📚 Year 10 CAIE Statistics: Case Study Workout | Year 10 CAIE 统计：案例分析实战演练
In CAIE Statistics, mastering theoretical concepts is only half the battle. Real exam success comes from applying those concepts to real-world scenarios through case studies. This article walks you through a comprehensive case study workout, where we analyse mathematics test scores of two classes and explore the relationship between revision time and marks. By following each step, you will sharpen your data-handling, interpretation, and evaluation skills — exactly what examiners look for.
在CAIE统计学中，掌握理论概念只是成功的一半。真正的考试成功来自于通过案例研究将概念应用于实际场景。本文将带你进行一次全面的案例分析实战演练，我们将分析两个班级的数学考试成绩，并探究复习时间与分数之间的关系。通过逐步跟进，你将锻炼数据处理、解读和评估能力——这正是考官所看重的。
1. Introduction to the Case Study | 案例简介
A mathematics teacher wants to compare the end-of-term test performance of her two Year 10 classes, Class A and Class B. She collects the test scores (out of 100) from all 20 students in each class. Later, she also records the revision time (in hours) and corresponding test marks of 12 randomly selected students across both classes to see if there is a link between effort and achievement. This case study will allow us to practise organising data, calculating averages and spreads, drawing box plots, and working with scatter diagrams.
一位数学老师想比较她所教的两个十年级班级（A班和B班）的期末考试成绩。她收集了每个班级20名学生的考试分数（满分100分）。随后，她还记录了从两个班级中随机选取的12名学生的复习时间（小时）和对应的考试分数，以观察努力与成绩之间是否存在联系。这个案例将让我们练习整理数据、计算平均数和离散程度、绘制箱线图以及处理散点图。
2. Data Collection and Types | 数据收集与类型
The raw test scores for Class A are: 45, 52, 58, 61, 63, 66, 68, 70, 72, 75, 78, 80, 82, 85, 88, 90, 92, 95, 98, 100. Class B scores: 50, 55, 55, 60, 62, 65, 67, 67, 70, 72, 74, 76, 78, 80, 82, 85, 88, 90, 92, 96. Both sets are quantitative, continuous data (even though recorded as integers). Revision times (hours) are also continuous quantitative data: 3, 4, 5, 5, 6, 7, 8, 9, 10, 11, 12, 15; the corresponding marks are: 55, 58, 63, 65, 70, 72, 76, 80, 82, 85, 90, 95.
A班的原始考试分数为：45, 52, 58, 61, 63, 66, 68, 70, 72, 75, 78, 80, 82, 85, 88, 90, 92, 95, 98, 100。B班分数：50, 55, 55, 60, 62, 65, 67, 67, 70, 72, 74, 76, 78, 80, 82, 85, 88, 90, 92, 96。两组数据都是定量的连续数据（尽管以整数记录）。复习时间（小时）也是连续定量数据：3, 4, 5, 5, 6, 7, 8, 9, 10, 11, 12, 15；对应的分数为：55, 58, 63, 65, 70, 72, 76, 80, 82, 85, 90, 95。
In any case study, always identify whether variables are discrete or continuous, because this affects your choice of display (e.g., histogram vs bar chart). The scores are continuous, so we use histograms and cumulative frequency curves for grouped data.
在任何案例研究中，都要先识别变量是离散的还是连续的，因为这会影响你选择的图表类型（例如直方图与条形图）。分数是连续的，因此我们对分组数据使用直方图和累积频数曲线。
3. Organising Data: Frequency Distributions | 数据整理：频数分布
To handle the class scores efficiently, we can group them into equal intervals. For Class A, using intervals 40–49, 50–59, 60–69, 70–79, 80–89, 90–100 gives a clear picture. The frequency table below summarises the data.
为了高效处理班级分数，我们可以将其分成等距区间。对于A班，使用区间40–49, 50–59, 60–69, 70–79, 80–89, 90–100可以清晰地呈现数据。下面的频数表对数据进行了汇总。
| Score Interval (Class A) | Frequency |
|---|---|
| 40–49 | 1 |
| 50–59 | 2 |
| 60–69 | 4 |
| 70–79 | 5 |
| 80–89 | 4 |
| 90–100 | 4 |
From this table, you can see the modal class is 70–79. Grouping helps us draw histograms and cumulative frequency diagrams, which are essential for finding the median and quartiles from grouped data.
从这张表可以看出，众数所在区间是70–79。分组有助于我们绘制直方图和累积频数图，这对于从分组数据中求中位数和四分位数至关重要。
4. Visual Representation: Histograms and Cumulative Frequency | 可视化：直方图与累积频数
For continuous data, we plot a histogram where the area of each bar is proportional to frequency. Since our intervals are equal, the bar heights represent the frequencies. A cumulative frequency curve for Class A can be drawn by plotting upper interval boundaries against cumulative totals (1, 3, 7, 12, 16, 20).
对于连续数据，我们绘制直方图，每个条形的面积与频数成正比。由于我们的区间是等距的，条形的高度就代表了频数。A班的累积频数曲线可以通过将区间上限与累积总数（1, 3, 7, 12, 16, 20）对应描点来绘制。
From the cumulative frequency curve, we can estimate the lower quartile (n/4 = 5th value), median (n/2 = 10th value), and upper quartile (3n/4 = 15th value). Reading from the curve for Class A: Q₁ ≈ 64, median ≈ 76, Q₃ ≈ 89. These estimates are close to our exact calculations later.
从累积频数曲线中，我们可以估算出下四分位数（n/4 = 第5个值）、中位数（n/2 = 第10个值）和上四分位数（3n/4 = 第15个值）。从A班曲线上读出：Q₁ ≈ 64，中位数 ≈ 76，Q₃ ≈ 89。这些估算值与我们稍后的精确计算很接近。
5. Central Tendency: Mean, Median, Mode | 集中趋势：均值、中位数、众数
Let’s calculate the exact measures for Class A and Class B. For Class A, the total sum of scores = 1518, so mean = 1518 / 20 = 75.9. The median is the average of the 10th and 11th ordered values: (75 + 78) / 2 = 76.5. There is no repeated value, so the data set has no simple mode; the modal class is 70–79.
我们来计算A班和B班的精确度量值。对于A班，分数总和=1518，因此均值=1518/20=75.9。中位数是第10和第11个有序数据的平均值：(75+78)/2=76.5。数据中没有重复值，所以没有简单的众数；众数所在区间是70–79。
For Class B, sum = 1464, mean = 73.2. The 10th value is 72 and the 11th is 74, so median = 73. The mode is 55 and 67, each appearing twice. Comparing the means, Class A has a higher average, but we need to check the spread to understand consistency.
对于B班，总和=1464，均值=73.2。第10个值是72，第11个是74，因此中位数=73。众数是55和67，各出现两次。比较均值，A班平均分更高，但我们还需要查看离散程度以了解一致性。
6. Dispersion: Range, Quartiles, IQR | 离散程度：极差、四分位数、四分位距
Dispersion tells us how spread out the marks are. For Class A, range = 100 – 45 = 55. Using the formula for ungrouped data: position of Q₁ = ¼(n+1) = 5.25th value, so Q₁ = 63 + 0.25×(66 – 63) = 63.75. Q₃ = 15.75th value, Q₃ = 88 + 0.75×(90 – 88) = 89.5. IQR = 89.5 – 63.75 = 25.75.
离散程度告诉我们分数的分散情况。对于A班，极差=100-45=55。使用未分组数据公式：Q₁的位置=¼(20+1)=5.25个值，所以Q₁=63+0.25×(66-63)=63.75。Q₃是第15.75个值，Q₃=88+0.75×(90-88)=89.5。IQR=89.5-63.75=25.75。
For Class B, range = 96 – 50 = 46. Q₁ = 62 + 0.25×(65 – 62) = 62.75. Q₃ = 82 + 0.75×(85 – 82) = 84.25. IQR = 84.25 – 62.75 = 21.5. Class A has a greater IQR, indicating more variability in scores, even though its mean is higher.
对于B班，极差=96-50=46。Q₁=62+0.25×(65-62)=62.75。Q₃=82+0.75×(85-82)=84.25。IQR=84.25-62.75=21.5。A班的IQR更大，表明分数变异性更高，尽管其均值更高。
7. Box Plots for Comparison | 箱线图比较
A box plot (box-and-whisker diagram) provides a powerful visual comparison. For Class A: Min=45, Q₁=63.75, Median=76.5, Q₃=89.5, Max=100. Class B: Min=50, Q₁=62.75, Median=73, Q₃=84.25, Max=96. By drawing both boxes on the same scale, we immediately spot that Class A’s median is higher but its longer box and whiskers reflect greater spread.
箱线图（盒须图）提供了强有力的视觉比较。A班：最小值=45，Q₁=63.75，中位数=76.5，Q₃=89.5，最大值=100。B班：最小值=50，Q₁=62.75，中位数=73，Q₃=84.25，最大值=96。在相同尺度上绘制两个箱形，我们可以立刻发现A班的中位数更高，但其更长的箱形和须线反映出更大的分散性。
Always remember to check for outliers. An outlier is usually defined as a value less than Q₁ – 1.5×IQR or greater than Q₃ + 1.5×IQR. For Class A, lower fence = 63.75 – 1.5×25.75 = 25.125, upper fence = 89.5 + 38.625 = 128.125; no outliers exist. Class B similarly has no outliers.
一定要记得检查离群值。离群值通常定义为小于Q₁ – 1.5×IQR或大于Q₃ + 1.5×IQR的值。对于A班，下截断点=63.75 – 1.5×25.75=25.125，上截断点=89.5+38.625=128.125；没有离群值。B班同样没有离群值。
8. Probability and Relative Frequency | 概率与相对频率
We can use relative frequency to estimate probabilities. If a student is chosen at random from Class A, what is the probability that their score is above 80? There are 8 such scores (82,85,88,90,92,95,98,100), so P(score > 80) = 8/20 = 0.4. For Class B, P(score > 80) = 6/20 = 0.3. The relative frequency shows a higher chance of earning a top mark in Class A.
我们可以用相对频率来估计概率。如果从A班随机抽取一名
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