GCSE WJEC Statistics: Unit Test Mock Paper Walkthrough | GCSE WJEC 统计：单元测试模拟卷解析

📚 GCSE WJEC Statistics: Unit Test Mock Paper Walkthrough | GCSE WJEC 统计：单元测试模拟卷解析

Practising with mock papers is one of the most effective ways to prepare for your GCSE WJEC Statistics exam. This walkthrough will guide you through a typical unit test, breaking down common question types and showing you the best approaches to tackle them confidently.

使用模拟卷进行练习是准备GSCE WJEC统计考试最有效的方法之一。本解析将带领你完成一份典型单元测试，分解常见题型，并展示自信应对它们的最佳方法。

1. Understanding the Mock Paper Structure | 了解模拟卷结构

A typical WJEC GCSE Statistics unit test covers data handling, averages, probability, sampling, and diagram interpretation. The mock paper we are analysing here consists of both short-answer and multi-step questions, testing not only your calculation skills but also your ability to explain statistical reasoning. Understanding the mark allocation helps you manage your time – for instance, questions that ask for interpretation usually carry more marks. Before diving into each question, always scan the entire paper to identify which topics appear.

典型的WJEC GCSE统计单元测试涵盖数据处理、平均数、概率、抽样和图表解读。我们正在分析的这份模拟卷既包含简答题，也包含多步骤问题，不仅考验你的计算能力，还考验你解释统计推理的能力。了解分数分配有助于你管理时间——例如，要求解释的问题通常分值更高。在深入每道题之前，务必快速浏览整份试卷，确定出现了哪些主题。

2. Question 1: Types of Data | 题目1：数据类型

This opening question often asks you to classify data as qualitative or quantitative, and further as discrete or continuous. For example, state whether each variable is qualitative or quantitative, and if quantitative, state whether it is discrete or continuous: a) Number of pets in a household; b) Colour of cars in a car park; c) Height of students in a class; d) Shoe size. Qualitative data are non-numerical categories like ‘colour’. Quantitative data are numerical. Discrete quantitative data can only take distinct values (usually counts), while continuous data can take any value within a range. Thus, (a) number of pets is quantitative discrete; (b) colour of cars is qualitative; (c) height is quantitative continuous; (d) shoe size is discrete because sizes come in set steps, not all possible decimal values. Always justify your choice – a common mistake is treating shoe size as continuous.

这类开场题通常要求你将数据分类为定性或定量，并进一步分为离散或连续。例如，判断以下变量是定性还是定量，如果是定量，指明是离散还是连续：a) 家庭宠物数量；b) 停车场汽车颜色；c) 班级学生身高；d) 鞋码。定性数据是非数值的类别，如 ‘颜色’。定量数据是数值型数据。离散定量数据只能取特定值（通常是计数），而连续数据可以取范围内的任意值。因此，(a) 宠物数量是定量离散数据；(b) 汽车颜色是定性数据；(c) 身高是定量连续数据；(d) 鞋码是离散的，因为鞋码是按固定步长设定的，并非所有小数值都可能出现。务必说明理由——一个常见错误是将鞋码当作连续数据。

3. Question 2: Averages from Frequency Tables | 题目2：从频率表求平均数

A grouped frequency table shows ages of participants: 10 ≤ a < 15 (f=4), 15 ≤ a < 20 (f=7), 20 ≤ a < 25 (f=6), 25 ≤ a < 30 (f=3). To estimate the mean, first find the midpoint of each class: 12.5, 17.5, 22.5, 27.5. Multiply each midpoint by its frequency (fx): 12.5×4=50, 17.5×7=122.5, 22.5×6=135, 27.5×3=82.5. Sum these products: 50+122.5+135+82.5 = 390. Total frequency Σf = 20. Estimated mean = 390/20 = 19.5. The formula is:

x̄ = Σfx / Σf

This is an estimate because we assume all values in a class are at the midpoint. Write down the fx column clearly to gain method marks even if the final answer is slightly off.

一个分组频数表显示参与者年龄：10 ≤ a < 15 (f=4)、15 ≤ a < 20 (f=7)、20 ≤ a < 25 (f=6)、25 ≤ a < 30 (f=3)。要估算平均数，首先找出每组组中值：12.5, 17.5, 22.5, 27.5。将每个组中值乘以频数(fx)：12.5×4=50, 17.5×7=122.5, 22.5×6=135, 27.5×3=82.5。求和：50+122.5+135+82.5 = 390。总频数 Σf = 20。估算平均数 = 390/20 = 19.5。公式为：

x̄ = Σfx / Σf

这是一个估计值，因为我们假设组内所有数值都位于组中值。清晰写出 fx 列，即使最终答案稍有偏差也能拿到步骤分。

4. Question 3: Measures of Spread – Range and Interquartile Range | 题目3：离散程度——全距和四分位距

Given the data set: 13, 18, 14, 12, 20, 25, 15, 11, 19, 22. Arrange in ascending order: 11, 12, 13, 14, 15, 18, 19, 20, 22, 25. Range = maximum − minimum = 25 − 11 = 14. To find the interquartile range (IQR), we need Q1 and Q3. For 10 values, Q1 is at position (10+1)/4 = 2.75, so average of 2nd and 3rd values: (12+13)/2 = 12.5. Q3 is at position 3(10+1)/4 = 8.25, average of 8th and 9th values: (20+22)/2 = 21. IQR = Q3 − Q1 = 21 − 12.5 = 8.5. The IQR measures the spread of the middle 50% and is less affected by outliers than the range.

给定数据集：13, 18, 14, 12, 20, 25, 15, 11, 19, 22。按升序排列：11, 12, 13, 14, 15, 18, 19, 20, 22, 25。全距 = 最大值 – 最小值 = 25 – 11 = 14。为求四分位距(IQR)，需要 Q1 和 Q3。10个数时，Q1 位置为 (10+1)/4 = 2.75，取第2和第3值的平均数：(12+13

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