Year 11 WJEC Statistics: Unit Test Mock Paper Walkthrough | WJEC 十一年级统计：单元测试模拟卷解析

📚 Year 11 WJEC Statistics: Unit Test Mock Paper Walkthrough | WJEC 十一年级统计：单元测试模拟卷解析

This walkthrough takes you through a typical WJEC GCSE Statistics unit test mock paper, breaking down each question type and revealing the key steps to secure full marks. We cover everything from sampling and data representation to probability, binomial distribution and the normal distribution, with detailed solutions and exam-focused commentary.

这份解析将带你完成一份典型的 WJEC GCSE 统计单元测试模拟卷，逐一拆解题型并揭示取得满分的核心步骤。内容涵盖抽样、数据呈现、概率、二项分布和正态分布等，配有详细解答与应试点评。

1. Overview of the Mock Paper and Key Topics | 模拟卷概述与核心考点

The mock paper is designed to mirror the structure of the actual Unit test: a mix of short-answer questions and structured problems that assess statistical literacy, calculation and interpretation. The main topics tested include data collection methods, stem-and-leaf diagrams, cumulative frequency, box plots, measures of central tendency and dispersion, probability with tree diagrams, the binomial distribution and the normal distribution.

模拟卷旨在贴合真实单元测试的结构：包含简答题和结构化问题，考查统计素养、计算与解读能力。主要考点涵盖数据收集方法、茎叶图、累积频率、箱线图、集中趋势量与离散量数、概率与树状图、二项分布以及正态分布。

2. Question Breakdown: Sampling Methods | 题目解析：抽样方法

Question: A school has 400 boys and 600 girls. The head teacher wants to survey homework habits using a sample of 50 students. Explain how a stratified sample could be obtained and why it is more suitable than a simple random sample.

题目：某校有 400 名男生和 600 名女生。校长想用 50 名学生作为样本调查作业习惯。请说明如何获得分层样本，并解释为什么分层抽样比简单随机抽样更合适。

Solution: First, compute the numbers from each group proportional to the whole. Boys: (400 / 1000) × 50 = 20; Girls: (600 / 1000) × 50 = 30. Then, within the boys’ stratum, use random sampling to select 20 boys, and likewise randomly select 30 girls from the girls’ stratum. A stratified sample guarantees that both genders are represented in the correct proportions, reducing bias and improving the precision of estimates compared with a simple random sample that might, by chance, include too few boys or girls.

解答：首先按比例计算各组的样本量。男生：(400/1000)×50 = 20；女生：(600/1000)×50 = 30。然后在男生层内用随机抽样选出 20 名男生，同样在女生层内随机选出 30 名女生。分层抽样能保证两性以正确比例被代表，降低偏倚，提高估计精度，而简单随机样本可能偶然包含过少男生或女生。

3. Question Breakdown: Stem-and-Leaf Diagrams and Averages | 题目解析：茎叶图与平均数

Data set (marks out of 50): 12, 15, 18, 21, 21, 23, 23, 25, 26, 30, 32, 34, 41

数据集（满分 50 分）：12, 15, 18, 21, 21, 23, 23, 25, 26, 30, 32, 34, 41

Stem-and-leaf (key: 1|2 means 12): 1 | 2 5 8; 2 | 1 1 3 3 5 6; 3 | 0 2 4; 4 | 1. To find the median, quartiles and IQR: n = 13. Median = 7th value = 23. Lower half: 12,15,18,21,21,23 → Q₁ = (18+21)/2 = 19.5. Upper half: 23,25,26,30,32,34 → Q₃ = (26+30)/2 = 28. IQR = 28 − 19.5 = 8.5. The mean can be calculated as (sum of all values)/13 ≈ 24.2. The stem-and-leaf diagram reveals the shape and spread without losing the original data.

茎叶图（图例：1|2 表示 12）：1 | 2 5 8; 2 | 1 1 3 3 5 6; 3 | 0 2 4; 4 | 1。求中位数、四分位数与四分位距：n = 13。中位数 = 第 7 个值 = 23。下半组：12,15,18,21,21,23 → Q₁ = (18+21)/2 = 19.5。上半组：23,25,26,30,32,34 → Q₃ = (26+30)/2 = 28。IQR = 28 − 19.5 = 8.5。均值计算为总和/13 ≈ 24.2。茎叶图能显示分布形态与离散情况，且不丢失原始数据。

4. Question Breakdown: Cumulative Frequency and Box Plots | 题目解析：累积频率与箱线图

Grouped frequency table: 0 ≤ x < 10: 5, 10 ≤ x < 20: 12, 20 ≤ x < 30: 20, 30 ≤ x < 40: 8, 40 ≤ x < 50: 5. Construct a cumulative frequency table: upper boundaries 10,20,30,40,50 and cumulative frequencies 5, 17, 37, 45, 50. Plot the points and draw a smooth curve. Median is at cumulative frequency 25 → about 22. Q₁ at 12.5 → about 15, Q₃ at 37.5 → about 31. From these, a box plot can be drawn with min = 0, Q₁ = 15, median = 22, Q₃ = 31, max = 50. Always label axes and include a meaningful title.

分组频数表：0 ≤ x < 10: 5, 10 ≤ x < 20: 12, 20 ≤ x < 30: 20, 30 ≤ x < 40: 8, 40 ≤ x < 50: 5。构建累积频率表：上界 10,20,30,40,50，累积频数 5, 17, 37, 45, 50。描点并绘制平滑曲线。中位数位于累积频数 25 处 → 约 22。Q₁ 在 12.5 → 约 15，Q₃ 在 37.5 → 约 31。据此可绘制箱线图：最小值 0，Q₁=15，中位数=22，Q₃=31，最大值=50。务必标注坐标轴并给出有意义的标题。

5. Question Breakdown: Probability and Tree Diagrams | 题目解析：概率与树状图

Question: A bag contains 4 red and 2 blue balls. Two balls are drawn without replacement. Find the probability that (a) both are red, (b) at least one is red. Solution: Draw a tree diagram. First draw: P(R)=4/6=2/3, P(B)=2/6=1/3. Second draw: after a red, P(R)=3/5, P(B)=2/5; after a blue, P(R)=4/5, P(B)=1/5. P(RR) = (2/3)×(3/5)=6/15=2/5. At least one red = 1 − P(BB) = 1 − (1/3)×(1/5) = 1 − 1/15 = 14/15. Highlight that without replacement, the denominators change, and a tree diagram systematically tracks all outcomes.

题目：袋中有 4 个红球和 2 个蓝球。不放回地抽取两球。求 (a) 两个都是红的概率，(b) 至少一个红球的概率。解答：绘制树状图。第一次抽：P(R)=4/6=2/3，P(B)=2/6=1/3。第二次抽：抽到红球后，P(R)=3/5，P(B)=2/5；抽到蓝球后，P(R)=4/5，P(B)=1/5。P(RR) = (2/3)×(3/5)=6/15=2/5。至少一个红 = 1 − P(BB) = 1 − (1/3)×(1/5) = 1 − 1/15 = 14/15。强调不放回时，分母会变化，树状图能系统追踪所有结果。

6. Question Breakdown: Standard Deviation and Comparison | 题目解析：标准差与比较

Sets of marks: Group A: 52, 55, 58, 61, 64; Group B: 48, 54, 60, 66, 72. Both have mean 58, but their spreads clearly differ. For sample standard deviation we use s = √[ Σ(x − x̄)² / (n − 1) ]. Group A deviations: −6, −3, 0, 3, 6; squared sum = 36+9+0+9+36 = 90; s = √(90/4) = √22.5 ≈ 4.74. Group B deviations: −10, −4, 2, 8, 14; squared sum = 100+16+4+64+196 = 380; s = √(380/4) = √95 ≈ 9.75. The much larger standard deviation for Group B confirms that its marks are more widely scattered, which is relevant when comparing consistency.

分数集：A组：52, 55, 58, 61, 64；B组：48, 54, 60, 66, 72。两组的均值均为 58，但离散状况明显不同。样本标准差公式为 s = √[ Σ(x − x̄)² / (n − 1) ]。A组离差：−6, −3, 0, 3, 6；平方和 = 36+9+0+9+36 = 90；s = √(90/4) = √22.5 ≈ 4.74。B组离差：−10, −4, 2, 8, 14；平方和 = 100+16+4+64+196 = 380；s = √(380/4) = √95 ≈ 9.75。B组标准差远大于A组，说明 B 组分数更分散，这在比较一组数据的稳定性时非常重要。

7. Question Breakdown: Binomial Distribution | 题目解析：二项分布

Question: A multiple-choice test has 10 questions, each with 4 options. A student guesses every answer. Find the probability of getting (a) exactly 6 correct, (b) at least 8 correct. Solution: This is binomial with n = 10, p = 0.25. P(X = r) = 10Cr × (0.25)^r × (0.75)^(10 − r). For exactly 6: 10C6 = 210, (0.25)⁶ ≈ 0.000244, (0.75)⁴ ≈ 0.3164, product ≈ 0.0162. For at least 8: sum P(8) + P(9) + P(10). 10C8 = 45, 10C9 = 10, 10C10 = 1. P(8) ≈ 45×0.00001526×0.5625 ≈ 0.000386, P(9) ≈ 10×0.000003815×0.75 ≈ 2.86×10⁻⁵, P(10) ≈ 1×9.54×10⁻⁷ ≈ 9.54×10⁻⁷. Total ≈ 0.000415. Very low chance, as expected for guessing.

题目：一份单选题测试有 10 道题，每题 4 个选项。一名学生全靠猜测作答。求 (a) 恰好猜对 6 题的概率，(b) 至少猜对 8 题的概率。解答：符合二项分布，n = 10，p = 0.25。P(X = r) = 10Cr × (0.25)^r × (0.75)^(10 − r)。恰好 6 题：10C6 = 210，(0.25)⁶ ≈ 0.000244，(0.75)⁴ ≈ 0.3164，乘积 ≈ 0.0162。至少 8 题：P(8)+P(9)+P(10)。10C8=45, 10C9=10, 10C10=1。P(8) ≈ 45×0.00001526×0.5625 ≈ 0.000386，P(9) ≈ 10×0.000003815×0.75 ≈ 2.86×10⁻⁵，P(10) ≈ 9.54×10⁻⁷，总和 ≈ 0.000415。全靠猜测得到高分的概率极低，符合直觉。

8. Question Breakdown: The Normal Distribution | 题目解析：正态分布

Question: Exam scores are normally distributed with mean μ = 60 and standard deviation σ = 10. Find (a) the proportion of students scoring above 75, (b) the minimum score for the top 10%. Solution: (a) Standardise: z = (75 − 60) / 10 = 1.5. From tables, P(Z < 1.5) = 0.9332, so P(X > 75) = 1 − 0.9332 = 0.0668. (b) For the top 10%, we need z such that P(Z > z) = 0.10, i.e. P(Z < z) = 0.90. The table gives z ≈ 1.28. Then x = μ + zσ = 60 + 1.28×10 = 72.8. A student needs about 73 marks to be in the top 10%. Always remember to sketch a normal curve and shade the relevant area.

题目：某次考试分数服从正态分布，均值 μ = 60，标准差 σ = 10。求 (a) 分数超过 75 的学生比例，(b) 进入前 10% 的最低分数。解答：(a) 标准化：z = (75 − 60) / 10 = 1.5。查表得 P(Z < 1.5

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