📚 Year 10 Edexcel Statistics: Unit Test Mock Paper Analysis | Year 10 Edexcel 统计：单元测试模拟卷解析
This article walks through a typical Year 10 Edexcel Statistics unit test, providing a detailed breakdown of each question type, model answers, and common pitfalls. Designed to mirror the style and coverage of a real end‑of‑unit assessment, the mock paper includes data types, sampling, charts, averages, box plots, cumulative frequency, probability and scatter graphs. Use this analysis to consolidate your understanding and improve your exam technique.
本文带你全面解析一份典型的 Year 10 Edexcel 统计单元测试卷，逐一拆解各题型、提供标答并揭示常见错误。模拟卷在题型与考点上高度还原真实单元测验，涵盖数据类型、抽样、图表、平均数、箱线图、累积频率、概率和散点图。利用这份精讲夯实概念、优化应试策略。
1. Mock Paper Overview and Topic Distribution | 模拟试卷概览与考点分布
The mock paper contains nine questions totalling 50 marks, designed to be completed in 60 minutes. The topics are weighted as follows: data types and sampling (6 marks), frequency tables and bar charts (6 marks), stem‑and‑leaf diagrams (5 marks), averages and range (7 marks), quartiles and box plots (8 marks), cumulative frequency (8 marks), probability (4 marks) and scatter graphs (6 marks). Each question tests both calculation skills and the ability to interpret statistical information.
模拟卷共 9 道题，满分 50 分，建议用时 60 分钟。各主题分值分布为：数据类型与抽样（6 分）、频率表与柱状图（6 分）、茎叶图（5 分）、平均数与极差（7 分）、四分位数与箱线图（8 分）、累积频率（8 分）、概率（4 分）及散点图（6 分）。每道题均同时考查计算能力和统计信息的解读能力。
2. Identifying Data Types | 识别数据类型
Mock Question 1 (4 marks): Classify each as qualitative, discrete quantitative or continuous quantitative. (a) Favourite colour of a car. (b) Number of siblings. (c) Volume of milk in a carton (ml). (d) Score in a video game (0–100 scale).
模拟题 1（4 分）：判断下列各项为定性、离散定量还是连续定量。(a) 汽车最喜欢的颜色。(b) 兄弟姐妹人数。(c) 一盒牛奶的容量（毫升）。(d) 电子游戏得分（0–100 评分）。
Answer (a): Qualitative — ‘colour’ is a non‑numerical category. (b): Discrete quantitative — number of siblings is a countable whole number. (c): Continuous quantitative — volume is measured and can take any value within a range. (d): Discrete quantitative — the score uses integer steps and is counted, not measured on a continuous scale. (Note: in some contexts an integer rating can be treated as qualitative if the numbers are just labels, but here it represents a count of points.)
答案 (a)：定性——‘颜色’是非数值类别。(b)：离散定量——兄弟姐妹数是可数的整数。(c)：连续定量——容量经测量获得，可在区间内取任意值。(d)：离散定量——游戏得分取整数值，属于计数而非连续测量。（注意：某些情景下整数评分若仅作为标签可视为定性，但此处代表累积分数，故为离散定量。）
Common mistake: treating all numerical data as continuous. Always ask: ‘Can it take any value, or is it limited to fixed steps?’
常见错误：将所有数值数据都当作连续定量。务必自问：‘它能取任意值，还是限定在固定间隔内？’
3. Choosing an Appropriate Sampling Method | 选择适当的抽样方法
Mock Question 2 (6 marks): A football club has 800 adult members (500 male, 300 female). The club wants a sample of 80 members regarding new kit designs. (a) Describe how to obtain a stratified sample, and state how many males and females should be in the sample. (b) Give one advantage of stratified sampling over simple random sampling in this situation.
模拟题 2（6 分）：某足球俱乐部有 800 名成人会员（男性 500 人，女性 300 人）。俱乐部希望抽取 80 名会员调查新球衣设计。(a) 描述如何获得分层样本，并说出样本中男性和女性各应多少人。(b) 就本情景给出分层抽样相比于简单随机抽样的一个优点。
Answer (a): Stratified sampling: divide the population into distinct groups (strata) — here by gender. The sample size from each group is proportional to the group size in the population. For males: (500/800) × 80 = 50 males. For females: (300/800) × 80 = 30 females. Then select 50 males randomly from the male list and 30 females randomly from the female list.
答案 (a)：分层抽样：将总体分成不同组别（层）——此处按性别分层。每层样本数与总体中该层大小成比例。男性：(500/800)×80 = 50 人；女性：(300/800)×80 = 30 人。然后分别从男性名单和女性名单中随机抽取 50 名和 30 名。
(b): Advantage — stratified sampling guarantees that both genders are represented exactly in proportion to the population, which may give a more representative view on kit preferences than a simple random sample (which might, by chance, over‑ or under‑represent one gender).
(b)：优点——分层抽样确保两种性别都能严格按总体比例得到代表，在球衣偏好上能比简单随机抽样（可能偶然过多或过少包含某一性别）给出更具代表性的意见。
4. Frequency Tables and Bar Charts | 频率表与柱状图
Mock Question 3 (6 marks): The number of pets owned by 20 families is recorded: 0, 1, 2, 1, 0, 2, 3, 1, 0, 0, 2, 1, 1, 3, 0, 2, 2, 1, 0, 2. (a) Complete the frequency table. (b) Draw a bar chart to represent the data. (c) What is the modal number of pets?
模拟题 3（6 分）：记录了 20 个家庭拥有宠物数量：0, 1, 2, 1, 0, 2, 3, 1, 0, 0, 2, 1, 1, 3, 0, 2, 2, 1, 0, 2。(a) 完成频率表。(b) 画出柱状图。(c) 宠物数量的众数是多少？
Answer: The frequency table: Number of pets 0: frequency 6; 1: frequency 6; 2: frequency 6; 3: frequency 2. The bar chart must have vertical bars with gaps between them (discrete data), axes labelled and scaled correctly. The mode is 0, 1 and 2 (all occur 6 times — the data is multimodal).
答案：频率表：宠物数 0：频数 6；1：频数 6；2：频数 6；3：频数 2。柱状图必须画垂直柱且柱间有适当间隔（离散数据），坐标轴需标注并正确设定刻度。众数为 0、1 和 2（均出现 6 次，数据为多峰）。
Exam tip: do not draw a histogram — bar charts are for discrete or categorical data with gaps between bars. Histograms are for continuous grouped data with no gaps.
考试提示：切勿画成直方图——柱状图用于离散或类别数据，柱间留有空隙；直方图用于连续分组数据且柱间无空隙。
5. Stem‑and‑Leaf Diagrams | 茎叶图
Mock Question 4 (5 marks): The times (seconds) for 15 pupils to solve a puzzle are: 23, 31, 19, 25, 36, 22, 18, 40, 33, 27, 21, 35, 28, 24, 30. (a) Draw an ordered stem‑and‑leaf diagram. (b) Find the median time. (c) State the range.
模拟题 4（5 分）：15 名学生解谜题所用时间（秒）为：23, 31, 19, 25, 36, 22, 18, 40, 33, 27, 21, 35, 28, 24, 30。(a) 画出有序茎叶图。(b) 求时间的中位数。(c) 指出极差。
Answer (a): Ordered stem‑and‑leaf with key 1|9 = 19 s: 1 | 8 9; 2 | 1 2 3 4 5 7 8; 3 | 0 1 3 5 6; 4 | 0. (b) With 15 values, median is the 8th value = 25 seconds. (c) Range = 40 − 18 = 22 seconds.
答案 (a)：有序茎叶图，图例 1|9 = 19 秒：1 | 8 9；2 | 1 2 3 4 5 7 8；3 | 0 1 3 5 6；4 | 0。(b) 共 15 个数据，中位数为第 8 个值 = 25 秒。(c) 极差 = 40 − 18 = 22 秒。
Remember: always include a key, order the leaves, and count accurately to find the median from the stem‑and‑leaf.
切记：务必写出图例、将叶排序，并准确数位找中位数。
6. Mean, Median, Mode and Range | 平均数、中位数、众数与极差
Mock Question 5 (7 marks): The test marks for 12 students are: 14, 18, 12, 17, 19, 15, 12, 20, 16, 14, 18, 13. (a) Calculate the mean. (b) Find the median. (c) State the mode. (d) Work out the range. (e) Explain why the median might be more useful than the mean if an extra student scored 50.
模拟题 5（7 分）：12 名学生的测试成绩：14, 18, 12, 17, 19, 15, 12, 20, 16, 14, 18, 13。(a) 计算平均数。(b) 求中位数。(c) 指出众数。(d) 计算极差。(e) 若增加一名得 50 分的学生，为何中位数可能比平均数更有用？
Answers: (a) Sum = 14+18+12+17+19+15+12+20+16+14+18+13 = 188, mean = 188 ÷ 12 = 15.67 (to 2 d.p.). (b) Ordered: 12, 12, 13, 14, 14, 15, 16, 17, 18, 18, 19, 20; median = (15+16)/2 = 15.5. (c) Mode = 12, 14, 18 (all appear twice). (d) Range = 20 − 12 = 8. (e) The score 50 is an outlier — it would pull the mean up significantly, making it unrepresentative of the typical mark. The median would only shift slightly and remains a better measure of central tendency for skewed data.
答案：(a) 总和 = 188，平均数 = 188 ÷ 12 = 15.67（保留两位小数）。(b) 排序后中位数 = (15+16)/2 = 15.5。(c) 众数为 12、14 和 18（均出现两次）。(d) 极差 = 20 − 12 = 8。(e) 50 分是一个异常值——它会大幅度拉高平均数，使其不具有代表性。中位数仅会略微移动，因此对于偏态分布是更好的集中趋势度量。
7. Quartiles and Box Plots | 四分位数与箱线图
Mock Question 6 (8 marks): The heights (cm) of 10 boys in Year 10 are: 152, 160, 155, 172, 165, 158, 168, 163, 170, 159. (a) Find the median, lower quartile (Q₁), upper quartile (Q₃) and interquartile range (IQR). (b) Draw a box plot. (c) Another box plot for girls has median 161, IQR 12 and range 25. Compare the two distributions.
模拟题 6（8 分）：10 名 Year 10 男生身高（cm）：152, 160, 155, 172, 165, 158, 168, 163, 170, 159。(a) 求中位数、下四分位数 (Q₁)、上四分位数 (Q₃) 和四分位距 (IQR)。(b) 画箱线图。(c) 女生组的箱线图中位数 161，IQR 12，极差 25。比较两组分布。
Answer: Ordered: 152, 155, 158, 159, 160, 163, 165
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