Case Study Practice: CAIE Year 11 Statistics | CAIE Year 11 统计案例分析实战演练

📚 Case Study Practice: CAIE Year 11 Statistics | CAIE Year 11 统计案例分析实战演练

Statistical case studies bring data to life. In your CAIE Year 11 course, you need to not only perform calculations but also interpret results in context. This revision guide provides two complete case studies with step-by-step analysis, covering data handling, averages, spread, statistical diagrams, probability, correlation, and time series. Mastering these scenarios will strengthen your ability to tackle exam questions with confidence.

统计案例研究让数据变得生动。在 CAIE Year 11 课程中，你不仅需要计算，还需要在具体情境中解释结果。这篇复习指南提供了两个完整的案例研究，逐步分析，涵盖数据处理、平均值、离散度、统计图、概率、相关性和时间序列。掌握这些场景将增强你自信应对考试问题的能力。

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1. Introduction to Statistical Case Studies | 统计案例研究介绍

A case study in statistics is a detailed investigation of a real-world scenario using data. You might be given a set of numbers, a table, or a graph, and asked to summarise, compare, predict, or test ideas. The key is to follow a logical process: collect or identify data, organise it using tables and diagrams, calculate summary statistics, and finally interpret your findings in words.

统计案例研究是使用数据对现实世界场景进行详细调查。你可能会被给出一组数字、一张表格或一张图表，并被要求总结、比较、预测或检验想法。关键是要遵循逻辑过程：收集或识别数据，用表格和图表进行整理，计算汇总统计量，最后用文字解释你的发现。

In Year 11, typical tasks include constructing grouped frequency tables, drawing bar charts, histograms, cumulative frequency graphs, scatter diagrams, and box-and-whisker plots. You will also calculate averages, measures of spread, probabilities from tables, and even Spearman’s rank correlation. This article uses two extended examples to demonstrate these skills.

在 Year 11 中，典型任务包括构造分组频数表、绘制条形图、直方图、累积频率图、散点图和箱线图。你还需要计算平均值、离散量度、从表格计算概率，甚至斯皮尔曼秩相关系数。本文用两个扩展案例来展示这些技能。

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2. Case 1: Collecting and Organising Test Scores | 案例一：收集与整理考试成绩

Let us begin with a data set of Mathematics test scores (out of 50) from 30 Year 11 students. The raw scores are: 23, 35, 28, 42, 37, 29, 45, 48, 31, 27, 38, 41, 44, 33, 39, 34, 46, 25, 36, 40, 32, 30, 47, 26, 43, 24, 49, 22, 50, 21.

让我们从30名 Year 11 学生的数学测试成绩（满分50）开始。原始成绩如下：23, 35, 28, 42, 37, 29, 45, 48, 31, 27, 38, 41, 44, 33, 39, 34, 46, 25, 36, 40, 32, 30, 47, 26, 43, 24, 49, 22, 50, 21。

To organise this data, we create a grouped frequency table. Since the scores range from 21 to 50, we can use three class intervals: 20−29, 30−39, and 40−50. Note that for continuous data, the true class boundaries should be 19.5−29.5, 29.5−39.5, and 39.5−50.5 to avoid gaps between groups.

为了整理这些数据，我们创建一个分组频数表。由于分数范围从21到50，我们可以使用三个组距：20−29、30−39 和 40−50。请注意，对于连续数据，真正的组边界应为 19.5−29.5、29.5−39.5 和 39.5−50.5，以避免组间间隙。

Class Interval	Tally	Frequency, f
20−29	|||| ||||	10
30−39	|||| |||| |	11
40−50	|||| ||||	9

From the table we see that the modal class is 30−39 with 11 students. A simple bar chart or histogram can be drawn from this table, with frequency on the vertical axis and score on the horizontal.

从表中我们看到众数所在组是 30−39，有 11 名学生。从该表可绘制简单的条形图或直方图，纵轴为频数，横轴为分数。

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3. Measures of Central Tendency and Spread for Test Scores | 考试成绩的集中趋势和离散度量

Now we calculate the mean, median, mode, range, and interquartile range from the raw data. First, sum all scores: 23+35+28+…+21+50 = 1085. With 30 students, the mean is

现在我们根据原始数据计算平均值、中位数、众数、极差和四分位距。首先求和：23+35+28+…+21+50 = 1085。有30名学生，平均值为

Mean, x̄ = Σx / n = 1085 / 30 ≈ 36.17

To find the median, we list the scores in ascending order: 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50. Since n=30 is even, the median is the average of the 15th and 16th values: (35+36)/2 = 35.5.

为求中位数，我们将分数升序排列：21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50。由于 n=30 是偶数，中位数为第15和第16个值的平均数：(35+36)/2 = 35.5。

The mode is the most frequent score. In this data set, several scores appear only once; no value repeats noticeably, but the modal class was 30−39. For individual values, the data is roughly uniform, so we can say there is no obvious single mode.

众数是出现最频繁的分数。在这个数据集中，许多分数只出现一次；没有明显重复的值，但众数所在组是 30−39。对于单个值，数据大致均匀，因此可以说没有明显的单一众数。

For spread, the range = maximum − minimum = 50 − 21 = 29. To find the interquartile range (IQR), we locate Q1 and Q3. The lower half consists of the first 15 values: median of lower half (8th value) is 28. The upper half median (23rd value) is 43. Thus IQR = 43 − 28 = 15. This tells us the middle 50% of scores spans 15 marks.

离散度方面，极差 = 最大值 − 最小值 = 50 − 21 = 29。为求四分位距 (IQR)，我们找到 Q1 和 Q3。下半部分包含前 15 个值，其中位数（第8个值）为 28。上半部分中位数（第23个值）为 43。因此 IQR = 43 − 28 = 15。这说明中间 50% 的分数跨度为 15 分。

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4. Cumulative Frequency and Box Plots | 累积频率与箱线图

A cumulative frequency table helps us estimate the median and quartiles graphically and draw a box plot. Using the same class intervals and boundaries, we add a cumulative frequency column.

累积频率表帮助我们用图形方式估计中位数和四分位数，并绘制箱线图。使用相同的组距和边界，我们添加累积频率列。

Score (Boundaries)	Frequency, f	Cumulative Frequency
19.5 ≤ x < 29.5	10	10
29.5 ≤ x < 39.5	11	21
39.5 ≤ x ≤ 50.5	9	30

Plotting cumulative frequency against upper class boundaries gives an S-shaped curve. From the graph, you can estimate the median (at cumulative frequency 15) as roughly 34.5, consistent with our calculation. The lower quartile (at 7.5) is about 26, and the upper quartile (at 22.5) is about 42. These values are approximate because we used grouped data.

将累积频率相对上组边界描点会得到 S 形曲线。从图中，你可以估计中位数（累积频率 15 处）约为 34.5，与我们的计算一致。下四分位数（7.5 处）约为 26，上四分位数（22.5 处）约为 42。这些值是近似的，因为我们使用了分组数据。

A box-and-whisker plot can summarise the five-number summary: minimum = 21, Q1 ≈ 26, median = 35.5, Q3 ≈ 42, maximum = 50. The box shows the middle 50%, and whiskers extend to the extremes. This quickly reveals that scores are slightly positively skewed and there are no outliers.

箱线图可以概括五数总结：最小值 = 21，Q1 ≈ 26，中位数 = 35.5，Q3 ≈ 42，最大值 = 50。箱体表示中间 50%，触须延伸到极值。这可以迅速揭示分数略呈正偏态，且没有异常值。

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5. Probability Calculations from Data | 从数据中计算概率

Probability questions often appear in case studies. For example, if one student is selected at random, what is the probability that their score is greater than 40? There are 9 scores in the 40−50 group (including 50). Thus P(score > 40) = 9/30 = 0.3. Alternatively, using raw data, we count 12 scores above 40 (41 to 50) giving 12/30 = 0.4. Be careful: ‘greater than 40’ means 41,42,…,50, so P = 10/30 = 0.333… actually we have scores 41,42,43,44,45,46,47,48,49,50 = 10 numbers. Let us verify: 41-50 inclusive is 10 values, so P = 10/30 = 1/3 ≈ 0.333.

概率问题常出现在案例研究中。例如，如果随机挑选一名学生，其分数大于 40 的概率是多少？40−50 组有 9 个频数（包括 50）。因此 P(分数 > 40) = 9/30 = 0.3。或者用原始数据，数出大于 40 的有 10 个（41 到 50），所以 P = 10/30 = 1

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