability in Tree Diagrams | 树状图中条件概率的错误应用
Many students lost marks on the probability question because they treated probabilities on the second set of branches as unconditional. The mark scheme emphasised that probabilities after the first event are always conditional on the preceding outcome. A typical mistake was writing P(second event) directly from a branch without referring to the intersection calculation, e.g. confusing P(A ∩ B) with the label on the branch P(B|A).
许多学生在概率题上失分，是因为他们把第二组分支上的概率当作无条件概率。评分方案强调，第一次事件之后的分支概率始终以前一结果的发生为条件。一个典型的错误是直接从分支上读取 P(第二个事件) 而不涉及交集的运算，例如混淆 P(A ∩ B) 与分支上标注的 P(B|A)。
Always identify the condition: a branch label after the first outcome is of the form P(second event | first event).
始终要识别条件：第一个结果之后的分支标签的形式是 P(第二个事件 | 第一个事件)。
Use the multiplication rule correctly: P(A ∩ B) = P(A) × P(B|A), not P(A) × P(B) unless independence is given or proven.
正确使用乘法规则：P(A ∩ B) = P(A) × P(B|A)，而非 P(A) × P(B)，除非题目已经给出或证明了独立性。
When drawing your own tree, label probabilities clearly and include the conditional symbol to avoid confusion.
自己绘制树状图时，要清楚地标注概率并使用条件符号，以免混淆。
2. Incorrect Use of the Normal Distribution Table: Wrong Tail Reading | 正态分布表使用错误：读取了错误的尾端
A recurrent error in the January 2021 paper was reading the standard normal table backwards. Candidates often looked up a probability in the body of the table and reported the corresponding z-value without adjusting for whether the required probability was in the lower tail or upper tail. The mark scheme deducted marks when a positive z was given where a negative one was needed, or when the complement was not taken.
2021 年 1 月试卷中反复出现的一个错误是反向读取标准正态分布表。考生经常在表体中查找概率并报告对应的 z 值，却没有根据所需概率处于下尾还是上尾进行调整。当题目需要负 z 值而考生给出正 z 值，或者没有取补概率时，评分方案均会扣分。
Sketch a normal curve and shade the required region before using the table – this small step reduces tail errors dramatically.
使用表格前先画出一条正态曲线并涂上所求区域——这个小步骤能显著减少尾端错误。
Remember: if P(Z < z) = p and p < 0.5, then z is negative. Use symmetry: z(P) = –z(1 – P).
记住：如果 P(Z < z) = p 且 p < 0.5，则 z 为负数。利用对称性：z(P) = –z(1 – P)。
For 'greater than' probabilities, do P(Z > a) = 1 – Φ(a); don't read Φ(a) directly as the tail probability.
对于“大于”概率，用 P(Z > a) = 1 – Φ(a)；不要直接把 Φ(a) 当作尾概率读取。
3. Omitting the Continuity Correction in Normal Approximation to Binomial | 二项分布正态近似中遗漏连续性修正
curacy penalty. Many candidates simply standardised the given value without adjusting the boundary. For example, when approximating P(X ≤ 15) with a binomial variable, they used P(Z < (15 – μ)/σ) instead of P(Z < (15.5 – μ)/σ).
评分方案对近似计算的题目清楚表明，不使用连续性修正而直接套用正态分布会导致精确度扣分。许多考生只是对给定的数值进行标准化，却未调整边界。例如，对二项变量近似计算 P(X ≤ 15) 时，他们用了 P(Z < (15 – μ)/σ) 而非 P(Z < (15.5 – μ)/σ)。
Adjust discrete boundaries: P(X ≤ n) → P(X < n + 0.5); P(X ≥ n) → P(X > n – 0.5).
调整离散边界：P(X ≤ n) → P(X < n + 0.5)，P(X ≥ n) → P(X > n – 0.5)。
For P(X < n), treat it as P(X ≤ n – 1), then apply the corresponding continuity correction.
对于 P(X < n)，将其视作 P(X ≤ n – 1)，然后再应用相应的连续性修正。
The mark scheme often awards a method mark for stating 'continuity correction' even if the final answer is slightly off – always show it.
即使最终答案稍有偏差，评分方案通常也会给提及“连续性修正”的方法分——所以务必展示出来。
4. Confusing Variance and Standard Deviation in Statistical Calculations | 统计计算中混淆方差与标准差
Several questions in Unit 5 required candidates to compute variance, standard deviation, or the standard error of the mean. A damaging mistake was entering the standard deviation into a formula that needed the variance, or vice versa. For instance, when calculating a confidence interval for the mean, some used σ instead of σ² when dividing by n, which gave a completely wrong interval width.
第五单元有多道题目要求计算方差、标准差或均值的标准误。一个杀伤力很大的错误是把标准差代入需要方差的公式，或反之。比如在计算均值的置信区间时，有些人用 σ 而非 σ² 除以 n，导致区间宽度完全错误。
Standard error of the mean is σ/√n, not σ²/√n and not σ/n. Write down definitions before substituting.
均值的标准误是 σ/√n，不是 σ²/√n，也不是 σ/n。代入前先把定义写下来。
When given a variance as a number, identify it clearly: 'σ² = 16, so σ = 4'. This prevents accidental double-rooting or squaring.
当题目给出一个方差数值时，要明确识别：“σ² = 16，故 σ = 4”。这样可以避免无意中再次开方或平方。
Check units: variance has squared units of the original data, standard deviation does not – use this to sense-check your intermediate values.
检查单位：方差的单位是原始数据单位的平方，标准差则不是——利用这一点检验中间值的合理性。
5. Hypothesis Testing Pitfalls: Misinterpreting the p-value and Significance | 假设检验误区：误读p值与显著性
The January 2021 paper included a hypothesis test where many candidates miscompared the p-value with the significance level. A typical false conclusion stated 'accept H₀' when the p-value was greater than α, or incorrectly declared a significant result when the test statistic barely exceeded the critical value. The mark scheme insisted on 'do not reject H₀' rather than 'accept H₀', and required a conclusion in context.
2021年1月试卷中有一道假设检验题，许多考生错误地将p值与显著性水平进行比较。常见的错误结论是当 p 值大于 α 时声称“接受 H₀”，或者当检验统计量刚刚超过临界值时错误地宣称结果显著。评分方案坚持使用“不拒绝 H₀”而非“接受 H₀”，并要求在具体情境下得出结论。
If p-value > significance level, the conclusion is 'There is insufficient evidence to reject H₀', not 'H₀ is true'.
若 p 值 > 显著性水平，结论应为“没有足够证据拒绝 H₀”，而不是“H₀ 为真”。
Always relate the conclusion back to the original problem: 'Therefore the mean weight has not significantly changed' rather than just 'Do not reject H₀'.
始终将结论联系回原题情境：“因此，平均重量未发生显著变化”，而不只是“不拒绝 H₀”。
In a two-tailed test, double the probability when using tables unless the critical region method is defined symmetrically.
在双侧检验中，使用表格时概率要加倍，除非已对称地定义了临界域。
6. Mismanaging Tied Ranks in Spearman's Rank Correlation | 斯皮尔曼等级相关中相等等级的处理不当
When calculating Spearman' the correction formula. The mark scheme penalised those who used the simplified formula rₛ = 1 – 6Σd²/[n(n² – 1)] without checking for ties, or who assigned ranks inconsistently.
在计算斯皮尔曼等级相关系数时，许多考生虽然正确分配了相等等级，但却忽略了存在相同等级而不使用修正公式。评分方案对那些未检查是否有相同等级就直接使用简化公式 rₛ = 1 – 6Σd²/[n(n² – 1)] 的考卷，或是等级分配不一致的考卷进行了扣分。
Always look for repeated values and choose the mid-rank: e.g. two values tied for 4th and 5th both get rank 4.5.
始终检查重复数值并取中间等级：例如两个数值并列第四和第五，则均给等级 4.5。
If ties exist, the mark scheme expects you to either use the full Pearson formula on ranks or apply the tie-correction to Spearman's formula.
若存在相同等级，评分方案要求你对等级使用完全的皮尔逊公式，或对斯皮尔曼公式进行相同等级修正。
Show the table of values, differences d, and d² clearly; many method marks are awarded for layout even if arithmetic slips occur.
清晰展示数值、差值 d 和 d² 的表格；即使计算有误，清晰的排版也常能获得方法分。
7. Errors in the Product Moment Correlation Coefficient Formula | 积矩相关系数公式使用错误
ently misused the sum notation, writing Σxy incorrectly as Σx × Σy, or forgetting to square the sums in the denominator. The mark scheme expected the correct substitution into the given formula without algebraic shortcuts.
2021年1月试卷中考积矩相关系数（PMCC）的题目暴露了公式运用的错误。考生常将 Σxy 错误地写成 Σx × Σy，或在分母中忘记将求和结果进行平方。评分方案要求正确代入给定的公式，不得偷工减料地做代数简化。
Recall the PMCC formula: r = [nΣxy – (Σx)(Σy)] / √[ (nΣx² – (Σx)²)(nΣy² – (Σy)²) ]. Copy it carefully from the formula booklet.
记住 PMCC 公式：r = [nΣxy – (Σx)(Σy)] / √[ (nΣx² – (Σx)²)(nΣy² – (Σy)²) ]。仔细从公式册中抄写。
Don't cancel Σx² and (Σx)² – they are fundamentally different quantities. Calculate each sum term separately and show substitutions.
不要消去 Σx² 和 (Σx)²——它们是完全不同的量。分别计算每一个求和项并展示代入过程。
Even a minor slip in one sum can lead to a correlation outside [-1, 1]; use this as a sanity check.
哪怕一个求和项出现微小失误，也会导致相关系数超出 [-1, 1]；可利用这一点进行合理性检查。
8. Problems with Histograms: Area Misinterpretation and Frequency Density | 直方图问题：面积误读与频率密度
that frequency is proportional to area. The mark scheme required candidates to use frequency density = frequency / class width, and to compute totals or probabilities from rectangular areas, not just heights.
涉及直方图的数据呈现题目让很多学生感到困难，因为他们把条形高度与频率混淆，而忽略了频率与面积成正比。评分方案要求考生使用 频率密度 = 频率 / 组距，并通过矩形面积而非仅仅通过高度来计算总数或概率。
stogram, frequency = frequency density × class width. Always compute the area of each bar.
在直方图中，频率 = 频率密度 × 组距。务必计算每个条形的面积。
When a question asks for a proportion or probability, divide the relevant area by the total area of the histogram.
当题目要求计算比例或概率时，用相关面积除以直方图的总面积。
Check the vertical axis label: if it says 'Frequency density', a bar of height 4 in a class of width 5 represents a frequency of 20, not 4.
检查纵轴标签：如果写的是“频率密度”，那么宽度为5的组距中高度为4的条形对应的频率是20，而非4。
9. Misapplication of Independent Events Formula | 独立事件公式的错误应用
The probability section revealed that many learners misused the test for independence. Instead of checking whether P(A ∩ B) equals P(A)P(B), they assumed independence based on the word 'random' or because events 'looked' unrelated. The mark scheme accepted only numerical verification or a statement that events were not independent when the equality failed.
概率部分揭示出许多学习者错误地使用了独立性检验。他们没有检查 P(A ∩ B) 是否等于 P(A)P(B)，而是根据“随机”一词或事件“看起来”不相关就假设独立性成立。评分方案只接受通过数值验证，或当等式不成立时声明事件不独立的做法。
To demonstrate independence, compute P(A) × P(B) and compare with the given P(A ∩ B). If equal, they are independent; otherwise, they are not.
要证明独立性，计算 P(A) × P(B) 并与给出的 P(A ∩ B) 相比较。若相等，则独立；否则不独立。
Do not write 'independent because the events don't affect each other' unless you can support it with a numerical argument – the mark scheme typically rejects such reasoning.
不要写“独立是因为事件互不影响”，除非你能用数值论据支持——评分方案通常会拒绝此类推理。
When constructing tree diagrams for independent trials, branch probabilities remain unchanged; clearly state that independence justifies this.
为独立试验绘制树状图时，分支概率保持不变；要明确指出正是独立性保证了这一点。
10. Discrete Random Variables: Forgetting the Sum of Probabilities Must Equal 1 | 离散随机变量：忽略概率之和必须为1
ep ΣP(X = x) = 1, or incorrectly solved the resulting equation, then used the wrong value of k in later parts for expectation and variance. The mark scheme awarded method marks for setting up the equation and follow-through marks if the error was consistent.
一道关于离散随机变量的标准题目要求考生求出概率分布中的未知常数 k。很多人遗漏了 ΣP(X = x) = 1 的步骤，或是解方程出错，然后在后续的期望值和方差部分继续使用错误的 k 值。评分方案对建立方程给予方法分，如果错误具有一致性，还可获得后续继续给予的后续分。
Always write 'sum of probabilities = 1' and form an equation. Solve it carefully – basic algebra errors here cost many marks.
始终写下“概率之和 = 1”并建立方程。仔细求解——此处的基础代数错误会损失大量分数。
Once k is found, verify by recalculating the sum; if it does not equal 1, check your working.
求出 k 后，通过重新计算总和进行验证；假如和不等于1，检查运算过程。
(X) = Σ[x·P(X=x)], use the corrected k. If you know your k is wrong, state it and continue – the mark scheme often allows error carried forward.
计算期望值 E(X) = Σ[x·P(X=x)] 时，使用修正后的 k。若你知道自己的 k 是错的，请说明并继续——评分方案通常允许错误携带。
11. Stem-and-Leaf Diagrams: Key and Ordering Errors | 茎叶图：图例与排序错误
Another common source of lost marks was the stem-and-leaf diagram. Candidates sometimes omitted the key, used inconsistent leaf units, or failed to order the leaves. The mark scheme explicitly required a valid key showing place value and leaves in ascending order for full marks.
另一个常见的失分点是茎叶图。考生有时遗漏图例、使用了不一致的叶单位，或未将叶按顺序排列。评分方案明确要求提供有效的说明位值的图例并按升序排列叶片才能获得满分。
State a clear key, e.g. '3 | 5 means 3.5 cm' or '12 | 4 represents 124', depending on the data.
写明清晰的图例，例如根据数据写成“3 | 5 表示 3.5 厘米”或“12 | 4 代表 124”。
After writing the stem, list leaves in increasing order. An unordered diagram loses accuracy marks.
写下茎后，将叶片按升序排列。未排序的图会损失准确度分。
If a back-to-back stem plot is used, place leaves for one group on the left and the other on the right, maintaining the order from the stem outward.
若使用背靠背茎叶图，将一组叶片放在左侧，另一组放在右侧，并保持从茎向外的顺序。
12. Reading and Interpreting Box Plots: Quartile Confusion | 箱形图的读取与解读：四分位数混淆
entified outliers. The mark scheme required correct identification of the minimum, lower quartile, median, upper quartile, and maximum, as well as the interquartile range (IQR) and the rule outlier < Q₁ – 1.5×IQR or > Q₃ + 1.5×IQR.
箱形图出现在数据比较部分，有几名考生将箱体的边界与四分位数 Q₁、Q₂、Q₃ 混淆，或未能正确识别异常值。评分方案要求正确识别最小值、下四分位数、中位数、上四分位数和最大值，以及四分位距（IQR）和异常值判定规则：小于 Q₁ – 1.5×IQR 或大于 Q₃ + 1.5×IQR。
Label the box plot with the five-number summary: min, Q₁, median, Q₃, max. Whiskers extend to the last data point within the fences.
用五数概括法标注箱形图：最小值、Q₁、中位数、Q₃、最大值。须线延伸至围栏内的最后一个数据点。
s beyond the whiskers. Mention the 1.5×IQR criterion in your working to gain method marks.
异常值作为须线之外的单独点绘出。在解答过程中提及 1.5×IQR 准则可获方法分。
When comparing two distributions, use the box plots to comment on central location (median) and spread (IQR, range) – do not speculate about shape beyond obvious skewness.
比较两个分布时，利用箱形图评论中心位置（中位数）和离散程度（IQR、极差）——在明显偏态之外不要过度推测分布形状。
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