📚 GCSE Eduqas Statistics Unit Test Mock Paper Walkthrough | GCSE Eduqas 统计单元测试模拟卷解析
This walkthrough provides detailed solutions and examiner insights for a GCSE Eduqas Statistics unit test mock paper. Each question is broken down step by step, highlighting common mistakes and essential revision points.
本解析为GCSE Eduqas统计单元测试模拟卷提供详细解答和考官见解。每道题逐步分解，强调常见错误和关键复习要点。
1. Identifying Sampling Methods | 识别抽样方法
Mock Question 1: A school has 600 students distributed across year groups 7 to 11. The numbers are: Year 7: 120, Year 8: 130, Year 9: 140, Year 10: 110, Year 11: 100. Explain how to select a stratified sample of 60 students. State one advantage of stratified sampling compared to simple random sampling.
模拟题1：某校有600名学生，分布在7至11年级。各年级人数为：7年级120名，8年级130名，9年级140名，10年级110名，11年级100名。解释如何选取60名学生的分层样本。说明分层抽样相比简单随机抽样的一个优点。
To select a stratified sample, first calculate the sample size for each stratum by multiplying the total sample size by the stratum proportion.
要选取分层样本，首先将总样本量乘以每个层的比例，计算各层的样本大小。
; Year 9: 60 × (140/600) = 14; Year 10: 60 × (110/600) = 11; Year 11: 60 × (100/600) = 10.
7年级：60 × (120/600) = 12；8年级：60 × (130/600) = 13；9年级：60 × (140/600) = 14；10年级：60 × (110/600) = 11；11年级：60 × (100/600) = 10。
Then use simple random sampling within each year group, such as numbering students and using a random number generator to select the required number.
然后在每个年级内使用简单随机抽样，例如给每个学生编号并使用随机数生成器选出所需人数。
Advantage: Stratified sampling ensures proportional representation of all year groups, reducing the risk of an unrepresentative sample that simple random sampling might produce.
优点：分层抽样确保所有年级按比例代表，减少了简单随机抽样可能产生不具代表性样本的风险。
Common mistake: Confusing stratified sampling with quota sampling. Quota sampling does not involve random selection within groups.
常见错误：混淆分层抽样与配额抽样。配额抽样在各组内不进行随机选择。
2. Designing a Questionnaire | 设计问卷
Mock Question 2: A student writes: "Don't you agree that school meals are too expensive and unhealthy?" Criticise this question and write an improved version.
模拟题2：一位学生写道：「你不觉得学校餐食太贵又不健康吗？」批评该问题并写出改进版。
The question is leading – it pushes the respondent towards a particular answer by using "Don't you agree". It is also double-barrelled, asking about cost and health at the same time.
该问题具有引导性——使用「你不觉得」将受访者推向一个特定答案。它还是双重问题，同时询问价格和健康。
Respondents might think the meals are expensive but healthy, and they cannot express that accurately. No response options are provided.
受访者可能认为餐食虽贵但健康，而他们无法准确表达这一点。没有提供回答选项。
Improved version: "How would you rate the value for money of school meals?" with options Very good, Good, Fair, Poor, Very poor; and a separate question "How would you rate the healthiness of school meals?" with the same scale.
改进版：「您如何评价学校餐食的性价比？」选项为非常好、好、一般、差、非常差；以及一个单独的问题「您如何评价学校餐食的健康程度？」使用相同量表。
This improves neutrality and allows each aspect to be measured separately, yielding analysable data.
这改善了中立性，并使每个方面得以单独测量，产生可分析的数据。
3. Stem-and-Leaf Diagrams and Averages | 茎叶图与平均值
Mock Question 3: Given the stem-and-leaf diagram (key: 2|5 = 25) for test scores: 1 | 2 8; 2 | 3 5 7; 3 | 0 4 5 9; 4 | 1. Find the mean, median, mode and range.
模拟题3：给定茎叶图（key: 2|5 = 25）表示测试成绩: 1 | 2 8; 2 | 3 5 7; 3 | 0 4 5 9; 4 | 1。求平均值、中位数、众数和极差。
First list all values: 12, 18, 23, 25, 27, 30, 34, 35, 39, 41. There are 10 data values.
首先列出所有数值：12, 18, 23, 25, 27, 30, 34, 35, 39, 41。共有10个数据值。
Mean = (12+18+23+25+27+30+34+35+39+41) ÷ 10 = 284 ÷ 10 = 28.4.
平均值 = (12+18+23+25+27+30+34+35+39+41) ÷ 10 = 284 ÷ 10 = 28.4。
Median is the average of the 5th and 6th values: (27+30) ÷ 2 = 28.5.
中位数是第5与第6个值的平均：(27+30) ÷ 2 = 28.5。
Mode: no repeated values, so there is no mode. Range = 41 − 12 = 29.
众数：无重复值，故无众数。极差 = 41 − 12 = 29。
Be careful to include all stems and leaves correctly; a common error is misreading the key.
注意正确包含所有茎和叶；常见错误是误读图例。
4. Mean and Standard Deviation from Grouped Data | 分组数据的平均值与标准差
Mock Question 4: The table shows hours of homework per week. Estimate the mean and standard deviation.
| Hours | 0-4 | 5-9 | 10-14 | 15-19 | 20-30 |
|---|---|---|---|---|---|
| Frequency | 8 | 14 | 10 | 6 | 2 |
表显示每周作业小时数。估计平均值和标准差。
10=120, 17×6=102, 25×2=50. Sum fx = 386. Total frequency = 40. Mean = 386/40 = 9.65 hours.
找出组中点(x)：2, 7, 12, 17, 25。乘以频数f：2×8=16, 7×14=98, 12×10=120, 17×6=102, 25×2=50。Σfx = 386。总频数 = 40。平均值 = 386/40 = 9.65 小时。
iation, compute fx²: 2²×8=32, 7²×14=686, 12²×10=1440, 17²×6=1734, 25²×2=1250. Σfx² = 5142.
对于标准差，计算fx²：2²×8=32, 7²×14=686, 12²×10=1440, 17²×6=1734, 25²×2=1250。 Σfx² = 5142。
σ = √[ Σfx²/Σf − (mean)² ] = √[ 5142/40 − 9.65² ] = √[ 128.55 − 93.1225 ] = √35.4275 ≈ 5.95 hours.
Remember: this estimate assumes data are evenly spread within each interval.
请记住：该估计假定数据在每个区间内均匀分布。
5. Drawing and Interpreting Histograms | 绘制和解读直方图
Mock Question 5: Use the same grouped data with unequal class widths. Complete the frequency density table and describe how to draw the histogram.
模拟题5：使用不等组距的分组数据。完成频数密度表并描述如何绘制直方图。
10-14 width=5, 15-19 width=5, 20-30 width=11. Frequency density = frequency ÷ class width.
组距：0-4宽度=5，5-9宽度=5，10-14宽度=5，15-19宽度=5，20-30宽度=11。频数密度 = 频数 ÷ 组距。
FD for 0-4: 8/5=1.6; 5-9: 14/5=2.8; 10-14: 10/5=2.0; 15-19: 6/5=1.2; 20-30: 2/11≈0.18.
频数密度：0-4: 8/5=1.6; 5-9: 14/5=2.8; 10-14: 10/5=2.0; 15-19: 6/5=1.2; 20-30: 2/11≈0.18。
Draw axes: horizontal for hours (with suitable scale), vertical for frequency density. For each class, draw a bar with width equal to the class interval and height equal to its frequency density. No gaps between bars.
绘制轴：横轴为小时（合适刻度），纵轴为频数密度。对每个组，绘制宽度等于组距、高度等于频数密度的条形。条形之间不留空隙。
Area of each bar represents frequency. This correctly displays data when class widths differ.
每个条形的面积代表频数。这样在组距不同时能正确显示数据。
6. Cumulative Frequency Graphs and Quartiles | 累计频率图与四分位数
Mock Question 6: The cumulative frequency table for test marks is given. Draw the graph and estimate the median, lower quartile and interquartile range.
| Mark ≤ | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Cumulative frequency | 5 | 18 | 38 | 48 | 50 |
模拟题6：给出测试分数的累计频率表。绘制图形并估计中位数、下四分位数和四分位距。
Plot upper class boundaries (10, 20, 30, 40, 50) against cumulative frequency. Join points with a smooth curve.
绘制组上限（10,20,30,40,50）与累计频率的点，用光滑曲线连接。
Median position = 50/2 = 25th value. From graph, median ≈ 26 marks.
中位数位置 = 50/2 = 第25个值。从图中得出中位数 ≈ 26 分。
17 marks. Upper quartile position = 37.5th value, UQ ≈ 34 marks. IQR = UQ − LQ ≈ 17 marks.
下四分位数位置 = 50/4 = 第12.5个值。下四分位数 ≈ 17分。上四分位数位置 = 第37.5个值，上四分位数 ≈ 34分。四分位距 = 34 − 17 ≈ 17分。
Linearity between points must be assumed; use a ruler for reading off precisely.
必须假设点间呈线性；使用直尺精确读取。
7. Box Plots and Outliers | 箱线图与异常值
Mock Question 7: Two classes took a test. Class A: min 10, LQ 30,
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