Year 10 Eduqas Statistics: Unit Test Mock Paper Walkthrough | Eduqas 十年级统计：单元测试模拟卷解析

📚 Year 10 Eduqas Statistics: Unit Test Mock Paper Walkthrough | Eduqas 十年级统计：单元测试模拟卷解析

This walkthrough takes you through a full Unit Test mock paper for Year 10 Eduqas GCSE Statistics. We break down each question type, covering data types, sampling, charts, averages, spread, probability and correlation, with clear step-by-step solutions and exam tips. Use this guide to identify key skills and avoid common mistakes.

本解析将带你完整梳理一份十年级 Eduqas GCSE 统计单元测试模拟卷。我们逐一拆解各类题型，涵盖数据类型、抽样、图表、集中趋势与离散度量、概率及相关性，并提供清晰的分步解答与考试建议。使用本指南，你可以掌握核心技能并规避常见错误。

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1. Mock Paper Overview | 模拟卷概览

This unit test reflects the style and demand of Eduqas GCSE Statistics papers. It contains a mix of multiple-choice style short questions, structured data response tasks and two extended problems. The total mark is 50, to be completed in 45 minutes, with command words such as ‘state’, ‘calculate’, ‘interpret’ and ‘compare’ used throughout.

本单元测试模拟了 Eduqas GCSE 统计试卷的风格与要求。试卷包含类似选择题的简答题、结构化数据回应任务及两道拓展题，总分 50 分，需要在 45 分钟内完成，全卷使用了“陈述”“计算”“解读”“比较”等指令词。

Question	Topic (EN)	主题 (CN)	Marks
1	Data types & classification	数据类型与分类	4
2	Sampling methods (stratified)	抽样方法（分层抽样）	5
3	Bar charts and pie charts	条形图与饼图	6
4	Mean, median, mode	平均数、中位数、众数	7
5	Range and interquartile range	极差与四分位距	6
6	Basic probability	基础概率	5
7	Scatter graphs & correlation	散点图与相关	5
8	Combined data tasks (frequency table)	综合数据任务（频数表）	7
9	Interpret and compare	解读与比较	5

Candidates are expected to show all working clearly and to use correct statistical vocabulary such as ‘discrete’, ‘continuous’, ‘representative’, ‘skew’ and ‘correlation’. The mark scheme rewards method marks even if the final answer is inaccurate.

考生应清晰展示所有计算步骤，并使用正确的统计术语，如“离散”“连续”“代表性”“偏态”“相关”等。评分方案即使最终答案有误，只要方法正确也会给步骤分。

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2. Data Types: Qualitative vs Quantitative | 数据类型：定性数据与定量数据

Mock Question 1: A questionnaire asks students to state the number of brothers they have, their favourite subject, the time they spend on homework each night, and their opinion of school dinners (rated ‘Excellent, Good, Fair, Poor’). For each variable, classify the data type as qualitative or quantitative. For quantitative variables, state whether they are discrete or continuous. [4 marks]

模拟题 1：一份问卷要求学生填写以下信息：兄弟姊妹的个数、最喜欢的科目、每晚做作业的时间以及对学校午餐的评价（选项为“极好、好、一般、差”）。请对每个变量进行分类，指出是定性数据还是定量数据；对于定量数据，进一步说明是离散型还是连续型。[4 分]

Number of brothers: quantitative, discrete. It is a countable numerical value that can only take whole numbers (0, 1, 2, …).

兄弟姊妹个数：定量数据，离散型。这是可数的数值，只能取整数值（0, 1, 2 …）。

Favourite subject: qualitative (categorical). The responses are names of subjects and cannot be measured numerically.

最喜欢的科目：定性数据（分类数据）。回答是科目的名称，无法用数值测量。

Time spent on homework: quantitative, continuous. Time can take any value within a range and can be measured to any degree of accuracy (e.g. 1.5 hours, 2.25 hours).

做作业的时间：定量数据，连续型。时间可以在一个区间内取任意值，并且可以按照任意精度测量（例如 1.5 小时、2.25 小时）。

Opinion of school dinners: qualitative, but also ordinal. Although the categories have a natural order, the data are not numerical, so they are treated as qualitative. Accept ‘qualitative’ or ‘ordinal categorical’.

对学校午餐的评价：定性数据，但也可视为有序分类数据。尽管类别有自然顺序，但由于不是数值，仍按定性数据处理。答案为“定性”或“有序分类”皆可得分。

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3. Sampling: Stratified Sampling Calculation | 抽样：分层抽样计算

Mock Question 2: A school has 820 students: 420 boys and 400 girls. The headteacher wants to survey a stratified sample of 80 students about after-school activities. Calculate how many boys and how many girls should be included in the sample. Show your working. [5 marks]

模拟题 2：一所学校共有 820 名学生，其中男生 420 人，女生 400 人。校长希望通过分层抽样选取 80 名学生进行课后活动调查。计算样本中应包含男生和女生各多少人，并写出计算过程。[5 分]

We use the formula: number in sample for stratum = (stratum size / total population) × overall sample size. For boys: (420 / 820) × 80. First work out 420 ÷ 820 = 21/41. Then 21/41 × 80 = 1680 ÷ 41 ≈ 40.9756. Since we cannot have a fraction of a person, we round to the nearest whole number: 41 boys.

使用公式：分层抽样人数 =（层内个体数 ÷ 总体总数）× 样本容量。男生部分： (420 ÷ 820) × 80。先计算 420 ÷ 820 = 21/41，再计算 21/41 × 80 = 1680 ÷ 41 ≈ 40.9756。由于不能抽取非整数人数，四舍五入得到 41 名男生。

For girls: (400 / 820) × 80. 400 ÷ 820 = 40/82 = 20/41. Then 20/41 × 80 = 1600 ÷ 41 ≈ 39.0244, which rounds to 39 girls. Check: 41 + 39 = 80, which matches the required sample size.

女生部分： (400 ÷ 820) × 80。400 ÷ 820 = 20/41，20/41 × 80 = 1600 ÷ 41 ≈ 39.0244，四舍五入得 39 名女生。检验：41 + 39 = 80，与所需样本容量一致。

Always state the rounding explicitly and verify that the sum equals the target sample size. If rounding causes a mismatch, adjust one of the groups to match exactly.

必须明确写出四舍五入过程并核对总和是否等于目标样本量。若因四舍五入产生偏差，应调整其中一个组别的人数以完全匹配。

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4. Charts: Bar Charts and Pie Charts | 图表：条形图与饼图

Mock Question 3: The table shows the favourite sport of 40 students. Football: 12, Netball: 10, Swimming: 8, Tennis: 6, Other: 4. (a) Draw a bar chart for this data. (b) Explain why a pie chart would be suitable, and calculate the angle for Tennis. [6 marks]

模拟题 3：下表列出了 40 名学生最喜欢的体育运动：足球 12 人，无挡板篮球 10 人，游泳 8 人，网球 6 人，其他 4 人。(a) 根据数据绘制条形图。(b) 解释为何饼图适合于此数据，并计算网球扇形的角度。[6 分]

For (a), a bar chart must have the categories along the horizontal axis, frequencies along the vertical axis, equal-width bars with gaps, and labels. The vertical axis should start at zero and be evenly scaled. We can describe: the highest bar is Football (12), followed by Netball (10), Swimming (8), Tennis (6) and Other (4). All bars are separated because the data are categorical.

(a) 部分：条形图的横轴为类别，纵轴为频数，条宽相等且条间留有空隙，并有标题和坐标轴标签。纵轴应从零开始、均匀刻度。我们可以描述：最高的条形是足球（12），其次为无挡板篮球（10）、游泳（8）、网球（6）和其他（4）。由于数据是分类数据，所有条形彼此分离。

For (b), a pie chart is appropriate because it shows the proportion of each category out of the whole. The total angle is 360°. Tennis frequency = 6, total = 40. Angle = (6 / 40) × 360° = 0.15 × 360° = 54°. Always write the formula, substitute numbers and give the unit °.

(b) 部分：饼图合适，因为它能直观显示每个类别占整体的比例。总角度为 360°。网球频数为 6，总和为 40。角度 = (6 ÷ 40) × 360° = 0.15 × 360° = 54°。务必写出公式、代入数值并标明单位 °。

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5. Measures of Central Tendency: Mean, Median, Mode | 集中趋势度量：平均数、中位数、众数

Mock Question 4: The ages of participants in a youth club are: 12, 15, 14, 10, 18, 15, 13. Find the mean, median and mode of this data set. [7 marks]

模拟题 4：一个青年俱乐部成员的年龄数据如下：12, 15, 14, 10, 18, 15, 13。求该数据集的平均数、中位数和众数。[7 分]

Mean: Add all values: 12 + 15 + 14 + 10 + 18 + 15 + 13 = 97. There are n = 7 values. Mean = 97 / 7 ≈ 13.857. Use the symbol x̄ and show the formula x̄ = Σx / n.

平均数：所有数值相加：12 + 15 + 14 + 10 + 18 + 15 + 13 = 97。数据个数 n = 7。平均数 = 97 ÷ 7 ≈ 13.857。使用符号 x̄，并写出公式 x̄ = Σx / n。

x = 97 / 7 = 13.857 (to 3 d.p.)

Median: First order the data: 10, 12, 13, 14, 15, 15, 18. The median is the middle value, which is the 4th value in an ordered list of 7 items. Median = 14. Explain that half the values are below 14 and half are above.

中位数：先将数据排序：10, 12, 13, 14, 15, 15, 18。中位数是位于中间的值，在 7 个数据中为第 4 个值。中位数 = 14。可以解释为有一半的数值小于 14，一半大于 14。

Mode: The value that appears most often is 15 (appears twice, while all others appear once). So mode = 15. If there were multiple modes, we would list them all. This data set has one mode.

众数：出现次数最多的数值是 15（出现两次，其余均出现一次）。因此众数为 15。若有多个众数，应全部列出。本数据集只有一个众数。

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6. Measures of Spread: Range and Interquartile Range | 离散度量：极差与四分位距

Mock Question 5: Using the same data set (10, 12, 13, 14, 15, 15, 18), find the range and the interquartile range (IQR). Show your method clearly. [6 marks]

模拟题 5：针对同一数据集（10, 12, 13, 14, 15, 15, 18），求极差和四分位距（IQR），并清晰展示计算过程。[6 分]

Range = Maximum – Minimum = 18 – 10 = 8. The range is very sensitive to extreme values, but it gives a quick measure of total spread.

极差 = 最大值 – 最小值 = 18 – 10 = 8。极差对极端值非常敏感，但它能快速反映数据的总离散程度。

To find the IQR, first locate the quartiles. Median (Q₂) = 14. The lower half (before median) is 10, 12, 13. The median of this lower half is Q₁ = 12. The upper half (after median) is 15, 15, 18, giving Q₃ = 15. IQR = Q₃ – Q₁ = 15 – 12 = 3. The IQR represents the spread of the middle 50% of the data and is not affected by outliers.

计算四分位距需要先确定四分位数。中位数（Q₂）= 14。前一半数据（中位数之前）为 10, 12, 13，其中位数为 Q₁ = 12。后一半数据为 15, 15, 18，其中位数为 Q₃ = 15。IQR = Q₃ – Q₁ = 15 – 12 = 3。IQR 表示中间 50% 数据的分散程度，且不受异常值影响。

In unequal splits, you may need to exclude the median itself when forming the halves. Here with n=7, we exclude the 4th value (the median) to get two groups of three numbers each. Always show this step clearly.

当数据个数为奇数时，在划分为两半时通常应剔除中位数本身。此例 n=7，剔除第 4 个值（中位数），得到两组各三个数据。务必清晰展示这一步骤。

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7. Basic Probability from a Word | 基础概率：基于单词的计算

Mock Question 6: The letters of the word STATISTICS are written on individual cards and placed in a bag. One card is chosen at random. Find the probability that the card shows (a) the letter S, (b) a vowel, (c) a consonant. [5 marks]

模拟题 6：将单词 STATISTICS 的每个字母分别写在卡片上，放入袋中。随机抽取一张卡片。求抽到的卡片显示 (a) 字母 S， (b) 元音字母，(c) 辅音字母的概率。[5 分]

First count the total number of letters: S T A T I S T I C S. There are 10 letters. Frequencies: S = 3, T = 3, A = 1, I = 2, C = 1. Vowels are A and I. Consonants are S, T, C.

首先统计字母总数：S T A T I S T I C S，共 10 个字母。各字母频数：S=3，T=3，A=1，I=2，C=1。元音字母为 A 和 I，辅音字母为 S、T、C。

(a) P(S) = number of S / total = 3 / 10.

(a) P(S) = S 的个数 / 总数 = 3 / 10。

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