📚 Year 11 CIE Statistics: Unit Test Mock Paper Walkthrough | Year 11 CIE 统计：单元测试模拟卷解析
Welcome to this detailed walkthrough of a typical Year 11 CIE Statistics unit test mock paper. We will break down the most common question types, provide step‑by‑step solutions, and expose the key statistical concepts and errors that can cost you marks. Whether you are sitting IGCSE Statistics 0480 or simply revising the core syllabus, this guide will strengthen your exam technique and deepen your understanding.
欢迎阅读这份典型的 Year 11 CIE 统计单元测试模拟卷的详细解析。我们将拆解最常见的题型，提供逐步求解过程，并揭示可能导致失分的核心统计概念与常见错误。不论你正准备 IGCSE Statistics 0480 考试还是巩固基础知识，本指南都将强化你的应试技巧并加深理解。
1. Data Types and Classification | 数据类型与分类
Question: A school survey records the following information for each student: (a) preferred learning style (visual, auditory, kinaesthetic), (b) number of books read last month, (c) time spent on homework per week in hours, (d) satisfaction rating from 1 to 5. Classify each variable as qualitative, quantitative discrete or quantitative continuous. Give a brief reason in each case.
问题：一项学校调查记录了每位学生的以下信息：(a) 偏好学习方式（视觉、听觉、动觉），(b) 上月阅读书籍数量，(c) 每周作业时间（小时），(d) 满意度评分（1 至 5）。将每个变量分类为定性、定量离散或定量连续，并简要说明理由。
Solution:
解答：
Preferred learning style is qualitative because it describes a non‑numerical attribute or category. Even if we code the styles with numbers, the data remain categorical.
偏好学习方式是定性变量，因为它描述的是非数值的属性或类别。即使我们用数字编码这些学习方式，数据本身仍然是分类型。
Number of books read is quantitative discrete. It arises from counting, takes only whole‑number values (0, 1, 2, …) and cannot be meaningfully subdivided.
阅读书籍数量是定量离散变量。它来自计数，只能取整数值（0, 1, 2, …），且无法进行有意义的细分。
Time spent on homework measured in hours is quantitative continuous. Time is measured on a continuous scale; a student could report 4.5 hours, and the variable can take any value within a realistic interval.
每周作业时间（小时）是定量连续变量。时间是在连续尺度上测量的；学生可能报告 4.5 小时，该变量可取其合理区间内的任意数值。
Satisfaction rating 1‑5 is quantitative discrete. Although it is often treated as ordinal in social sciences, in IGCSE statistics such a scale is treated as discrete numerical data because the rating takes only fixed integer values and arithmetic operations such as calculating a mean make sense.
满意度评分（1‑5）是定量离散变量。尽管在社会科学中它常被视为顺序变量，但在 IGCSE 统计中这类评分被视为离散数值数据，因为评分只能取固定的整数值，且计算均值等运算具有意义。
2. Frequency Distributions and Histograms | 频数分布与直方图
Question: The grouped frequency table shows the waiting times, t seconds, for 40 customers at a checkout.
| Waiting time, t (seconds) | Frequency |
|---|---|
| 20 ≤ t < 30 | 5 |
| 30 ≤ t < 40 | 10 |
| 40 ≤ t < 50 | 12 |
| 50 ≤ t < 70 | 8 |
| 70 ≤ t < 100 | 5 |
(a) Explain why frequency density must be used to construct a histogram for these data. (b) Calculate the frequency density for the interval 50 ≤ t < 70. (c) Describe one key feature of the histogram that would be observed if waiting times are generally short with a few extreme delays.
问题：分组频数表显示了 40 位顾客在收银台的等待时间 t（秒）。(a) 解释为何必须用频率密度绘制这些数据的直方图。(b) 计算区间 50 ≤ t < 70 的频率密度。(c) 若等待时间普遍较短但有少数极端延迟，描述直方图的一个关键特征。
Solution:
解答：
(a) The class widths are not equal; the last two intervals have widths of 20 and 30 seconds while the first three have width 10. In a histogram, the area of each bar represents frequency. If we plotted frequency directly on the vertical axis, wider intervals would appear disproportionately tall and mislead the eye. Frequency density (= frequency ÷ class width) corrects this by ensuring that area ∝ frequency.
(a) 组距并不相等；最后两个区间的宽度分别为 20 秒和 30 秒，而前三个区间的宽度为 10 秒。在直方图中，每个条形的面积代表频数。若直接在纵轴上标绘频数，较宽的区间会显得不成比例地高，产生视觉误导。频率密度（= 频数 ÷ 组距）通过保证面积与频数成正比来纠正这一点。
(b) For 50 ≤ t < 70, class width = 70 − 50 = 20 seconds. Frequency = 8. Therefore frequency density = 8 ÷ 20 = 0.4.
(b) 对于 50 ≤ t < 70，组距 = 70 − 50 = 20 秒。频数为 8。因此频率密度 = 8 ÷ 20 = 0.4。
(c) The histogram would be highly positively skewed; there would be a tall bar on the left for short waiting times and a long tail of very low frequency‑density bars stretching to the right, indicating the few extreme delays.
(c) 直方图会呈现明显的正偏态；左侧等待时间短的条形会很高，而右侧延伸出一条频率密度很低的“长尾”，反映出少数极端的延迟。
3. Measures of Central Tendency | 集中趋势的度量
Question: The hourly wages (£) of nine workers are: 8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00, 35.00 (manager). (a) Calculate the mean, median and mode. (b) The manager’s salary is an outlier. Which measure of central tendency best represents the typical worker’s wage? Justify your choice.
问题：九名工人的时薪（英镑）为：8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00, 35.00（经理）。(a) 计算均值、中位数和众数。(b) 经理的薪资是一个异常值。哪一个集中趋势的度量最能代表普通工人的薪资？说明理由。
Solution:
解答：
(a) Sum = 8.50 + 9.00 + 9.25 + 9.50 + 9.50 + 10.00 + 10.50 + 11.00 + 35.00 = 112.25. Mean = 112.25 ÷ 9 ≈ £12.47. Ordered list: 8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00, 35.00. Median is the 5th value = £9.50. Mode = £9.50 (appears twice).
(a) 总和 = 8.50 + … + 35.00 = 112.25。均值 = 112.25 ÷ 9 ≈ £12.47。排序后：8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00, 35.00。中位数为第 5 个值 = £9.50。众数 = £9.50（出现两次）。
(b) The outlier £35.00 inflates the mean to £12.47, which does not reflect the majority. The median (£9.50) is unaffected by the extreme value and lies near the centre of the bulk of the data. The mode is also £9.50, but the median is generally preferred in skewed distributions. Therefore the median best represents the typical wage.
(b) 异常值 £35.00 将均值拉高到 £12.47，不能反映大多数工人的情况。中位数（£9.50）不受极端值影响，处于大部分数据的中心位置。众数也是 £9.50，但在偏态分布中通常中位数更为可靠。因此中位数最能代表普通薪资。
4. Measures of Dispersion: Range, IQR and Standard Deviation | 离散程度的度量：极差、四分位距与标准差
Question: Using the same wage data (£): 8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00, 35.00. (a) Find the range and the interquartile range (IQR). (b) Calculate the standard deviation for the eight workers excluding the manager, i.e. the values 8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00. Comment on how the outlier would affect the standard deviation.
问题：使用同样的薪资数据（£）：8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00, 35.00。(a) 计算极差和四分位距（IQR）。(b) 计算除经理外八位工人的标准差，即数值 8.50, 9.00, 9.25, 9.50, 9.50, 10.00, 10.50, 11.00。说明异常值会如何影响标准差。
Solution:
解答：
(a) Range = maximum − minimum = 35.00 − 8.50 = £26.50. For IQR: ordered values as before. n = 9, so Q1 is at position (9+1)/4 = 2.5th; Q1 = (9.00+9.25)/2 = £9.125. Q3 is at 3(9+1)/4 = 7.5th; Q3 = (10.50+11.00)/2 = £10.75. IQR = Q3 − Q1 = 10.75 − 9.125 = £1.625.
(a) 极差 = 最大值 − 最小值 = 35.00 − 8.50 = £26.50。四分位距：数据已排序。n = 9，Q1 位于第 (9+1)/4 = 2.5 个位置；Q1 = (9.00+9.25)/2 = £9.125。Q3 位于第 3(9+1)/4 = 7.5 个位置；Q3 = (10.50+11.00)/2 = £10.75。IQR = 10.75 − 9.125 = £1.625。
(b) Excluding 35.00, the eight wages have mean = (8.50+9.00+9.25+9.50+9.50+10.00+10.50+11.00) ÷ 8 = 77.25 ÷ 8 ≈ £9.65625. Deviations squared: (8.50-9.656)²=1.337, (9.00-9.656)²=0.431, (9.25-9.656)²=0.165, (9.50-9.656)²=0.0244 (×2), (10.00-9.656)²=0.118, (10.50-9.656)²=0.711, (11.00-9.656)²=1.806. Sum of squares ≈ 4.617. Variance = 4.617 ÷ 8 ≈ 0.577. Standard deviation σ = √0.577 ≈ £0.76.
(b) 除去 35.00 后，八位工人的薪资均值为 (8.50+…+11.00) ÷ 8 = 77.25 ÷ 8 ≈ £9.65625。偏差平方：(8.50-9.656)²=1.337, … , (11.00-9.656)²=1.806。平方和 ≈ 4.617。方差 = 4.617 ÷ 8 ≈ 0.577。标准差 σ = √0.577 ≈ £0.76。
If the manager were included, the standard deviation would be pulled dramatically upward (to about £8.07) because the squared deviation of 35.00 from the mean of £12.47 is enormous. The range and standard deviation are very sensitive to outliers, whereas the IQR remains small and resistant.
若包含经理薪资，标准差会被大幅拉高（升至约 £8.07），因为 35.00 与均值 £12.47 的偏差平方极大。极差和标准差对异常值非常敏感，而 IQR 保持较小的值，具有较强的抗干扰性。
5. Cumulative Frequency and Box Plots | 累积频数与箱线图
Question: The grouped frequency table below shows the time, m minutes, taken by 50 students to complete a puzzle.
| Time (m minutes) | Frequency |
|---|---|
| 0 ≤ m < 5 | 6 |
| 5 ≤ m < 10 | 14 |
| 10 ≤ m < 15 | 18 |
| 15 ≤ m < 20 | 更多咨询请联系16621398022（同微信）
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