Year 10 WJEC Statistics: Unit Test Mock Paper Walkthrough | WJEC 十年级统计：单元测试模拟卷解析

📚 Year 10 WJEC Statistics: Unit Test Mock Paper Walkthrough | WJEC 十年级统计：单元测试模拟卷解析

Welcome to this full walkthrough of a WJEC Year 10 Statistics unit test mock paper. Designed to mirror the structure and style of the actual Unit 1 assessment, this resource provides step‑by‑step model answers, commentary on common pitfalls, and examiner‑style tips. Each question tackles core topics such as data types, sampling, diagrams, probability, and the binomial distribution, helping you build confidence ahead of your test.

欢迎阅读这份 WJEC 十年级统计单元测试模拟卷的完整解析。本文模拟真实单元测试（Unit 1）的题型与风格，提供逐步示范答案、常见错误点评和阅卷老师式的答题技巧。每道题目涵盖数据类型、抽样、图表、概率和二项分布等核心内容，帮助你在考前建立信心。

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1. Classifying Data and Designing Questions | 数据分类与问题设计

Question: A survey asks for ‘age’, ‘favourite colour’, ‘number of pets’ and ‘time spent on homework per day (in minutes)’. For each variable, state whether it is qualitative, quantitative discrete or quantitative continuous.

问题：一份问卷收集了“年龄”“最喜欢的颜色”“宠物数量”“每天做作业花费的时间（分钟）”。指出每个变量属于定性变量、离散型定量变量还是连续型定量变量。

Answer: ‘Favourite colour’ is qualitative (non‑numeric category). ‘Number of pets’ is quantitative discrete (countable, takes whole number values). ‘Age’ could be continuous if recorded exactly, but in many questionnaires it is recorded as age in whole years, making it discrete; however, precision matters – if asked as ‘age in years’, treat it as discrete. ‘Time in minutes’ is quantitative continuous because it can take any value within a range. Examiners reward precise justification.

答案：“最喜欢的颜色”是定性变量（非数值分类）。“宠物数量”是离散型定量变量（可数，取整数值）。“年龄”如果精确记录可以是连续变量，但问卷中通常按整岁收集，因此可作为离散变量；考试时若有疑惑应根据题目情境判断。“时间（分钟）”是连续型定量变量，因为它可以在一个范围内取任意值。阅卷人看重精确的说明理由。

Common mistake: mixing up discrete and continuous when a variable is rounded. Tip: ask yourself whether the variable is measured or counted.

常见错误：当变量被四舍五入时，混淆离散和连续。建议：问自己该变量是被测量出来的还是计数出来的。

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2. Simple Random and Stratified Sampling | 简单随机抽样与分层抽样

Question: A school has 1200 students: 360 in Year 7, 300 in Year 8, 280 in Year 9, 160 in Year 10 and 100 in Year 11. The headteacher wants a sample of 80 students. Describe how to select a stratified sample by year group, and calculate the number of students needed from Year 10.

问题：一所学校有 1200 名学生：七年级 360 人，八年级 300 人，九年级 280 人，十年级 160 人，十一年级 100 人。校长希望抽取一个 80 名学生的样本。描述如何按年级进行分层抽样，并计算十年级需要抽取的学生人数。

Answer: Stratified sampling divides the population into distinct strata (year groups) and selects a random sample from each in proportion to its size. For Year 10, the fraction is 160/1200 = 2/15. Multiply by total sample size: 80 × (160/1200) = 80 × 2/15 = 160/15 = 10.67, so round to 11 students (round to nearest whole number). From within the Year 10 stratum, use a random number generator or lottery method to select 11 students, ensuring every Year 10 student has an equal chance. This method guarantees representation from each year group.

答案：分层抽样先将总体划分为互不重叠的层（年级），然后从每一层中按比例随机抽取样本。十年级的比例为 160/1200 = 2/15。乘以总样本量：80 × (160/1200) = 80 × 2/15 = 160/15 ≈ 10.67，四舍五入后抽取 11 名学生。在十年级这一层内，使用随机数生成器或抽签法选出 11 人，确保每位十年级学生有均等机会。该方法能保证每个年级都有代表被抽中。

Examiner note: Show working with fractions and explain rounding. Mentioning ‘equal chance’ and ‘proportional’ gains marks.

阅卷要点：展示分数计算过程并解释舍入。提及“均等机会”和“按比例”可以得分。

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3. Stem‑and‑Leaf Diagrams and Averages | 茎叶图与平均数

Question: The following are the pulse rates (beats per minute) of 15 students after exercise: 62, 65, 68, 71, 72, 72, 75, 78, 80, 82, 82, 85, 88, 91, 94. Construct an ordered stem‑and‑leaf diagram and find the median and mode.

问题：以下是 15 名学生运动后的脉搏率（次/分）：62, 65, 68, 71, 72, 72, 75, 78, 80, 82, 82, 85, 88, 91, 94。构建一张有序茎叶图，并找出中位数和众数。

Answer: Use stems for tens digits (6,7,8,9). Leaves are units digits, ordered. Diagram:

答案：以十位数作茎（6,7,8,9），个位数为叶，按顺序排列。图示：

6 | 2 5 8
7 | 1 2 2 5 8
8 | 0 2 2 5 8
9 | 1 4

Key: 6|2 means 62 bpm. The ordered leaves show distribution. For median (n=15), the 8th value is the median. Counting through the diagram: 62,65,68,71,72,72,75,78 → 8th value is 78. So median = 78 bpm. Mode is the most frequent value: both 72 and 82 appear twice, so the data is bimodal with modes 72 and 82.

图例：6|2 表示 62 次/分。有序的叶子展示了分布。中位数（n=15）是第 8 个数值。从茎叶图中顺数：62,65,68,71,72,72,75,78 → 第 8 个是 78，所以中位数 = 78 次/分。众数是出现次数最多的值：72 和 82 均出现两次，故这组数据为双峰分布，众数为 72 和 82。

Mark scheme tip: remember to include a key and order the leaves; stating ‘bimodal’ earns extra credit.

得分技巧：务必写出图例并让叶子有序；指出“双峰”可获得加分。

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4. Box‑and‑Whisker Plots and Comparisons | 盒须图与比较

Question: The five‑number summaries for test scores of two classes are: Class A: min=34, LQ=52, median=68, UQ=79, max=94. Class B: min=41, LQ=60, median=66, UQ=72, max=85. Draw box plots (on the same scale) and compare their spreads and central tendency.

问题：两个班级测验成绩的五数概括为：A 班 min=34, LQ=52, 中位数=68, UQ=79, max=94。B 班 min=41, LQ=60, 中位数=66, UQ=72, max=85。在同一尺度上绘制盒须图，并比较它们的离散度和集中趋势。

Answer: Draw a scale from 30 to 100. For each class, plot the box from LQ to UQ with median line inside; whiskers extend to min and max, provided no outliers (here none identified). Visually, Class A has a larger interquartile range (IQR = 79‑52 = 27) compared to Class B’s IQR = 72‑60 = 12, showing greater spread in the middle 50%. The range for A is 60, for B is 44. Class A’s median (68) is slightly higher than Class B’s (66), but their central tendency is similar. Class B’s scores are more consistent; the box is narrower and the whiskers shorter. Always support comparisons with numbers like IQR and median.

答案：画出 30 到 100 的尺度。每个班级绘制从下四分位数到上四分位数的矩形，中间用线段标出中位数；须线延伸到最小值和最大值（假设无异常值）。可以看出，A 班的四分位距 IQR = 79‑52 = 27，对比 B 班的 IQR = 72‑60 = 12，说明 A 班中间 50% 数据的离散度更大。全距 A 班 60，B 班 44。A 班中位数（68）略高于 B 班（66），但集中趋势接近。B 班成绩更稳定，盒子更窄且须线更短。答题时务必用 IQR 和中位数等具体数字支撑比较。

Graph details: Use a ruler; label median, LQ, UQ. When comparing, mention ‘consistency’, ‘spread’ and ‘average’.

作图细节：用直尺绘制；标注中位数、LQ、UQ。比较时要提及“一致性”“离散度”和“平均水平”。

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5. Cumulative Frequency Curves and Percentiles | 累积频率曲线与百分位数

Question: The grouped frequency table shows the mass (grams) of 80 apples. Use it to construct a cumulative frequency table and draw a cumulative frequency curve. Then estimate the median mass and the 90th percentile.

问题：分组频数表记录了 80 个苹果的质量（克）。据此构建累积频数表并画出累积频率曲线，然后估计中位数质量和第 90 百分位数。

Table: 100–110 (8), 110–120 (14), 120–130 (26), 130–140 (20), 140–150 (12).

Answer: Build cumulative frequency: ≤110: 8; ≤120: 22; ≤130: 48; ≤140: 68; ≤150: 80. Plot upper class boundaries against cumulative frequency and join with a smooth curve. Median is the value at cumulative frequency 40 (half of 80). From the curve, draw a line from 40 to the curve then down to the mass axis; reading gives approximately 127 g. The 90th percentile corresponds to 0.9×80 = 72. From 72 on CF axis, read across to mass, giving about 143 g. Always show construction lines on the graph.

答案：构建累积频数：≤110 为 8；≤120 为 22；≤130 为 48；≤140 为 68；≤150 为 80。将各组上限与累积频数描点，用平滑曲线连接。中位数对应累积频数 40（80 的一半）。从 40 画水平线交于曲线，再向下读质量轴，约为 127 g。第 90 百分位数对应 0.9×80 = 72，从累积频数 72 处读得质量约 143 g。作答时务必在图上保留作图线。

Exam tip: Label axes ‘Cumulative frequency’ and ‘Mass (g)’. The curve must start at the first upper boundary with CF=0 if data starts above 0, but here first boundary is 110 with CF=0 before? Actually cumulative frequency before the first class is 0 at lower boundary 100. Plot (100,0) for a complete curve. Detail: plot (100,0) for a true start.

考试技巧：坐标轴标注“累积频数”和“质量（克）”。曲线应从 (100,0) 开始，以完整反映起始累积频数为 0。

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6. Basic Probability Laws | 基础概率运算法则

Question: A fair six‑sided die is rolled. Event A: score is even. Event B: score is a prime number. List the outcomes, then find P(A), P(B), P(A ∩ B), and P(A ∪ B). Show that the addition law holds.

问题：掷一枚公平的六面骰子。事件 A：点数为偶数；事件 B：点数为质数。列出结果，求 P(A)、P(B)、P(A ∩ B) 和 P(A ∪ B)，并验证加法法则成立。

Answer: Sample space S = {1,2,3,4,5,6}. A = {2,4,6} so P(A)=3/6=1/2. B = {2,3,5} (prime numbers) so P(B)=3/6=1/2. A ∩ B = {2} so P(A ∩ B)=1/6. A ∪ B = {2,3,4,5,6} so P(A ∪ B)=5/6. Using the addition law: P(A ∪ B) = P(A)+P(B)‑P(A ∩ B) = 1/2+1/2‑1/6 = 1‑1/6 = 5/6. The law holds.

答案：样本空间 S = {1,2,3,4,5,6}。A = {2,4,6}，故 P(A)=3/6=1/2。B = {2,3,5}（质数），故 P(B)=3/6=1/2。A ∩ B = {2}，P(A ∩ B)=1/6。A ∪ B = {2,3,4,5,6}，P(A ∪ B)=5/6。根据加法法则：P(A ∪ B) = P(A)+P(B)‑P(A ∩ B) = 1/2+1/2‑1/6 = 1‑1/6 = 5/6，验证一致。

Common error: forgetting that 1 is not prime. Remember primes have exactly two distinct factors; 1 is excluded.

常见错误：忘记 1 不是质数。质数有且仅有两个不同的正因子，因此 1 不是质数。

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7. Tree Diagrams and Conditional Probability | 树形图与条件概率

Question: A bag contains 5 red and 3 blue beads. Two beads are drawn at random without replacement. Draw a tree diagram to show all probabilities, and find the probability that the second bead is blue given that the first was red. Also find the probability that both beads are the same colour.

问题：袋中有 5 颗红珠和 3 颗蓝珠。随机不放回地抽取两颗珠子。画出树形图标明所有概率，并求在第一颗为红珠的条件下第二颗为蓝珠的概率，以及两颗珠子同色的概率。

Answer: First draw: P(Red)=5/8, P(Blue)=3/8. Second draw: if first Red, remaining 4R,3B → P(Red|1st Red)=4/7, P(Blue|1st Red)=3/7. If first Blue, remaining 5R,2B → P(Red|1st Blue)=5/7, P(Blue|1st Blue)=2/7. Probability (2nd blue | 1st red) = 3/7. For both same colour: (both red) = (5/8)×(4/7)=20/56; (both blue) = (3/8)×(2/7)=6/56. Sum = 26/56 = 13/28. The tree diagram must label branches with probabilities and final outcomes.

答案：第一次抽取：P(红)=5/8，P(蓝)=3/8。第二次：若第一次为红，剩余 4 红 3 蓝 → P(红|1st 红)=4/7，P(蓝|1st 红)=3/7。若第一次为蓝，剩余 5 红 2 蓝 → P(红|1st 蓝)=5/7，P(蓝|1st 蓝)=2/7。给定第一次红，第二次蓝的概率 = 3/7。同色概率：（均红）= (5/8)×(4/7)=20/56；（均蓝）= (3/8)×(2/7)=6/56；总和 = 26/56 = 13/28。树形图必须标注各分支概率及最终结果。

Check: Sum of joint probabilities = 20/56+15/56+15/56+6/56 = 56/56 = 1. Excellent check for accuracy.

检查：联合概率之和 = 20/56+15/56+15/56+6/56 = 1，证明计算无误。

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8. Introduction to the Binomial Distribution | 二项分布入门

Question: The probability that a seed germinates is 0.7. Ten seeds are planted independently. Using the binomial formula, find the probability that exactly 7 seeds germinate. (You may use the binomial coefficient (10 choose 7) = 120.)

问题：一颗种子发芽的概率为 0.7。独立种植 10 颗种子。使用二项分布公式计算恰好有 7 颗种子发芽的概率。（已知二项式系数 10 选 7 = 120。）

Answer: Let X ~ B(10, 0.7). P(X = 7) = (10C7) × (0.7)⁷ × (0.3)³ = 120 × (0.7)⁷ × (0.3)³. Compute powers: (0.7)⁷ = 0.7^7 = 0.0823543 (approx), (0.3)³ = 0.027. Product: 120 × 0.0823543 × 0.027 = 120 × 0.0022235661 ≈ 0.2668. So P(X = 7) ≈ 0.267 (to 3 decimal places). Interpretation: about a 26.7% chance of exactly 7 seeds germinating.

答案：设 X ~ B(10, 0.7)。P(X = 7) = (10C7) × (0.7)⁷ × (0.3)³ = 120 × (0.7)⁷ × (0.3)³。计算幂：0.7⁷ ≈ 0.0823543，0.3³ = 0.027。乘积：120 × 0.0823543 × 0.027 ≈ 120 × 0.0022235661 ≈ 0.2668，即约 0.267（保留三位小数）。解释：恰有 7 颗发芽的概率约为 26.7%。

Formula reminder: P(X = r) = ⁿCᵣ × pʳ × (1‑p)ⁿ⁻ʳ. Ensure you clearly define n and p. If using tables, locate n=10, p=0.7, r=7. Values in WJEC tables typically give 0.2668 directly. Show substitution before final answer.

公式提示：P(X = r) = ⁿCᵣ × pʳ × (1‑p)ⁿ⁻ʳ。务必明确定义 n 和 p。若查表，核验 n=10, p=0.7, r=7 确认。考试时先代入公式再给出最终结果。

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9. Relative Risk and Two‑Way Tables | 相对风险与双向表

Question: A study observes whether taking vitamin C affects occurrence of colds. The results: 35 out of 100 taking vitamin C caught a cold; 55 out of 100 not taking vitamin C caught a cold. Construct a 2×2 table, calculate the risk for each group, and find the relative risk of catching a cold for the vitamin group compared to the no‑vitamin group. Interpret your result.

问题：一项研究观察服用维生素 C 是否影响感冒发生。结果：服用维生素 C 的 100 人中有 35 人感冒；未服用的 100 人中有 55 人感冒。构建 2×2 表格，计算各组的风险，并求出服用维生素组相对于未服用组感冒的相对风险，解释结果。

Answer: Table:

	Cold	No Cold	Total
Vitamin C	35	65	100
No Vitamin	55	45	100

Risk for vitamin group = 35/100 = 0.35. Risk for no‑vitamin group = 55/100 = 0.55. Relative risk (RR) = 0.35 / 0.55 = 0.636 (approx). Interpretation: The risk of catching a cold for those taking vitamin C is 0.64 times the risk for those not taking it, i.e. a 36% lower risk. Values below 1 indicate a protective effect.

答案：如上表。维生素组风险 = 35/100 = 0.35；未服用组风险 = 55/100 = 0.55。相对风险 RR = 0.35 / 0.55 ≈ 0.636。解释：服用维生素 C 者患感冒的风险是未服用者的 0.64 倍，即风险降低约 36%。RR 小于 1 提示存在保护效应。

Important: Relative risk does not imply causation; it measures association. Always state ‘times the risk’ rather than percentage change unless specified.

要点：相对风险不能直接推断因果关系，它衡量关联程度。通常表述为“风险的多少倍”，除非题目要求，否则谨慎使用百分比变化。

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10. Quality Control Charts and Warning Limits | 质量控制图与警戒限

Question: A machine fills bottles with a target mean of 500 ml. Based on past data, the standard deviation of the filling process is 3 ml. Samples of size 4 are taken every hour. Calculate the warning limits (mean ± 2 standard errors) and action limits (mean ± 3 standard errors) for the sample mean. Sketch the control chart framework.

问题：一台灌装机目标均值为 500 ml。根据历史数据，灌装过程的标准差为 3 ml。每小时抽取容量为 4 的样本。计算样本均值的警戒限（均值 ± 2 倍标准误）和行动限（均值 ± 3 倍标准误），并画出控制图框架。

Answer: Standard error of the mean (SE) = σ / √n = 3 / √4 = 3 / 2 = 1.5 ml. Warning limits: 500 ± 2×1.5 = 500 ± 3, i.e. 497 ml and 503 ml. Action limits: 500 ± 3×1.5 = 500 ± 4.5, i.e. 495.5 ml and 504.5 ml. On a control chart, plot a horizontal target line at 500, upper and lower warning lines at 503 and 497, and upper and lower action lines at 504.5 and 495.5. Points falling outside warning limits suggest the process may be drifting; points outside action limits indicate the process needs immediate adjustment.

答案：均值标准误 SE = σ / √n = 3 / √4 = 1.5 ml。警戒限为 500 ± 3，即 497 ml 和 503 ml。行动限为 500 ± 4.5，即 495.5 ml 和 504.5 ml。在控制图上画出目标中心线 500，上下警戒线 503 和 497，上下行动线 504.5 和 495.5。样本点超出警戒限提示过程可能漂移；超出行动限说明过程必须立即调整。

Concept link: Control charts use the Normal approximation; here 2 and 3 standard errors correspond to approximate 95% and 99.7% coverage. In WJEC, you may be asked to identify patterns or suggest corrective actions.

概念衔接：控制图基于正态近似，±2 和 ±3 标准误分别对应约 95% 和 99.7% 的覆盖范围。在 WJEC 考试中，你可能需要识别模式或提出纠正措施。

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11. Handling Evaluation and Limitations | 评价与局限性分析

Question: Discuss limitations of the sampling method used in Question 2 and suggest how the survey design could be improved.

问题：讨论第 2 题中抽样方法的局限性，并提出如何改进调查设计。

Answer: The stratified sample ensures proportional

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