Year 10 CAIE Statistics: Case Study Mastery in Action | Year 10 CAIE 统计：案例分析实战演练

📚 Year 10 CAIE Statistics: Case Study Mastery in Action | Year 10 CAIE 统计：案例分析实战演练

In this article, we will work through a full statistical investigation based on a real-world scenario. A class of 30 Year 10 students took a Mathematics test and a Science test, both marked out of 50. We will explore how to organise, present, analyse and interpret the data using the key tools from the CAIE Statistics syllabus. Follow the steps carefully — you can use the same approach in your own exam case studies.

本文将基于一个真实场景进行一次完整的统计调查。某 Year 10 班级的 30 名学生参加了一次数学测试和一次科学测试，满分均为 50 分。我们将利用 CAIE 统计课程中的关键工具，探究如何整理、呈现、分析和解读这些数据。请仔细按照步骤学习，你可以在自己的考试案例研究中采用同样的方法。

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1. Introducing the Case Study | 案例介绍

The teacher recorded the marks of each student and wants to answer several questions: How did the class perform overall? Were the Science marks more spread out than the Mathematics marks? Is there a relationship between the two sets of marks? What is the chance that a randomly chosen student scored above 35 in Mathematics? To answer these, we will apply a range of statistical techniques.

老师记录了每名学生的成绩，并希望回答以下几个问题：班级的整体表现如何？科学成绩的离散程度是否比数学更大？两组成绩之间是否存在关系？随机挑选一名学生，其数学成绩高于 35 分的概率是多少？为了回答这些问题，我们将运用一系列统计方法。

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2. Collecting and Presenting Raw Data | 收集与呈现原始数据

First, we need to organise the raw marks. The table below shows the data for all 30 students. Each row gives a student’s Mathematics mark and Science mark.

首先我们需要整理原始成绩。下表列出了全部 30 名学生的数据，每一行分别给出该生的数学成绩和科学成绩。

Student	Maths (x)	Science (y)
1	32	24
2	45	38
3	28	26
4	50	42
5	37	33
6	41	36
7	22	19
8	33	28
9	48	40
10	26	22
11	39	34
12	44	39
13	20	18
14	35	30
15	42	37
16	31	27
17	47	44
18	24	20
19	38	32
20	29	25
21	46	41
22	34	31
23	27	23
24	43	35
25	49	43
26	36	29
27	30	26
28	40	34
29	25	21
30	33	28

Reading directly from this list is useful, but for deeper analysis we prefer to group the marks into intervals. This reduces the number of categories and helps us see patterns more clearly.

直接阅读这份列表是有用的，但为了进行更深入的分析，我们倾向于将成绩分组到区间内。这样可以减少类别数量，并帮助我们更清晰地看到整体模式。

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3. Grouped Frequency Table | 分组频数表

For the Mathematics marks, we can choose equal class intervals of width 5, starting from 20. The resulting frequency table is shown below.

对于数学成绩，我们可以选择宽度为 5 的等距组区间，从 20 分开始。生成的频数表如下所示。

Marks (Maths)	Frequency (f)
20 – 24	4
25 – 29	5
30 – 34	6
35 – 39	5
40 – 44	5
45 – 50	5

Always check that the sum of frequencies equals the total number of data values — here 4+5+6+5+5+5 = 30. The grouped table shows that the marks are fairly evenly spread, with a slight peak in the 30–34 interval.

务必检查频数总和是否等于数据的总个数，这里是 4+5+6+5+5+5 = 30。分组表显示成绩分布相对均匀，在 30–34 区间出现了轻微的高峰。

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4. Visualising Data: Histogram | 数据可视化：直方图

Since the class intervals are equal, we can draw a histogram with frequency on the vertical axis. Each bar’s height represents the frequency in that interval. The histogram reveals the shape of the distribution at a glance.

由于组区间等距，我们可以绘制一个以频数为纵轴的直方图。每个柱子的高度代表该区间内的频数。直方图可以让我们一眼看出分布形状。

Imagine the bars: the interval 20–24 has height 4, 25–29 has height 5, and so on. There is no strong skew — the distribution is roughly symmetric with a slight left tail. This tells us that very low marks are less common than marks near the middle.

想象一下这些柱子：20–24 区间高度为 4，25–29 区间为 5，依此类推。没有明显的偏斜——分布大致对称，左侧尾部略长。这告诉我们，极低的分数不如中间分数常见。

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5. Measures of Central Tendency | 集中趋势的度量

To describe a typical Maths mark we can calculate the mean, median and mode from the raw data.

为了描述典型的数学成绩，我们可以根据原始数据计算平均数、中位数和众数。

Mean: Sum of all Maths marks = 1050. Number of students = 30.

Mean = 1050 / 30 = 35

平均数：所有数学成绩之和为 1050，学生人数 30。

平均数 = 1050 / 30 = 35

Median: Arrange all 30 marks in ascending order. Since there is an even number of values, the median is the average of the 15th and 16th values. Ordered data: 20,22,24,25,26,27,28,29,30,31,32,33,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50. The 15th value is 35, and the 16th is 36.

Median = (35 + 36) / 2 = 35.5

中位数：将全部 30 个分数按升序排列。由于数值个数为偶数，中位数是第 15 和第 16 个数值的平均值。排序后数据：20,22,24,25,26,27,28,29,30,31,32,33,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50。第 15 个值为 35，第 16 个值为 36。

中位数 = (35 + 36) / 2 = 35.5

Mode: Look for the most frequent mark. From the raw data, the mark 33 appears twice; all other marks appear once. So the data set has a single mode.

Mode = 33

众数：找出出现次数最多的分数。在原始数据中，33 分出现了两次，其他所有分数各出现一次。因此该数据集只有一个众数。

众数 = 33

The three measures are close together, which confirms the distribution is quite balanced. The mean and median are both around 35, so a typical student scored just above half marks.

这三个度量值很接近，证实了分布相当均衡。平均数与中位数都在 35 左右，因此一名典型学生的得分略高于满分的一半。

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6. Measures of Spread | 离散程度的度量

Spread tells us how varied the marks are. For the Maths marks we can find the range, quartiles and interquartile range (IQR).

离散程度告诉我们成绩的变化有多大。对于数学成绩，我们可以计算极差、四分位数和四分位数间距 (IQR)。

Range = Maximum − Minimum = 50 − 20 = 30. This shows a wide spread from the lowest to the highest score.

极差 = 最大值 − 最小值 = 50 − 20 = 30。这表明从最低分到最高分存在较大的范围。

To find the lower quartile (Q₁) and upper quartile (Q₃), we split the ordered data into two halves. The lower half consists of the first 15 values: 20,22,24,25,26,27,28,29,30,31,32,33,33,34,35. The median of this half is the 8th value, so Q₁ = 29. The upper half contains values from 36 to 50; its median is the 8th value in that half, which is 43. Therefore Q₃ = 43.

要找出下四分位数 (Q₁) 和上四分位数 (Q₃)，我们将排序后的数据分成两半。下半部分包含前 15 个值：20,22,24,25,26,27,28,29,30,31,32,33,33,34,35。该部分的中位数是第 8 个值，因此 Q₁ = 29。上半部分包含 36 至 50 的值；它里面的中位数是第 8 个值，即 43。因此 Q₃ = 43。

IQR = Q₃ − Q₁ = 43 − 29 = 14

The IQR of 14 points means the middle 50% of students have Maths marks spanning 14 marks. This gives a better picture of typical spread than the range, because it is not affected by extreme values.

四分位数间距为 14 分，意味着中间 50% 学生的数学成绩跨越了 14 分的区间。这比极差更能反映典型的离散程度，因为它不受极端值的影响。

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7. Cumulative Frequency and Quartiles | 累积频数与四分位数

Another way to find quartiles is to draw a cumulative frequency diagram. We first build a cumulative frequency table from the grouped frequency table, adding an extra row for the lower boundary 20 (with cumulative frequency 0).

另一种寻找四分位数的方法是绘制累积频数图。我们首先从分组频数表中构建一个累积频数表，并添加一行下限 20（累积频数为 0）。

Marks ≤	Cumulative Frequency
20	0
25	4
30	4+5=9
35	9+6=15
40	15+5=20
45	20+5=25
50	25+5=30

Plot the upper boundary of each interval on the horizontal axis against the cumulative frequency on the vertical axis, then join the points with a smooth curve. Reading off at cumulative frequencies 7.5 (for Q₁) and 22.5 (for Q₃) gives approximately 27 and 42. The small difference from the ungrouped values is due to grouping.

将每个区间的上边界作为横轴、累积频数作为纵轴描点，然后用平滑曲线连接。在累积频数 7.5（对应 Q₁）和 22.5（对应 Q₃）处读数，大约得到 27 和 42。与未分组数据计算值的微小差异是由分组造成的。

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8. Box-and-Whisker Plots | 箱线图

Using the five-number summary (min = 20, Q₁ = 29, median = 35.5, Q₃ = 43, max = 50) we can draw a box plot for the Maths marks. The classroom can also construct a box plot for the Science marks for comparison. Suppose the Science five-number summary is: min = 18, Q₁ = 24, median = 30, Q₃ = 35, max = 44.

利用五数概括（最小值 = 20，Q₁ = 29，中位数 = 35.5，Q₃ = 43，最大值 = 50），我们可以绘制数学成绩的箱线图。课堂上同时也可以为科学成绩绘制箱线图以进行比较。假设科学成绩的五数概括为：最小值 = 18，Q₁ = 24，中位数 = 30，Q₃ = 35，最大值 = 44。

The box plot visually shows that the Maths marks have a higher median and a wider spread than Science. This answers the teacher’s question about which subject had more variation.

箱线图直观地显示出，数学成绩的中位数更高，且离散程度也比科学更大。这回答了老师关于哪个科目变异更大的疑问。

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9. Scatter Diagrams and Correlation | 散点图与相关性

Is there a link between how well a student does in Maths and in Science? We can investigate this by plotting a scatter diagram with Maths on the x-axis and Science on the y-axis. Each point represents one student.

学生在数学和科学上的表现之间是否存在某种联系？我们可以通过绘制散点图来探究，以数学成绩为 x 轴、科学成绩为 y 轴。每个点代表一名学生。

When we plot the 30 points, they show an upward trend: students with higher Maths marks tend to have higher Science marks. This is called positive correlation. We describe the correlation as ‘strong positive’ because the points lie fairly close to an imaginary straight line.

当我们画出这 30 个点时，呈现出一种向上的趋势：数学成绩较高的学生，其科学成绩往往也较高。这称为正相关。由于这些点相当接近一条虚拟的直线，我们将这种相关描述为“强正相关”。

We must be cautious — correlation does not imply causation. Doing well in Maths does not automatically cause a high Science score, but the two are linked, perhaps through general ability or study habits.

我们必须小心——相关并不意味着因果。数学考得好并不会自动导致科学高分，但两者之间存在联系，或许是通过综合能力或学习习惯关联起来的。

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10. Introduction to Probability from Data | 从数据看概率入门

We can use the frequency table to estimate simple probabilities. For example, what is the probability that a randomly selected student scored more than 35 in Mathematics?

我们可以利用频数表来估计简单概率。例如，随机选取一名学生，其数学成绩高于 35 分的概率是多少？

From the raw data, we count how many marks are above 35: these are 36,37,38,39,40,41,42,43,44,45,46,47,48,49,50 — that is 15 students. So the experimental probability is 15/30 = 1/2.

从原始数据中，我们数一下有多少个分数大于 35：36,37,38,39,40,41,42,43,44,45,46,47,48,49,50——共 15 名学生。因此实验概率为 15/30 = 1/2。

P(Maths > 35) = 1/2

Similarly, the probability that a student scored between 25 and 35 (inclusive) can be found as 12/30 = 2/5. This is a great way to link data handling with probability.

类似地，某名学生成绩在 25 到 35 分之间（含）的概率为 12/30 = 2/5。这是将数据处理与概率联系起来的好方法。

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11. Interpreting Results and Drawing Conclusions | 结果解读与得出结论

Now we can answer the teacher’s questions clearly. The class performed relatively well overall, with a mean Maths mark of 35 and a median of 35.5. The Science marks were less spread out, indicating more consistent performance in Science. There is a strong positive correlation between the two subjects, suggesting that students who are strong in one tend to be strong in the other.

现在我们可以清楚地回答老师的问题了。班级整体表现相对不错，数学平均分为 35，中位数为 35.5。科学成绩的离散程度更小，表明科学成绩更稳定。两个科目之间存在强正相关，这意味着在某一科上较强的学生往往在另一科上也较强。

The probability of scoring above 35 in Maths was one‑half, and the box plot confirmed that the middle half of the class scored between 29 and 43. These findings can help the teacher decide where to focus revision sessions.

数学成绩高于 35 分的概率为一半，箱线图也证实了班级中间一半学生的分数在 29 到 43 分之间。这些发现有助于老师决定复习课的重点应放在哪里。

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12. Practice Tips for Exams | 考试实践技巧

When tackling a case study in your CAIE Statistics exam, always read the scenario carefully and identify what data is provided. Present a neatly labelled table or chart whenever possible. Calculate statistics step by step and remember to use the correct formula notation. For grouped data, use midpoints when finding the mean. Finally, round your answers to a sensible degree of accuracy — usually one or

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