📚 GCSE Cambridge Statistics: Unit Test Mock Paper Solutions | GCSE剑桥统计：单元测试模拟卷解析
This mock paper solutions article walks through a complete unit test designed for the Cambridge GCSE Statistics syllabus. It covers core topics including data classification, frequency distributions, measures of central tendency, probability, cumulative frequency, sampling, scatter graphs and weighted averages. By working through each question with full worked solutions and examiner tips, students can diagnose their understanding and sharpen exam technique.
本模拟卷解析文章将逐题讲解一套针对剑桥GCSE统计课程设计的单元测试卷。内容涵盖数据分类、频率分布、集中趋势度量、概率、累积频率、抽样方法、散点图和加权平均数等核心主题。通过完整的解题过程和考官点评，学生可以诊断自己的知识漏洞，提升应试技巧。
1. Mock Paper Overview | 模拟卷概述
This unit test paper consists of 8 questions with a total of 50 marks. The recommended time is 50 minutes. It is designed to assess the key statistical skills required at GCSE level: classifying data, constructing and interpreting diagrams, calculating summary statistics, applying probability rules, using cumulative frequency curves, selecting appropriate sampling methods, interpreting bivariate data, and working with weighted indices. The mark allocation is shown next to each question, and the solutions below mirror the style of Cambridge mark schemes. Always show full working to gain method marks even if the final answer is incorrect.
本单元测试卷由8道题组成，总分50分，建议用时50分钟。试卷旨在评估GCSE阶段的关键统计技能：数据分类、图表的绘制与解读、摘要统计量的计算、概率规则的应用、累积频率曲线的使用、合适抽样方法的选择、双变量数据的解释以及加权指数的运算。各题旁标注了分值，下面的解析风格与剑桥评分标准一致。答题时务必展示完整解题过程，即使最终答案有误，也可获得方法分。
2. Question 1: Classifying Data | 第1题：数据分类
Question: A student collects the following information from classmates: favourite sport, number of pets, and time taken to travel to school in minutes. Classify each variable as qualitative, discrete quantitative, or continuous quantitative. [3 marks]
题目：一名学生从同学那里收集了以下信息：最喜爱的运动、宠物数量、上学路上所用时间（分钟）。将每个变量分类为定性、离散定量或连续定量。[3分]
Favourite sport: This variable describes a non-numerical category, so it is qualitative data.
最喜爱的运动：该变量描述的是非数值类别，因此为定性数据。
Number of pets: This is a count taking whole-number values (0, 1, 2, ...), which makes it discrete quantitative data.
宠物数量：这是一个取整数值的计数（0, 1, 2, ...），因此为离散定量数据。
Travel time in minutes: Time is measured on a continuous scale; even though we record it to the nearest minute, it can take any value within an interval. Hence it is continuous quantitative data.
上学用时（分钟）：时间是在连续尺度上测量的；虽然我们以分钟为单位记录，但它在区间内可任意取值。因此是连续定量数据。
Mark scheme: One mark for each correct classification. Accept 'categorical' for qualitative and 'numerical' for quantitative with the discrete/continuous distinction made clear.
评分标准：每个正确分类得1分。可接受将qualitative表述为categorical，但必须明确区分离散与连续数值。
3. Question 2: Histogram with Unequal Class Widths | 第2题：组距不等的直方图
Question: The table below gives the masses, m kg, of 85 parcels delivered by a courier. Complete the table by calculating frequency densities, and draw a fully labelled histogram to represent the data. [6 marks]
题目：下表给出某快递公司85个包裹的质量（m kg）。通过计算频率密度完成表格，并绘制一个带完整标签的直方图来表示数据。[6分]
| Mass, m (kg) | Frequency | Class width | Frequency density |
|---|---|---|---|
| 0 < m ≤ 2 | 15 | 2 | 7.5 |
| 2 < m ≤ 4 | 24 | 2 | 12 |
| 4 < m ≤ 8 | 36 | 4 | 9 |
| 8 < m ≤ 12 | 10 | 4 | 2.5 |
First, find the class width for each interval by subtracting the lower boundary from the upper boundary. Then calculate frequency density using the formula:
首先，用区间上界减去下界得到每组的组距。然后使用公式计算频率密度：
Frequency density = Frequency ÷ Class width
频率密度 = 频数 ÷ 组距
equency density = 36 ÷ 4 = 9. For 8 < m ≤ 12: width = 4, frequency density = 10 ÷ 4 = 2.5.
对于0 < m ≤ 2：组距2，频率密度 = 15 ÷ 2 = 7.5。2 < m ≤ 4：组距2，频率密度 = 24 ÷ 2 = 12。4 < m ≤ 8：组距4，频率密度 = 36 ÷ 4 = 9。8 < m ≤ 12：组距4，频率密度 = 10 ÷ 4 = 2.5。
To draw the histogram: label the horizontal axis 'Mass (kg)' with a continuous scale from 0 to 12. Label the vertical axis 'Frequency density'. Draw adjacent rectangles whose widths equal the class widths and whose heights equal the frequency densities. The area of each rectangle is proportional to the frequency. Since the vertical axis starts at zero, no gaps should appear between bars. [A sketch is expected in the real paper.]
绘制直方图：横轴标注“质量(kg)”，刻度从0到12连续。纵轴标注“频率密度”。绘制相邻矩形，宽度等于组距，高度等于频率密度。每个矩形的面积与频数成正比。纵轴从零开始，条形之间不留空隙。[实际试卷中需作图。]
Mark scheme: 2 marks for correct frequency densities, 2 marks for correctly scaled and labelled axes, 2 marks for accurate bars with correct widths and heights. If frequency densities are wrong but the bars follow the candidate's calculation, award scale marks provisionally.
评分标准：频率密度正确得2分，坐标轴刻度与标签正确得2分，矩形宽度和高度正确得2分。如果频率密度计算错误但所绘条形与考生计算结果一致，可暂给坐标轴与绘图分。
4. Question 3: Mean, Median, Mode and Range | 第3题：平均数、中位数、众数和极差
Question: The numbers below show the goals scored by a hockey team in 7 matches: 2, 5, 3, 5, 1, 8, 4. (a) Calculate the mean number of goals. [2] (b) Find the median. [1] (c) State the mode. [1] (d) Work out the range. [1]
题目：以下数字显示了一支曲棍球队在7场比赛中的进球数：2, 5, 3, 5, 1, 8, 4。 (a) 计算平均进球数。[2] (b) 找出中位数。[1] (c) 指出众数。[1] (d) 计算极差。[1]
of goals = 2 + 5 + 3 + 5 + 1 + 8 + 4 = 28. Number of matches = 7. Mean = 28 ÷ 7 = 4 goals.
(a) 进球总数 = 2 + 5 + 3 + 5 + 1 + 8 + 4 = 28。比赛场次 = 7。平均数 = 28 ÷ 7 = 4 球。
(b) Arrange the data in order: 1, 2, 3, 4, 5, 5, 8. The median is the middle value, the 4th value, which is 4 goals.
(b) 将数据排序：1, 2, 3, 4, 5, 5, 8。中位数为中间值，即第4个值，为4球。
hat appears most often. The number 5 occurs twice, more than any other, so mode = 5 goals.
(c) 众数是出现次数最多的值。数字5出现两次，多于其他，所以众数为5球。
(d) Range = highest value - lowest value = 8 - 1 = 7 goals.
(d) 极差 = 最大值 - 最小值 = 8 - 1 = 7 球。
Examiner tip: For the mean, show the addition and division steps clearly to secure the method mark even if a calculation slip occurs. When finding the median, always sort the data first.
考官提示：计算平均数时，清晰展示加法和除法步骤，即使出现计算失误也能确保方法分。求中位数时，务必先排序。
5. Question 4: Probability with Overlapping Events | 第4题：含重叠事件的概率
Question: A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is (a) a Heart, (b) a King, (c) a Heart or a King. [4 marks]
题目：从一副标准的52张扑克牌中随机抽取一张。求抽到的牌是 (a) 红桃的概率，(b) K的概率，(c) 红桃或K的概率。[4分]
(a) There are 13 Hearts in a pack of 52. P(Heart) = 13/52 = 1/4.
(a) 一副牌中有13张红桃。P(红桃) = 13/52 = 1/4。
(b) There are 4 Kings. P(King) = 4/52 = 1/13.
(b) 有4张K。P(K) = 4/52 = 1/13。
(c) 'Heart or King' (King) - P(Heart and King). P(Heart and King) is the probability of drawing the King of Hearts = 1/52. So P(Heart or King) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13.
(c) “红桃或K”包括红桃K，如果简单相加P(红桃)和P(K)会将该牌重复计算。使用加法法则：P(红桃或K) = P(红桃) + P(K) - P(红桃和K)。P(红桃和K)即抽到红桃K的概率 = 1/52。因此P(红桃或K) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13。
King of Hearts is already in the Hearts), giving 16 favourable outcomes. So 16/52 = 4/13.
替代推理：共有13张红桃，再加上另外3张K（因为红桃K已包含在红桃中），共16个有利结果。所以16/52 = 4/13。
Mark allocation: (a) 1, (b) 1, (c) 2 marks – one for applying the general addition rule or listing outcomes, one for correct simplified answer.
分值分配： (a)1分, (b)1分, (c)2分——正确运用一般加法法则或正确列举结果得1分，结果化简正确得1分。
6. Question 5: Cumulative Frequency and Quartiles | 第5题：累积频率与四分位数
Question: The table shows the times taken by 60 students to complete a puzzle.
| Time, t (seconds) | Frequency |
|---|---|
| 0 < t ≤ 10 | 8 |
| 10 < t ≤ 20 | 14 |
| 20 < t ≤ 30 | 20 |
| 30 < t ≤ 40 | 12 |
| 40 < t ≤ 60 | 6 |
(a) Construct a cumulative frequency table. [2] (b) Draw a cumulative frequency graph. [3] (c) Use your graph to estimate the median and the interquartile range. [3]
题目：下表显示了60名学生完成一个谜题所用的时间。
| 时间, t (秒) | 频数 |
|---|---|
| 0 < t ≤ 10 | 8 |
| 10 < t ≤ 20 | 14 |
| 20 < t ≤ 30 | 20 |
| 30 < t ≤ 40 | 12 |
| 40 < t ≤ 60 | 6 |
(a) 构建累积频率表。[2] (b) 绘制累积频率图。[3] (c) 使用图形估计中位数和四分位距。[3]
54+6=60. Plot points at the upper boundaries: (10,8), (20,22), (30,42), (40,54), (60,60).
(a) 累积频率：≤10: 8; ≤20: 22; ≤30: 42; ≤40: 54; ≤60: 60。在上界处描点：(10,8), (20,22), (30,42), (40,54), (60,60)。
is is cumulative frequency from 0 to 60; the horizontal axis is time from 0 to 60 seconds.
(b) 用平滑曲线连接各描点。由于时间不可能≤0，将曲线起点与原点(0,0)相连。纵轴为累积频率，范围0到60；横轴为时间，0到60秒。
a well-drawn graph, median ≈ 25 seconds. Lower quartile position = 60/4 = 15, reading from graph ≈ 16.5 seconds. Upper quartile position = 3×60/4 = 45, reading ≈ 33 seconds. Interquartile range = 33 - 16.5 = 16.5 seconds. Answers may vary slightly depending on the curve's smoothness.
(c) 中位数位置 = 60/2 = 30。从纵轴30处作水平线与曲线相交，向下读横轴时间。绘制良好的图形中，中位数≈25秒。下四分位数位置=60/4=15，从图形读取≈16.5秒。上四分位数位置=3×60/4=45，读取≈33秒。四分位距 = 33 - 16.5 = 16.5秒。根据曲线平滑度，答案可能略有出入。
Mark scheme: (a) correct cumulative frequencies 2 marks; (b) scale, labels, points plotted correctly, smooth curve 3 marks; (c) median and IQR within an acceptable range 3 marks.
评分标准： (a)累积频率正确2分；(b)刻度、标签、描点正确，平滑曲线3分；(c)中位数和IQR在可接受范围内得3分。
7. Question 6: Sampling Methods | 第6题：抽样方法
Question: A school has 400 students spread across Years 7-11. The head teacher wants to survey a sample of 40 students about their views on school lunches. Suggest a suitable sampling method and explain how it would be carried out. Explain one advantage of your chosen method. [5 marks]
题目：某学校有400名学生，分布在7-11年级。校长希望抽取40名学生组成样本，调查他们对学校午餐的看法。建议一种合适的抽样方法并解释如何实施。说明所选方法的一个优点。[5分]
strata based on year group because views on lunches may differ by age. The number of students sampled from each year should be proportional to the year group' with 100 students samples 10; and so on. Within each year group, use simple random sampling, e.g. assign numbers and use a random number generator, to select the required number of students. One advantage is that the sample is more likely to be representative of the whole school because it reflects the year-group structure. This reduces sampling bias compared to a simple random sample that might, by chance, over-represent one year.
分层抽样是合适的方法。首先按年级将总体分层，因为不同年龄对午餐的看法可能不同。从各年级抽取的样本数与其人数成比例：若Year 7有80人，则样本数 = (80/400)×40 = 8人；Year 8 100人，抽10人；依此类推。在每个年级内，使用简单随机抽样，例如编号后借助随机数生成器选取所需人数。优点之一是该样本更可能代表全校，因为它反映了年级结构。与可能偶然过多抽取某年级的简单随机样本相比，这降低了抽样偏差。
Other accepted methods: quota sampling with careful control, but the proportional method is more rigorous. The explanation must include both the stratification and the random element to gain full marks.
其他可接受的方法：配额抽样（需严格管控），但比例法更严谨。答案须同时包含分层和随机要素才能获满分。
8. Question 7: Scatter Graph and Correlation | 第7题：散点图与相关性
Question: A teacher recorded the number of hours ten students spent revising and their test scores.
| Revision (h) | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 | 4.5 | 5.0 |
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