📚 IGCSE CIE Statistics: Mock Unit Test Walkthrough | IGCSE CIE 统计：单元测试模拟卷解析
This walkthrough breaks down a full-length IGCSE CIE Statistics mock unit test, explaining each question step by step to help you identify common pitfalls and consolidate your exam technique. The paper is designed to mirror the core topics prescribed in the 0409 syllabus, including data collection, representation, averages, dispersion, probability, correlation, normal distribution, and sampling.
本文逐题解析一份完整的IGCSE CIE 统计单元模拟卷，助你发现常犯错误并巩固应试技巧。试卷紧扣0409教学大纲的核心主题，涵盖数据收集、数据展示、平均数、离散程度、概率、相关性、正态分布及抽样方法。
1. Data Collection & Types | 数据收集与类型
Question: A researcher wants to know the average weekly screen time of teenagers in a city. She divides the city into five districts, then randomly selects two schools from each district and surveys all students in those schools. Identify the population, the sampling method used, and state one advantage and one disadvantage of this method.
问题：研究人员想了解某城市青少年每周平均屏幕使用时间。她将城市分为五个区，从每区随机抽取两所学校，并对所选学校的所有学生进行调查。请确定总体、所用抽样方法，并说明该方法的一个优点和一个缺点。
Answer & Explanation: The population is all teenagers in that city. The sampling method is stratified cluster sampling (or multistage sampling). The district division ensures representation across regions (strata), then schools are clusters, and all individuals within the selected clusters are surveyed. Advantage: It is less time-consuming and cheaper than a simple random sample of all teenagers, because travel and administration are limited to selected schools. Disadvantage: If the chosen schools are not typical, the results may be biased; students within a school may be more similar to each other than to the overall teenage population, inflating sampling error.
答案与解析：总体是该城市所有青少年。抽样方法为分层整群抽样（或多阶段抽样）。分区保证了区域代表性（层），学校为群，选中群内全部个体接受调查。优点：相比对所有青少年进行简单随机抽样，此法更省时、成本更低，因为只需前往选定学校开展调查。缺点：如果所选学校不具代表性，结果就会有偏差；同一学校内的学生彼此相似度可能高于整体青少年群体，从而增大抽样误差。
2. Diagrams & Frequency Distributions | 图表与频率分布
Question: The table below shows the number of books read by a group of students in a term. Draw a frequency polygon to represent the data and estimate the modal number of books.
问题：下表显示了一组学生在一学期内阅读的书籍数量。绘制频率多边形表示数据，并估计阅读书籍数量的众数。
| Number of books | 0-4 | 5-9 | 10-14 | 15-19 | 20-24 |
|---|---|---|---|---|---|
| Frequency | 6 | 14 | 18 | 10 | 2 |
Answer & Explanation: To draw a frequency polygon, plot the midpoint of each class interval against its frequency: 2, 7, 12, 17, 22. Connect the points with straight lines and add points at the start (midpoint before first class: -3) and end (midpoint after last class: 27) with frequency 0 to close the polygon. The modal class is 10-14 books because it has the highest frequency (18). The polygon’s peak will visually confirm this. The exact mode within the interval cannot be determined without further assumptions, so we report the modal class.
答案与解析：绘制频率多边形时，以每个组距的中点（2, 7, 12, 17, 22）为横坐标，对应频数为纵坐标描点，用直线连接。多边形两端须添加频率为0的点：首个组距前的中点-3和末个组距后的中点27。众数所在组为10-14本书，因其频数最高（18）。图形峰值也直观显示这一区间。在无额外假设的情况下，无法确定组内精确众数，因此通常报告众数所在组。
3. Measures of Central Tendency | 集中趋势测量
Question: Eleven students recorded their test scores: 45, 52, 55, 57, 60, 62, 65, 68, 70, 72, 95. Calculate the mean, median, and mode. Explain which measure best represents the typical score.
问题：11名学生的测验成绩为：45, 52, 55, 57, 60, 62, 65, 68, 70, 72, 95。计算平均数、中位数和众数，并说明哪个指标最能代表典型成绩。
Answer & Explanation: Mean = (45+52+55+57+60+62+65+68+70+72+95) / 11 = 701 / 11 ≈ 63.7. Median is the 6th value in ascending order: 62. Mode: all values are unique, so there is no mode. The score 95 is an outlier that pulls the mean upward. The median of 62 is resistant to the outlier and better represents the centre of the bulk of the data. Therefore, the median is the most appropriate measure here.
答案与解析：平均数 = (45+52+55+57+60+62+65+68+70+72+95) ÷ 11 = 701 ÷ 11 ≈ 63.7。中位数为升序排列第6个值：62。众数：所有数值均唯一，因此无众数。95为异常值，拉高了平均数。中位数62不受异常值影响，更能代表数据主体中心，因此中位数是最合适的集中趋势指标。
4. Measures of Dispersion | 离散程度测量
Question: For the same eleven scores, find the range, interquartile range (IQR), and standard deviation (using the formula for sample standard deviation). Interpret the IQR.
问题：针对上述11个成绩，求极差、四分位距（IQR）及样本标准差，并解释IQR的意义。
Answer & Explanation: Range = 95 – 45 = 50. To find IQR: Q₁ is the median of the first 5 values (45,52,55,57,60) = 55; Q₃ is the median of the last 5 values (65,68,70,72,95) = 70. IQR = Q₃ – Q₁ = 70 – 55 = 15. Standard deviation is calculated using Σ(x – x̄)²/(n-1). Deviations: -18.7, -11.7, -8.7, -6.7, -3.7, -1.7, 1.3, 4.3, 6.3, 8.3, 31.3. Squared sum ≈ 351.7 + 136.9 + 75.7 + 44.9 + 13.7 + 2.9 + 1.7 + 18.5 + 39.7 + 68.9 + 979.7 = 1734.3. Divide by 10: variance ≈ 173.43, so standard deviation ≈ √173.43 ≈ 13.2. The IQR of 15 indicates that the middle 50% of scores lie within a span of 15 marks, showing moderate spread around the median.
答案与解析：极差 = 95 – 45 = 50。求IQR：下四分位数Q₁为前5个数的中位数（45,52,55,57,60）= 55；上四分位数Q₃为后5个数的中位数（65,68,70,72,95）= 70。IQR = 70 – 55 = 15。样本标准差用 Σ(x – x̄)²/(n-1) 计算。离均差：-18.7, -11.7, -8.7, -6.7, -3.7, -1.7, 1.3, 4.3, 6.3, 8.3, 31.3。平方和≈1734.3，除以10得方差173.43，开方得标准差≈13.2。IQR为15，表明中间50%的成绩跨度为15分，围绕中位数的离散程度适中。
5. Basic Probability | 概率基础
Question: A bag contains 5 red, 3 blue, and 2 green marbles. One marble is drawn at random. Find the probability that it is red or blue. If a red marble is drawn and not replaced, what is the probability that a second marble drawn is blue?
问题：袋中有5颗红、3颗蓝、2颗绿弹珠。随机抽取一颗，求抽到红色或蓝色的概率。如果抽出一颗红色且不放回，则第二次抽到蓝色的概率是多少？
Answer & Explanation: Total marbles = 10. P(red or blue) = P(red) + P(blue) = 5/10 + 3/10 = 8/10 = 0.8. After removing one red marble, remaining marbles = 9, blue marbles still = 3. Therefore, P(second blue | first red) = 3/9 = 1/3 ≈ 0.333. This demonstrates conditional probability when events are not independent due to without-replacement sampling.
答案与解析：总弹珠数10。P(红或蓝) = P(红) + P(蓝) = 5/10 + 3/10 = 8/10 = 0.8。取出一个红球且不放回后，剩余9颗，蓝色仍为3颗，故 P(第二次蓝|第一次红) = 3/9 = 1/3 ≈ 0.333。这体现了不放回抽样时事件不独立下的条件概率。
6. Tree Diagrams & Conditional Probability | 树形图与条件概率
Question: The probability that a student revises for a test is 0.7. If they revise, the probability of passing is 0.8; if not, the probability of passing is 0.3. Draw a tree diagram and find the probability that a randomly chosen student passes the test. Given a student passed, what is the probability that they revised?
问题：学生复习测试的概率为0.7。若复习，通过概率为0.8；不复习，通过概率为0.3。画出树形图，求随机选一学生通过测试的概率。已知某学生通过，求其复习过的概率。
Answer & Explanation: Tree branches: Revise (0.7) → Pass (0.8), Fail (0.2); Not revise (0.3) → Pass (0.3), Fail (0.7). Total probability of passing = 0.7 × 0.8 + 0.3 × 0.3 = 0.56 + 0.09 = 0.65. For the reverse probability, use Bayes’ theorem: P(revise | pass) = P(revise and pass) / P(pass) = 0.56 / 0.65 ≈ 0.8615. The tree diagram provides a clear visual breakdown of compound events.
答案与解析：树分支：复习(0.7)→通过(0.8)、未通过(0.2)；不复习(0.3)→通过(0.3)、未通过(0.7)。总通过概率 = 0.7×0.8 + 0.3×0.3 = 0.56 + 0.09 = 0.65。逆概率用贝叶斯定理：P(复习|通过) = P(复习且通过) / P(通过) = 0.56 / 0.65 ≈ 0.8615。树形图清晰展示了复合事件的分解。
7. Correlation & Regression | 相关性与回归
Question: The table shows hours of study (x) and test scores (y) for six students. Calculate the product moment correlation coefficient (PMCC) and comment on the correlation. Then find the equation of the regression line of y on x.
| x (hours) | 2 | 3 | 5 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|
| y (score) | 41 | 48 | 62 | 70 | 76 | 83 |
问题：下表为六名学生的学习时数(x)与测验成绩(y)。计算积矩相关系数(PMCC)并评述相关性，进而求y对x的回归直线方程。
Answer & Explanation: Compute using PMCC formula r = Σ[(x – x̄)(y – ȳ)] / √[Σ(x – x̄)² Σ(y – ȳ)²]. x̄ = 34/6 ≈ 5.667, ȳ = 380/6 ≈ 63.333. Calculate cross-products and squares. After working through, r ≈ 0.997, indicating a very strong positive linear correlation. Regression line of y on x: y – ȳ = b(x – x̄), with b = Σ(x – x̄)(y – ȳ) / Σ(x – x̄)². The calculated slope b ≈ 6.4, intercept a = ȳ – b x̄ ≈ 27.2. Thus equation: y = 27.2 + 6.4x. This line can be used to predict scores from hours of study.
答案与解析：用PMCC公式 r = Σ[(x – x̄)(y – ȳ)] / √[Σ(x – x̄)² Σ(y – ȳ)²]。x̄ = 34÷6≈5.667，ȳ = 380÷6≈63.333。计算交叉乘积与平方项，得 r≈0.997，表明极强的正线性相关。y对x的回归线：斜率b = Σ(x – x̄)(y – ȳ) / Σ(x – x̄)² ≈ 6.4，截距 a = ȳ – b x̄ ≈ 27.2。方程为 y = 27.2 + 6.4x，可用于通过学习时数预测成绩。
8. Normal Distribution | 正态分布
Question: The weights of apples from an orchard are normally distributed with mean 150 g and standard deviation 12 g. Find the percentage of apples weighing between 138 g and 168 g. Also, determine the weight below which the lightest 10% of apples fall.
问题：某果园苹果重量服从正态分布，均值为150克，标准差12克。求重量在138克到168克之间的苹果所占百分比，并求最轻10%的苹果重量上限。
Answer & Explanation: Standardize: z₁ = (138 – 150)/12 = -1, z₂ = (168 – 150)/12 = 1.5. From standard normal tables, P(-1 < Z < 1.5) = P(Z<1.5) - P(Z<-1) = 0.9332 - 0.1587 = 0.7745, so about 77.5%. For the lightest 10%, find z such that P(Z < z) = 0.10. That z ≈ -1.2816. Weight = μ + zσ = 150 + (-1.2816)(12) ≈ 134.6 g. So apples lighter than about 135 g are in the bottom 10%.
答案与解析：标准化：z₁ = (138-150)/12 = -1，z₂ = (168-150)/12 = 1.5。查标准正态表，P(-1 < Z < 1.5) = 0.9332 - 0.1587 = 0.7745，约77.5%。最轻10%：求z使P(Z Question: A company has 1200 employees across five departments. Describe how to select a stratified sample of 100 employees, ensuring proportional representation from each department. Explain why a simple random sample might not be suitable. 问题：某公司有1200名员工，分属五个部门。说明如何抽取一个由100名员工组成的分层样本，确保各部门按比例代表。解释为何简单随机抽样可能不适用。 Answer & Explanation: First, determine the number of employees in each department. Suppose department sizes are: A=300, B=200, C=250, D=150, E=300. Compute sample allocation proportional to size: A= (300/1200)×100 = 25, B≈17, C≈21, D≈12.5→13, E=25, total 101 (adjust rounding to 100). Within each department, use simple random sampling to select the required number. Advantage over simple random sample: stratification ensures each department is represented in proportion to its size, reducing sampling error and guaranteeing that smaller departments are included. A simple random sample could by chance miss department D entirely or over-represent a particular department, leading to biased estimates. 答案与解析：首先统计各部门人数。假设A=300, B=200, C=250, D=150, E=300。按比例分配样本数：A=(300/1200)×100=25，B≈17，C≈21，D≈12.5→13，E=25，总101（调整舍入至100）。然后在每个部门内用简单随机抽样抽取所需人数。相比简单随机抽样，分层保证了各部门按比例代表，降低了抽样误差，也确保小部门被纳入。简单随机抽样可能偶然遗漏部门D或过度代表某部门，导致估计偏差。 Throughout this mock test, we reinforced key principles: always define the population clearly, choose the correct measure of central tendency based on data shape, understand how outliers affect mean and standard deviation, master tree diagrams for conditional probability, and remember to standardise when using normal tables. In the exam, show full working, label axes on graphs, and interpret results in context. Practice past papers under timed conditions to build speed and accuracy. 通过本次模拟卷，我们巩固了核心原则：清晰界定总体，根据数据分布形态选择正确的集中趋势指标，理解异常值对均值和标准差的影响，掌握条件概率的树形图方法，使用正态表时务必标准化。考试时，展示完整步骤，图表轴标签完整，结合语境诠释结果。定时刷历年真题，提升解题速度与准确率。 Published by TutorHao | Statistics Revision Series | aleveler.com 更多咨询请联系16621398022（同微信）
9. Sampling Methods | 抽样方法
10. Summary & Exam Tips | 总结与答题技巧
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导