📚 Year 12 Edexcel Statistics: Unit Test Mock Paper Analysis | Year 12 爱德思统计：单元测试模拟卷解析
Mock papers are an essential revision tool, and analysing them unlocks the deeper understanding required for top marks. This article walks through a representative Year 12 Edexcel Statistics unit test, dissecting typical questions across the syllabus and revealing examiner thinking.
模拟试卷是关键的复习工具，而深入解析它能帮助理解得分要点，冲击高分。本文围绕一份典型的 Year 12 爱德思统计单元测试，剖析大纲范围内各类型典型题目，并揭示出题人的命题思路。
1. Sampling Methods | 抽样方法题目解析
A mock question states: “A school has 1200 students in Years 7–13. The headteacher wants a sample of 80 students, ensuring each year group is fairly represented. Name a suitable sampling method and give one advantage of using it in this context.”
一个模拟题这样出：“某校七至十三年级共有1200名学生。校长想抽取80名学生的样本，同时确保每个年级组都有公平的代表性。请说出一种合适的抽样方法，并给出在本情境中使用该方法的一个优点。”
This scenario demands proportional representation of distinct strata – the year groups. Stratified sampling is the appropriate choice. The population is divided into strata by year, and a random sample is taken from each in proportion to its size.
该情境要求不同层（年级组）具有比例代表性，因此分层抽样是恰当的选择。总体按年级分层，再从各层中按比例随机抽取个体。
One advantage is that it guarantees each year group’s representation, eliminating the risk of an unbalanced sample that could distort findings about the whole school.
一个优点是它保证了每个年级组的代表性，消除了样本不平衡的风险，从而避免对整个学校的结论产生歪曲。
A common mistake is suggesting cluster sampling – taking whole tutor groups – but this does not guarantee proportional representation across years.
常见错误是提议整群抽样——抽取整个导师组——但这并不能保证跨年级的比例代表性。
2. Histograms and Frequency Density | 直方图与频率密度
A question presents a grouped frequency table for the distances travelled by 60 commuters, with unequal class widths. It asks students to draw the histogram and estimate the number of commuters who travelled less than 15 km.
某题给出60名通勤者出行距离的分组频数表，组距不等。要求学生画出直方图，并估计行程小于15公里的人数。
t be used on the vertical axis. The area of each bar is proportional to frequency. In a typical table with classes 0–5, 5–15, 15–30, students first calculate the frequency densities.
对于组距不等的分组，纵轴需使用频率密度 = 频数 ÷ 组距。每个直条的面积与频数成正比。在典型的0–5、5–15、15–30这类分组中，学生首先计算各组的频率密度。
To estimate the number travelling under 15 km, linear interpolation within the 5–15 class can be used: assume the data are evenly spread across the interval. If 20 commuters are in 5–15, and we need 5–15 km full bar gives 20, then 5–10 portion is half the width, so approximately 10 commuters. Adding the 0–5 frequency gives the total.
为估计15公里以下的人数，可在5–15组内进行线性插值：假设数据在区间内均匀分布。若5–15组有20人，则5–10部分占一半宽度，约10人，加上0–5组的频数即可得到总数。
Examiners want to see the bar for 15– rking shown clearly: (10/15) × frequency of that class, or similar proportional reasoning.
考官期望看到15–30组的直条绘制准确，并清楚地展示插值过程：例如该组宽度的比例乘以该组频数，或类似的比例推理。
3. Box Plots and Outlier Detection | 箱线图与异常值检测
he question asks students to draw a box plot and determine whether the smallest value, 0.09, is an outlier.
一组20个反应时间（秒）数据：Q₁ = 0.22，Q₂ = 0.28，Q₃ = 0.35。题目要求学生绘制箱线图，并判断最小值0.09是否为异常值。
3 = 0.025. Since 0.09 > 0.025, it is not an outlier. It should be drawn as the end of the lower whisker.
异常值界限的计算：下围栏 = Q₁ - 1.5 × IQR，上围栏 = Q₃ + 1.5 × IQR。本题IQR = 0.35 - 0.22 = 0.13，下围栏 = 0.22 - 1.5 × 0.13 = 0.025。由于0.09 > 0.025，所以不是异常值，应作为下须的端点绘制。
Students often forget the 1.5 multiplier or misidentify an outlier by comparing the value to the quartiles directly. Always show the calculation of IQR and fences for method marks.
学生常忘记1.5的乘数，或者直接将数值与四分位数比较而错误判断。务必展示IQR和围栏的计算过程以获取方法分。
IQR = Q₃ - Q₁ = 0.13 → Lower Fence = 0.22 - 1.5 × 0.13 = 0.025
IQR = Q₃ - Q₁ = 0.13 → 下围栏 = 0.22 - 1.5 × 0.13 = 0.025
4. Mean, Variance and Standard Deviation | 均值、方差与标准差
245, Σx² = 6235. It asks for the mean and an unbiased estimate of the population variance.
某题给出汇总统计量：n = 10，Σx = 245，Σx² = 6235。要求计算均值以及总体方差的无偏估计。
Sample mean x̄ = (6235 - 245²/10) / 9 = (6235 - 6002.5) / 9 = 232.5 / 9 ≈ 25.833. Standard deviation s = √25.833 ≈ 5.08.
样本均值 x̄ = Σx/n = 24.5。无偏方差估计 s² = (Σx² - (Σx)²/n) / (n - 1)。代入得：s² = (6235 - 245²/10) / 9 = (6235 - 6002.5) / 9 = 232.5 / 9 ≈ 25.833。标准差 s = √25.833 ≈ 5.08。
A common error is dividing by n instead of n - 1 for the unbiased variance. Remember: the formula with n - 1 removes the bias when estimating the population variance from a sample.
常见错误是在计算无偏方差时除以 n 而非 n - 1。切记：分母为 n - 1 的公式用于从样本估计总体方差时可消除偏差。
| Quantity | Formula |
| Sample Mean | x̄ = Σx/n |
| Unbiased Variance | s² = (Σx² - (Σx)²/n)/(n-1) |
| 量 | 公式 |
| 样本均值 | x̄ = Σx/n |
| 无偏方差 | s² = (Σx² - (Σx)²/n)/(n-1) |
5. Correlation and Linear Regression | 相关性与线性回归
o interpret this correlation. Part (b) requires the regression line equation, given that x̄ = 4.5, ȳ = 62, Sxx = 88.2, Sxy = 146.7.
一份模拟卷中包含12名学生复习时间与测试分数的散点图，相关系数 r = 0.82。 (a) 问要求解释这种相关性。(b) 问要求计算回归直线方程，已知 x̄ = 4.5, ȳ = 62, Sxx = 88.2, Sxy = 146.7。
Interpretation: there is a strong positive linear correlation between revision hours and test scores. As revision hours increase, the test score tends to increase. The correlation is not perfect, but it is strong.
解释：复习时间与测试分数之间存在强正线性相关。随着复习时间增加，测试分数往往也增加。相关性虽非完美，但较强。
The regression line of y on x has gradient b = Sxy/Sxx = 146.7 / 88.2 ≈ 1.663. Intercept a = ȳ - b x̄ .f.). This line can be used to predict test score from revision hours, but not vice versa.
y对x的回归直线斜率 b = Sxy/Sxx = 146.7 / 88.2 ≈ 1.663。截距 a = ȳ - b x̄ = 62 - 1.663 × 4.5 ≈ 54.52。因此回归方程为：y = 54.52 + 1.66x (3位有效数字)。该方程可用于根据复习时间预测测试分数，但不能反向预测。
Make sure to use the correct regression line (y on x, not x on y). The reliability of predictions decreases for extrapolation beyond the data range.
务必使用正确的回归直线（y对x，而非x对y）。超出数据范围进行外推时，预测的可靠性会下降。
6. Probability and Tree Diagrams | 概率与树形图
A typical question: “In a bag there are 5 red and 3 blue counters. A counter is drawn at random and not replaced. A second counter is then drawn. Find the probability that both counters are the same colour.”
一道典型题目：“袋中有5个红色和3个蓝色筹码。随机抽取一个不放回，接着抽取第二个。求两个筹码颜色相同的概率。”
(Red) = 5/8, P(Blue) = 3/8. Second draw probabilities change due to non-replacement. P(Red|Red) = 4/7, P(Blue|Red) = 3/7, P(Red|Blue) = 5/7, P(Blue|Blue) = 2/7.
画出树形图，第一层分支：P(红) = 5/8，P(蓝) = 3/8。由于不放回，第二层的概率改变：P(红|红) = 4/7，P(蓝|红) = 3/7，P(红|蓝) = 5/7，P(蓝|蓝) = 2/7。
th same colour = P(RR) + P(BB) = (5/8)×(4/7) + (3/8)×(2/7) = 20/56 + 6/56 = 26/56 = 13/28.
两球同色的概率 = P(红红) + P(蓝蓝) = (5/8)×(4/7) + (3/8)×(2/7) = 20/56 + 6/56 = 26/56 = 13/28。
Labelling each branch and multiplying along them is crucial. Many students lose marks by assuming replacement when not stated.
关键要标出每个分支，并按分支相乘。许多学生因在非放回情境中错误地假定为放回而失分。
7. Discrete Random Variables and Expectation | 离散随机变量与期望
The probability distribution of a discrete random variable X is given:
x: 1, 2, 3 ; P(X=x): 0.2, 0.5, 0.3
离散随机变量X的概率分布如下：
x: 1, 2, 3 ; P(X=x): 0.2, 0.5, 0.3
Find E(X) and Var(X).
求 E(X) 和 Var(X)。
3 = 0.2 + 2.0 + 2.7 = 4.9. Then Var(X) = E(X²) - [E(X)]² = 4.9 - 2.1² = 4.9 - 4.41 = 0.49.
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。于是 Var(X) = E(X²) - [E(X)]² = 4.9 - 2.1² = 4.9 - 4.41 = 0.49。
f asked for Var(3X + 2), remember Var(aX + b) = a² Var(X), so Var(3X+2) = 9 × 0.49 = 4.41.
务必检查概率之和为1。若要求计算 Var(3X + 2)，记住 Var(aX + b) = a² Var(X)，因此 Var(3X+2) = 9 × 0.49 = 4.41。
8. Binomial Distribution Calculations | 二项分布计算
A mock paper question: “X ~ B(15, 0.3). Find P(X ≤ 5) and P(X = 7).” Such calculations are required using calculator or tables, but working must be shown.
模拟卷题目：“X ~ B(15, 0.3)。求 P(X ≤ 5) 和 P(X = 7)。”此类计算允许使用计算器或表格，但必须展示运算过程。
ufficient to state the binomial formula and then give the computed value, e.g. 0.0811 (3 s.f.).
P(X = 7) = ¹⁵C₇ × (0.3)⁷ × (0.7)⁸。写出二项概率公式，再给出计算结果即可，例如0.0811（3位有效数字）。
ative probabilities. Many students incorrectly enter parameters; double-check that p is the success probability as defined in the question.
P(X ≤ 5) 需要对 x = 0 到 5 的概率求和。仔细列出累积概率。许多学生会输错参数；要再次确认 p 是题目所定义的成功概率。
The expected number of successes is np = 4.5; this can be used to sense-check the answer – P(X ≤ 5) should be somewhat greater than 0.5 since 5 is above the mean.
期望成功次数为 np = 4.5；可据此检验答案——由于5大于均值，P(X ≤ 5) 应略大于0.5。
9. Hypothesis Testing with Binomial Distribution | 利用二项分布进行假设检验
A classic Year 12 test item: “A manufacturer claims that at most 8% of its light bulbs are defective. A random sample of 20 bulbs finds 4 defectives. Test the manufacturer's claim at the 5% significance level.”
一道经典的Year 12测试题：“一家制造商声称其灯泡的次品率不超过8%。随机抽取20个灯泡发现4个次品。在5%显著性水平下检验该制造商的说法。”
Define p as the probability of a defective bulb. H₀: p = 0.08, H₁: p > 0.08. Under H₀, X ~ B(20, 0.08). Test statistic is the number of defectives in the sample, x = 4.
设 p 为灯泡为次品的概率。H₀: p = 0.08, H₁: p > 0.08。在 H₀ 下，X ~ B(20, 0.08)。检验统计量为样本中的次品数，x = 4。
Calculate P(X ≥ 4 | > 0.05, we do not reject H₀. There is insufficient evidence to contradict the manufacturer's claim.
计算 P(X ≥ 4 | p=0.08) = 1 - P(X ≤ 3)。查表或使用计算器：P(X ≤ 3) ≈ 0.9334，因此 p-值 = 0.0666。由于0.0666 > 0.05，我们不拒绝原假设。没有充分证据否定制造商的说法。
Always conclude in context: “There is insufficient evidence, at the 5% significance level, to say that the proportion of defective bulbs exceeds 8%.”
务必在上下文中给出结论：“在5%显著性水平下，无充分证据表明次品率超过8%。”
≈ 1 - 0.9830 = 0.0170 < 0.05. Since 4 is not in the critical region, the decision matches.
许多学生忘记陈述临界域。本检验的临界域为 X ≥ 5，因为 P(X ≥ 5) = 1 - P(X ≤ 4) ≈ 1 - 0.9830 = 0.0170 < 0.05。由于4不在临界域中，结论与p-值法一致。
10. Common Pitfalls and Exam Technique | 常见失分点与考试技巧
Mock analysis reveals frequent mistakes: confusing histograms with bar charts, using biased variance divisor n, misreading “at most” as “less than” in binomial inequalities, and omitting units or context in final answers.
通过模拟卷分析可发现常见错误：混淆直方图与条形图、使用有偏方差除数n、在二项分布中把“不超过”误读为“小于”、在最终答案中遗漏单位或上下文。
To avoid these, annotate the question: underline key words like “sample”, “population”, “unbiased”, “critical region”. Always draw diagrams where possible – tree diagrams, box plots, and scatter plots make reasoning visible and can earn method marks even if numerical answers go wrong.
为避免这些错误，请圈画关键词：“样本”“总体”“无偏”“临界域”。尽可能画图——树形图、箱线图和散点图能使推理过程可视化，即使数值答案出错也能挣得方法分。
Practice interpreting computer output or summary statistics quickly; many unit tests embed real-world contexts. Finally, check that calculated probabilities are between 0 and 1, and variances are non-negative.
练习快速解读计算机输出或汇总统计量；许多单元测试都嵌入真实情境。最后，检查计算出的概率是否在0到1之间，方差是否为非负值。
Time management is critical. Allocate about one minute per mark, and leave time to review the hypothesis test conclusion – examiners heavily penalise missing contextual conclusions.
时间管理至关重要。大致按每分钟完成1分的节奏分配时间，并留出时间检查假设检验的结论——考官对缺失上下文结论的扣分很重。
Published by TutorHao | Statistics Revision Series | aleveler.com
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