📚 GCSE CIE Statistics: Unit Test Mock Paper Walkthrough | GCSE CIE 统计：单元测试模拟卷解析
Mock papers are essential for mastering GCSE CIE Statistics. This walkthrough analyses a typical unit test mock paper, covering key topics such as data collection, measures of central tendency and spread, probability, and distributions. Each section explains common question types and effective solving strategies, helping you identify pitfalls and boost your exam confidence.
模拟卷是掌握 GCSE CIE 统计的关键。本文解析一份典型的单元测试模拟卷，涵盖数据收集、集中趋势与离散程度、概率及分布等重点内容。每个部分讲解常见题型和有效解题策略，帮助您识别易错点，提升考试信心。
1. Overview of the Unit Test Mock Paper | 模拟卷概览
The mock paper is designed to assess understanding of the entire Statistics unit. It typically comprises short-answer questions on sampling, data presentation, summary statistics, and longer problems involving probability distributions and interpretation. The total marks are 50, to be completed in 1 hour.
该模拟卷旨在评估对整个统计单元的理解。通常包括关于抽样、数据呈现、汇总统计的简答题，以及涉及概率分布和解释的较长时间题。总分 50 分，需在 1 小时内完成。
Familiarise yourself with the formula sheet provided by CIE. Questions often test not only calculations but also the ability to comment on reliability, bias, and real-life context. Use clear notation and always show working to gain method marks.
请熟悉 CIE 提供的公式表。题目不仅考查计算，还考查对可靠性、偏差和现实背景的评述能力。使用清晰的符号，并始终展示解题步骤以获取过程分。
2. Sampling Methods and Bias | 抽样方法与偏差
Question 1: A researcher wants to survey the eating habits of students in a school. She selects every 5th student from the register list. Identify the sampling method and discuss potential bias. (4 marks)
题目1： 研究者想调查学校学生的饮食习惯。她按学生名册每隔5人选择一人。请指出抽样方法并讨论可能的偏差。(4分)
This is systematic sampling. The register may be ordered alphabetically, which does not directly link to eating habits, so it could provide a representative snapshot. However, bias may arise if students with certain surnames share a common background or if some students are absent during registration, leading to under-representation. To improve, a stratified sample by year group or a simple random sample could be used.
这是系统抽样。名册可能是按字母顺序排列的，与饮食习惯没有直接关联，因此可以提供具有代表性的样本。但如果某些姓氏的学生有着相似的背景，或者点名时有学生缺席，就会导致某些群体被低代表，从而产生偏差。改进方法可采用按年级分层抽样或简单随机抽样。
Question 2: A company tests light bulb lifetimes by picking every 10th bulb off the production line. State one advantage and one disadvantage of this method. (2 marks)
题目2： 一家公司通过每隔10个灯泡抽取一个来检测寿命。请说明该方法的一个优点和一个缺点。(2分)
Advantage: It is quick and easy to implement, ensuring the sample is spread across the whole production period. Disadvantage: If there is a regular pattern in the production process (e.g. a faulty machine every 10th cycle), the sample could be biased, systematically missing or over-representing defective items.
优点：该方法快速简便，能确保样本覆盖整个生产周期。缺点：如果生产过程中存在规律性变化（例如每10个周期出现一次机器故障），样本可能会产生系统性偏差，漏掉或过多包含次品。
3. Frequency Tables and Histograms | 频数表与直方图
Question: The table shows the times (in minutes) for 40 students to complete a puzzle.
| Time (t min) | Frequency |
|---|---|
| 0 < t ≤ 5 | 6 |
| 5 < t ≤ 10 | 12 |
| 10 < t ≤ 15 | 14 |
| 15 < t ≤ 25 | 8 |
a) Construct a histogram. b) Estimate the mean time. (6 marks)
题目：下表为40名学生完成拼图所需时间（分钟）的分布。 a) 绘制直方图。 b) 估算平均时间。(6分)
For the histogram, calculate frequency density 12/5=2.4, 14/5=2.8, 8/10=0.8. Plot bars with these heights on a continuous scale labelled 'Time (minutes)' and 'Frequency density'. No gaps between bars.
绘制直方图时，计算频数密度 = 频数 / 组距。各组组距分别为5, 5, 5, 10。密度依次为：6/5=1.2, 12/5=2.4, 14/5=2.8, 8/10=0.8。用这些高度绘制矩形条，横轴标为“时间（分钟）”，纵轴标为“频数密度”。条形之间无间隔。
To estimate the mean, use midpoints: 2.5, 7.5, 12.5, 20. The formula is x̄ = Σ(f × x) / Σf 5)+(12×7.5)+(14×12.5)+(8×20) = 15+90+175+160 = 440. Σf = 40. Mean = 440 / 40 = 11 minutes.
估算均值时，使用组中值：2.5, 7.5, 12.5, 20。公式为 x̄ = Σ(f × x) / Σf。f × x 总和 = (6×2.5)+(12×7.5)+(14×12.5)+(8×20) = 15+90+175+160 = 440。总频数=40。均值 = 440 / 40 = 11 分钟。
x̄ = Σfx / Σf = 440/40 = 11
4. Measures of Central Tendency | 集中趋势的度量
Another question gives the list: 5, 7, 8, 9, 9, 11, 14. Find the mean, median, and mode. Comment on which is most useful.
另一题目给出数据：5, 7, 8, 9, 9, 11, 14。求均值、中位数和众数，并评论哪一个最有用。
. Median is the 4th value: 9. Mode is 9. Since the data are reasonably symmetric with no outliers, the mean or median both represent the centre well. The mode adds little here. If the data included an extreme value like 50, the median would be more reliable.
均值 = 63/7 = 9。中位数为第4个值：9。众数为9。由于数据大致对称且无异常值，均值和中位数都能很好地代表中心。众数提供的附加信息较少。如果数据包含一个极端值如50，中位数将更可靠。
, the median lies in the 10 < t ≤ 15 class; if cumulative frequencies are 6, 18, 32, 40, the 20.5th value falls in that class, and linear interpolation gives median ≈ 10 + (20.5-18)/14 *5 = 10.89 min.
在频数表中，中位数位置通过 (n+1)/2 确定，分组数据可能需要插值。例如在拼图时间中，累积频数为6,18,32,40，第20.5个值落在10-15组，线性插值得中位数 ≈ 10 + (20.5-18)/14 ×5 ≈ 10.89分钟。
5. Measures of Spread: Range, IQR, Standard Deviation | 离散程度：极差、四分位距、标准差
an of upper half (9,11,14) → 11. So IQR = 11-7 = 4. This gives the spread of the middle 50%.
对于原始数据5,7,8,9,9,11,14：极差=14-5=9。下四分位数Q₁是下半部分(5,7,8)的中位数→7；上四分位数Q₃是上半部分(9,11,14)的中位数→11。因此IQR=11-7=4，反映中间50%数据的分散情况。
, -2, -1, 0, 0, 2, 5. Squared: 16,4,1,0,0,4,25. Sum=50. If treated as sample, variance s² = 50/(7-1) ≈ 8.33, so s = √8.33 ≈ 2.89. If population, σ² = 50/7 ≈ 7.14, σ ≈ 2.67. Check the question's requirement; CIE often uses the divisor n for a given dataset unless 'sample' is specified.
标准差衡量数据与均值的平均距离。按公式，均值为9。偏差：-4, -2, -1, 0, 0, 2, 5。平方和=50。若作为样本，方差 s² = 50/(7-1) ≈ 8.33，s ≈ 2.89。若作为总体，σ² = 50/7 ≈ 7.14，σ ≈ 2.67。注意题目要求；CIE 通常对给出的数据集使用除数 n，除非明确提及“样本”。
σ = √( Σ(x - x̄)² / n ) = √50/7 ≈ 2.67
6. Box Plots and Cumulative Frequency | 箱线图与累积频数
ers extend to the min and max within 1.5×IQR, with outliers marked separately. Always label the scale and give a key.
利用拼图时间的累积频数曲线，可以读出Q₁、中位数、Q₃。绘制箱线图用于比较两组数据。箱体表示IQR，须状线延伸至1.5倍IQR范围内的最小和最大值，异常值单独标记。务必标注刻度并添加图例。
In the mock, a question might ask: 'Compare the completion times for boys and girls using box plots.' Look at medians (central tendency), IQR and range (spread), and skewness. For instance, boys' median lower but greater IQR suggests more variability. Use comparative words: 'higher than', 'more spread out'.
模拟卷中可能有题：“使用箱线图比较男生和女生的完成时间。” 查看中位数（集中趋势）、IQR和极差（分散度）以及偏态。例如，男生中位数较低但IQR较大，说明变异性更大。使用比较性词语：“高于”、“更分散”。
7. Scatter Diagrams and Correlation | 散点图与相关性
cores tend to increase. Draw a line of best fit by eye, ensuring equal numbers of points above and below the line. Use it to estimate a score for 7 hours: read vertically from x=7 to the line, then horizontally to the y-axis.
散点图显示复习小时数与测验分数呈正相关。描述关系：随着复习时间增加，测验分数往往上升。目测绘制最佳拟合直线，使线上下的点数大致相等。用该线估计复习7小时的分数：从x=7垂直向上交于直线，再水平读y轴。
s linear correlation. CIE Statistics may require calculating r using the formula r = Sxy / √Sxx Syy ta, r = 0.89 suggests a strong relationship. Beware: correlation does not imply causation.
积差相关系数（PMCC）r 量化线性相关性。CIE 统计可能要求使用公式 r = Sxy / √Sxx Syy 计算，其中 Sxy = Σxy - (Σx Σy)/n。r 接近+1表明强正线性相关。对于复习数据，r=0.89 提示强相关。注意：相关不蕴含因果。
r = Sxy / √Sxx × Syy
8. Probability Basics and Tree Diagrams | 概率基础与树状图
A bag contains 4 red and 5 blue balls. Two balls are drawn without replacement. Draw a tree diagram to find the probability that both are the same colour.
袋中有4个红球和5个蓝球。不放回地抽取两个球。绘制树状图，求两个球颜色相同的概率。
First draw: P(Red)=4/9, P(Blue)=5/9. Second draw: if red taken, P(Red|Red)=3/8, P(Blue|Red)=5/8. If blue taken, P(Red|Blue)=4/8=1/2, P(Blue| B) = (4/9×3/8)+(5/9×1/2) = 12/72 + (5/9×4/8 → 5/9×1/2=5/18=20/72) = 12/72+20/72=32/72=4/9.
第一次抽取：P(红)=4/9，P(蓝)=5/9。第二次：若第一次为红，P(红|红)=3/8，P(蓝|红)=5/8。若第一次为蓝，P(红|蓝)=4/8=1/2，P(蓝|蓝)=4/8=1/2。同色概率：P(RR)+P(BB) = (4/9×3/8)+(5/9×1/2) = 12/72 + 20/72 = 32/72 = 4/9。
and add probabilities for combined events. For conditional probability questions, use P(A|B) = P(A∩B)/P(B). In the mock, they may ask for 'probability red in the second draw given first was blue' – simply read from the tree.
始终沿分支相乘，相加得到组合事件的概率。对于条件概率问题，使用 P(A|B) = P(A∩B)/P(B)。模拟卷中可能会问“已知第一个是蓝球，第二个是红球的概率” – 直接从树状图读取。
9. Binomial Distribution Applications | 二项分布应用
The conditions for a binomial distribution: fixed number of trials n, each trial independent, only two outcomes (success/failure), constant probability of success p. In the mock, a typical question: 'It is known that 20% of light bulbs are defective. A random sample of 10 bulbs is taken. Find the probability exactly 2 are defective.'
二项分布的条件：固定试验次数 n，每次独立，只有两种结果（成功/失败），每次成功的概率 p 恒定。模拟卷中典型题：“已知20%的灯泡有缺陷。随机抽取10个，求恰好有2个缺陷的概率。”
So P ≈ 45 × 0.04 × 0.1678 ≈ 0.302. Use the formula sheet or calculator. Expect follow-up: 'Find the probability at least 8 are non-defective', which means P(X ≤ 2) since defective defined as success.
使用 B(10, 0.2)：P(X=2) = ¹⁰C₂ × (0.2)² × (0.8)⁸。¹⁰C₂ = 45。计算后 P ≈ 0.302。需用公式表或计算器。可能追问：“至少8个无缺陷的概率”，即 P(X ≤ 2)，因成功定义为缺陷。
P(X = k) = ⁿCₖ pᵏ (1 - p)ⁿ⁻ᵏ
, variance = np(1-p)=10×0.2×0.8=1.6. Always state the distribution and parameters clearly.
二项分布均值：μ = np = 2，方差 = np(1-p) = 1.6。务必清晰写出分布和参数。
10. Normal Distribution and Standardisation | 正态分布与标准化
ormally distributed with μ=45 cm, σ=4 cm. Find the proportion of plants taller than 50 cm.
正态分布是对称的钟形曲线，由均值 μ 和标准差 σ 定义。一道题给出：植物高度服从正态分布，μ=45 cm, σ=4 cm。求高度超过50 cm的比例。
normal table or technology. P(Z > 1.25) = 1 - Φ(1.25) ≈ 1 - 0.8944 = 0.1056. So about 10.6% exceed 50 cm. If the question asks for probability between two values, find Φ(z₂) - Φ(z₁).
标准化：z = (50-45)/4 = 1.25。然后查标准正态表或使用技术工具。P(Z > 1.25) = 1 - Φ(1.25) ≈ 0.1056。所以约10.6%的植株超过50 cm。若问两值之间的概率，计算 Φ(z₂) - Φ(z₁)。
Inverse normal: given a percentage, find the corresponding x. E.g., top 5% ut-off: find z-value for 0.95, which is ≈1.645, then x = μ + zσ = 45 + 1.645×4 ≈ 51.58 cm.
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