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

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

Welcome to this step-by-step walkthrough of a GCSE AQA Statistics unit test mock paper. This mock paper has been designed to reflect the style and difficulty of the actual AQA exam. In this article, we will tackle ten representative questions, each targeting a key topic from the specification. By working through these problems, you will consolidate your understanding of data types, sampling, summary statistics, graphical representations, probability, and time series analysis. Whether you are preparing for a school unit test or revising for your final GCSE, this guide will boost your confidence and improve your problem-solving skills.

欢迎阅读这份 GCSE AQA 统计单元测试模拟卷的逐步解析。这份模拟卷旨在反映实际 AQA 考试的题型和难度。在本文中，我们将解决十个代表性题目，每个题目针对考纲中的一个关键主题。通过完成这些问题，你将巩固对数据类型、抽样、汇总统计、图形表示、概率和时间序列分析的理解。无论你是在准备学校的单元测试还是为最终 GCSE 复习，这份指南都将增强你的信心并提高解题能力。

1. Question 1: Classifying Data | 问题1：数据分类

Question: Classify each of the following variables as qualitative or quantitative. If quantitative, state whether it is discrete or continuous. (a) The height of a tree (b) Number of pets owned (c) Favourite music genre (d) Time spent on homework per week.

问题：将以下每个变量分类为定性或定量。如果是定量，说明它是离散还是连续。(a) 树的高度 (b) 拥有的宠物数量 (c) 最喜欢的音乐类型 (d) 每周花在作业上的时间。

Solution: (a) Height is numerical and can take any value within a range, so it is quantitative continuous. (b) Number of pets is numerical but can only be whole numbers; it is quantitative discrete. (c) Music genre describes categories, so it is qualitative (categorical). (d) Time is numerical and can be any value, so quantitative continuous. Always check whether a numerical variable can be measured on a continuous scale or counted in distinct steps.

解析：(a) 高度是数值，可以取范围内的任何值，因此是定量连续型。(b) 宠物数量是数值但只能是整数；是定量离散型。(c) 音乐类型描述类别，因此是定性（分类）。(d) 时间是数值，可以是任何值，因此是定量连续型。始终检查数值变量是否可以在连续尺度上测量还是只能以离散步骤计数。

2. Question 2: Sampling Methods | 问题2：抽样方法

Question: A school wants to find out students' opinions on school meals. There are 1200 students across five year groups, each with a different number of students. Describe a suitable sampling method and explain why it is better than a simple random sample in this context.

问题：一所学校想了解学生对校餐的意见。该校共有1200名学生，分布在五个年级，每个年级的学生人数不同。描述一种合适的抽样方法，并解释为什么在这种情况下它比简单随机抽样更好。

Solution: A stratified sample would be appropriate. First, the population is divided into strata by year group. Then a random sample is taken from each year group in proportion to its size. This ensures each year group is fairly represented. Compared to a simple random sample, stratified sampling guarantees that the sample reflects the demographic structure, reducing the risk of underrepresenting smaller year groups and making the results more reliable. A simple random sample might by chance miss out some year groups completely or overrepresent another.

解析：分层抽样是合适的。首先，将总体按年级分组。然后从每个年级按比例随机抽取样本。这确保了每个年级都能公平代表。与简单随机抽样相比，分层抽样保证了样本能反映人口结构，降低了较小年级代表性不足的风险，使结果更可靠。简单随机抽样可能偶然完全遗漏某些年级或过度代表另一些。

3. Question 3: Mean from a Grouped Frequency Table | 问题3：从分组频数表计算平均值

Question: The table below shows the ages of people at a local library on a Saturday morning. Calculate an estimate for the mean age.

问题：下表显示了周六早晨当地图书馆中人们的年龄。计算平均年龄的估计值。

, 35×6=210. Sum of fx = 20+120+300+210 = 650. Total frequency n = 4+8+12+6 = 30. Use the formula for the estimated mean.

解析：首先，找出每个组的中点：5、15、25、35。每个中点乘以相应频数得到fx：5×4=20, 15×8=120, 25×12=300, 35×6=210。∑fx = 20+120+300+210 = 650。总频数 n = 30。使用估计平均值公式。

Estimated mean = Σfx / n = 650 / 30 ≈ 21.7 years (to 1 d.p.)

Remember, because we use midpoints, this is an estimate and not the exact mean.

请记住，由于使用了组中点，这只是估计值，并非精确平均值。

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4. Question 4: Cumulative Frequency and Quartiles | 问题4：累积频率与四分位数

Question: The following frequency table shows the test scores of 50 students. Complete the cumulative frequency column and draw a cumulative frequency graph. Use your graph to estimate the median and the interquartile range.

问题：以下频数表显示了50名学生的测试分数。完成累积频率列并绘制累积频率图。利用你的图估计中位数和四分位距。

curve. The median is the score corresponding to the 25th value (50/2). From the graph, median ≈ 65. The lower quartile is the 12.5th value, LQ ≈ 56; the upper quartile is the 37.5th value, UQ ≈ 74. The interquartile range (IQR) = UQ - LQ ≈ 74 - 56 = 18.

解析：将上组界对应累积频率描点：(50,6), (60,16), (70,32), (80,44), (90,50)。用光滑曲线连接各点。中位数对应第25个值（50/2），从图中中位数≈65。下四分位数为第12.5个值，LQ≈56；上四分位数为第37.5个值，UQ≈74。四分位距 IQR = UQ - LQ ≈ 18。

5. Question 5: Box Plot and Comparison | 问题5：箱形图与比较

imum of 45, LQ=60, median=70, UQ=80, maximum=95. Draw both box plots on the same scale and compare the two distributions.

问题：利用问题4的汇总统计量（最小值=40, LQ=56, 中位数=65, UQ=74, 最大值=90）绘制箱形图。另一个班级的最小值为45, LQ=60, 中位数=70, UQ=80, 最大值=95。在同一尺度上绘制两个箱形图，并比较两个分布。

Solution: For the first class, draw a box from 56 to 74 with a line at 65, whiskers to 40 and 90. For the second class, box from 60 to 80, median line at 70, whiskers to 45 and 95. Comparing them: the second class has a higher median (70 vs 65), indicating better central performance. The IQR for the second class is slightly larger (20 vs 18), showing a bit more variation. Both distributions are roughly symmetric, but the second class is shifted to the right overall.

解析：第一个班级，从56到74画箱，中位线在65，触须延伸至40和90。第二个班级，从60到80画箱，中位线在70，触须至45和95。比较：第二个班级中位数更高（70对65），表明整体成绩更好。第二个班级的IQR略大（20对18），显示分散程度稍高。两个分布大致对称，但第二个班级整体右移。

6. Question 6: Scatter Diagram and Line of Best Fit | 问题6：散点图与最佳拟合线

Question: The table shows the number of hours ten students spent revising and their test scores. (a) Plot a scatter diagram. (b) Describe the correlation. (c) Draw a line of best fit by eye. (d) Use your line to estimate the test score for a student who revised 4.5 hours.

问题：该表显示了10名

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