📚 AS CIE Statistics: Case Study Practical Exercises | AS CIE 统计：案例分析实战演练
Welcome to this focused case study, where we explore a single dataset through the lens of AS CIE Statistics (9709). The 30 test scores (out of 50) recorded by a teacher will serve as our practical lab. By applying techniques from data representation to probability distributions, you will learn how exam questions weave multiple concepts together and how to approach them step by step.
欢迎来到这个集中的案例分析，我们将通过 AS CIE 统计学（9709）的视角探索一个数据集。一位教师记录的 30 个考试成绩（满分 50）将作为我们的实战实验室。通过应用从数据表示到概率分布的各种技巧，你将了解考题如何把多个概念编织在一起，以及如何逐步解决它们。
1. Understanding the Dataset | 理解数据集
Our raw data consists of the following scores: 12, 15, 18, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 44, 46, 48, 50. Although each score is a whole number, in statistical analysis we often treat boundaries as continuous to construct histograms and cumulative frequency curves.
我们的原始数据由以下分数组成：12, 15, 18, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 44, 46, 48, 50。尽管每个分数都是整数，但在统计分析中我们通常将边界视为连续值，以便构造直方图与累积频率曲线。
Before jumping into calculations, always clarify the context: the variable is ‘test score out of 50’, and the sample size n = 30. This small dataset is ideal for manual methods often tested on Paper 5 (S1).
在开始计算之前，务必要明确背景：变量是“满分 50 的考试成绩”，样本量 n = 30。这个小数据集非常适合 Paper 5 (S1) 中常考的手工方法。
2. Stem-and-Leaf Diagrams: Sorting Data | 茎叶图：数据排序
A stem-and-leaf plot simultaneously sorts the data and reveals its shape. We set the stem as the tens digit and the leaf as the units digit. With stems 1, 2, 3, 4, 5 we obtain the ordered display shown below.
茎叶图可同时对数据进行排序并揭示其形状。我们将十位数字设为茎，个位数字设为叶。使用茎 1、2、3、4、5，我们得到如下有序展示。
Stem-and-leaf plot:
1 | 2 5 8
2 | 0 1 2 3 4 5 5 6 7 8 9
3 | 0 1 2 3 4 5 6 7 8 9
4 | 0 2 4 6 8
5 | 0
From this plot we can immediately identify the minimum (12), maximum (50), and see that the bulk of the scores lie in the 20s and 30s. There is a slight positive skew, with a tail stretching towards the high 40s.
从该图我们可立即看出最小值（12）、最大值（50），并发现大部分分数集中在 20 多分和 30 多分。分布呈轻微正偏态，尾部延伸至 40 多分的高分段。
3. Box Plots and Outlier Detection | 箱形图与异常值检测
To construct a box plot we need the five‑number summary. Using the median position (n+1)/2 = 15.5, the median is the average of the 15th and 16th ordered values: (30 + 31)/2 = 30.5. For Q1, position = (30+1)/4 = 7.75, so Q1 = 23 + 0.75×(24−23) = 23.75. For Q3, position = 3×(30+1)/4 = 23.25, so Q3 = 38 + 0.25×(39−38) = 38.25.
要构建箱形图，我们需要五数概括。中位数位置为 (n+1)/2 = 15.5，中位数是第 15 与第 16 个有序值的平均数：(30+31)/2 = 30.5。对于下四分位数 Q1，位置 = (30+1)/4 = 7.75，则 Q1 = 23 + 0.75×(24−23) = 23.75。对于上四分位数 Q3，位置 = 3×(30+1)/4 = 23.25，则 Q3 = 38 + 0.25×(39−38) = 38.25。
The interquartile range IQR = Q3 − Q1 = 14.5. Fences for outliers: lower fence = Q1 − 1.5×IQR = 23.75 − 21.75 = 2; upper fence = Q3 + 1.5×IQR = 38.25 + 21.75 = 60. No scores fall outside [2, 60], so there are no outliers. The box plot will have whiskers reaching the minimum 12 and maximum 50.
四分位距 IQR = Q3 − Q1 = 14.5。异常值的界限：下界 = Q1 − 1.5×IQR = 23.75 − 21.75 = 2；上界 = Q3 + 1.5×IQR = 38.25 + 21.75 = 60。没有分数落在 [2, 60] 之外，因此不存在异常值。箱形图的触须将分别延伸至最小值 12 与最大值 50。
4. Histograms: Grouping for Distribution Shape | 直方图：分组展示分布形状
We group the scores into equal-width intervals: 10–19, 20–29, 30–39, 40–50. Since each class has the same width (10), the heights of the histogram bars can simply represent frequency. However, remembering the rule frequency = frequency density × class width is essential for unequal-width intervals.
我们将分数分组为等宽区间：10–19、20–29、30–39、40–50。因为每个组距相同（均为 10），直方图柱的高度可直接表示频数。但请记住，对于不等宽分组，必须使用规则 频数 = 频数密度 × 组距。
| Score | Frequency | Frequency density |
|---|---|---|
| 10–19 | 3 | 0.3 |
| 20–29 | 11 | 1.1 |
| 30–39 | 10 | 1.0 |
| 40–50 | 6 | 0.6 |
Notice that the area of each bar corresponds to frequency. The peak density occurs in the 20–29 interval, confirming the modal class. Such a histogram supports the earlier observation of a right‑skewed shape.
请注意，每个柱的面积对应频数。最高密度出现在 20–29 区间，确认了众数所在组。这样的直方图印证了先前右偏分布的观察。
5. Cumulative Frequency Curves and Percentile Estimation | 累积频率曲线与百分位数估计
Using upper class boundaries (19.5, 29.5, 39.5, 50.5) and cumulative frequencies (3, 14, 24, 30), we can plot a cumulative frequency curve. From the curve we can estimate any percentile by reading off the value corresponding to the desired cumulative frequency.
使用上组界（19.5, 29.5, 39.5, 50.5）与累积频数（3, 14, 24, 30），我们可以绘制累积频率曲线。通过该曲线，可以读取与所需累积频数对应的值，进而估计任何百分位数。
For example, the 80th percentile corresponds to a cumulative frequency of 0.80 × 30 = 24. Tracing across from 24 on the vertical axis, the curve gives approximately 39.5 (the upper boundary of the 30–39 class), indicating that 80% of students scored around 39.5 or below.
例如，第 80 百分位数对应累积频数 0.80 × 30 = 24。从纵轴 24 处追踪，曲线给出约为 39.5（即 30–39 组的上界），这表明约 80% 的学生分数在 39.5 或以下。
This graphical method is a core skill in S1: you must be able to draw smooth curves through points (0,0) and the upper boundaries, and then interpret percentiles, the median, and the inter‑percentile range.
这种图形化方法是 S1 中的核心技能：你必须能够通过 (0,0) 和上组界点绘制平滑曲线，然后解读百分位数、中位数以及百分位距。
6. Central Tendency: Mean, Median and Mode | 集中趋势：均值、中位数与众数
We calculate the mean using the formula x = Σx / n. Summing all 30 scores gives 930, so x = 930 / 30 = 31. The median we have already found: 30.5. For the mode, the score 25 occurs twice, but there is no strong peak, so we simply note the modal class is 20–29.
我们使用公式 x̄ = Σx / n 计算均值。所有 30 个分数的总和为 930，因此 x̄ = 930 / 30 = 31。中位数我们已求得：30.5。
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