📚 Comparing Data Sets with Averages and Range | 利用平均数、中位数、众数与极差比较数据集
In KS3 Cambridge Mathematics, data handling questions often ask you to compare two sets of data. A good comparison always uses at least one average and the range, and it must be written in the context of the original problem. This article revises the key skills, completes the frequency-table method, and shows you how to write full-mark, exam-style comparison answers.
在 KS3 剑桥数学中，数据处理题经常要求你比较两组数据。一个好的比较至少要使用一个平均数和极差，并且必须结合题目原本的情境来写。本文将复习关键技能，补全频数表方法，并示范如何写出能拿满分的、符合考试要求的比较答案。
1. Why Compare Data Sets? | 为什么要比较数据集
Comparison questions appear in Cambridge Lower Secondary Checkpoint, KS3, IGCSE and GCSE papers. They usually give you two groups - for example, two classes, two shops, two machines - and ask which is better, more consistent, or more reliable. Marks are awarded for the calculation, for the comparison, and for the conclusion in context.
比较题出现在剑桥 Lower Secondary Checkpoint、KS3、IGCSE 和 GCSE 试卷中。题目通常给出两组对象——例如两个班、两家商店、两台机器——并问哪一个更好、更稳定或更可靠。分数分别给在计算、比较，以及结合情境的结论上。
A complete answer has three steps. First, calculate the averages and the spread for both sets. Second, compare the numbers directly, using phrases such as "greater than" or "the same as". Third, interpret what the numbers mean in the real situation.
一个完整的答案有三个步骤。第一，算出两组数据的平均数和离散程度。第二，直接用数字比较，使用如 "大于" 或 "相同" 这样的表述。第三，解释这些数字在实际情境中意味着什么。
- Calculate: mean, median, mode, range | 计算：平均数、中位数、众数、极差
- Compare: state which is larger and by how much | 比较：说明哪个更大、大多少
- Conclude: link back to the question | 结论：回到题目本身
2. Mean, Median, Mode and Range Recap | 平均数、中位数、众数和极差回顾
The mean is found by adding all the data values and dividing by the number of values. The median is the middle value when the data are placed in ascending order. The mode is the value that occurs most often. The range is the largest value minus the smallest value.
平均数的求法是将所有数据值相加，再除以数据的个数。中位数是将数据按从小到大的顺序排列后位于中间位置的值。众数是出现次数最多的值。极差是最大值减去最小值。
n = Σx ÷ n Median = middle value Range = xₘₐₓ − xₘᵢₙ
so the mean is 39 ÷ 5 = 7.8. The median is 8, the mode is 8, and the range is 12 − 4 = 8.
对于数据集 4、7、8、8、12：所有数据之和为 39，所以平均数 = 39 ÷ 5 = 7.8；中位数为 8；众数为 8；极差 = 12 − 4 = 8。
39 ÷ 5 = 7.8 Median = 8 Mode = 8 Range = 12 − 4 = 8
the two middle values. For example, in the set 3, 5, 7, 9, the median is (5 + 7) ÷ 2 = 6.
如果数据的个数为偶数，中位数就是两个中间值的平均数。例如在数据集 3、5、7、9 中，中位数为 (5 + 7) ÷ 2 = 6。
| Measure | 统计量 | What it tells you | 它说明什么 | Best used when | 最适用情形 |
| Mean | 平均数 | The overall balance point | 整体平衡点 | Data are roughly symmetric | 数据大致对称 |
| Median | 中位数 | The typical middle value | 典型的中间值 | There are outliers or skew | 存在异常值或偏斜 |
| Mode | 众数 | The most common value | 最常见的值 | Data are categorical or discrete | 数据为分类或离散型 |
| Range | 极差 | The total spread | 整体离散程度 | A quick measure of consistency | 快速判断稳定性 |
3. Choosing the Right Average | 如何选择合适的平均数
You cannot always use the mean. The mean is pulled towards extreme values, so a single very large number can make it misleading. The median is not affected by one extreme value, which is why house prices and salaries are usually reported using the median.
你不能总是用平均数。平均数会被极端值拉扯，因此一个特别大的数值就可能让它变得有误导性。中位数不受单个极端值影响，这正是房价和工资通常用中位数来报道的原因。
The mode is the only average that works for non-numerical data, such as favourite colour or shoe brand. However, a data set can have no mode, one mode, or several modes, so the mode is often less useful for comparison.
众数是唯一适用于非数值数据的平均数，例如最喜欢的颜色或鞋的品牌。但一个数据集可能没有众数、有一个众数或有多个众数，所以众数在比较中往往用处较小。
In a comparison question, the safest approach is to quote the median as your average and the range as your measure of spread. If the question mentions "average" without specifying which one, you may use the mean, but you must state clearly which average you have chosen.
在比较题中，最稳妥的做法是引用中位数作为平均数、引用极差作为离散程度的度量。如果题目提到 "average"（平均数）但没有指定是哪一个，你可以用算术平均数，但必须清楚说明你选的是哪一个。
4. Calculating the Mean from a Frequency Table | 从频数表计算平均数
When data are shown in a frequency table, multiply each value by its frequency, add the products, then divide by the total frequency. The total frequency is Σf, and the sum of the products is Σ(fx).
当数据以频数表的形式给出时，将每个数值乘以它的频数，把所有乘积相加，再除以总频数。总频数为 Σf，所有乘积之和为 Σ(fx)。
Mean = Σ(fx) ÷ Σf
+ (4 × 1) = 3 + 10 + 6 + 4 = 23. So the mean is 23 ÷ 11 = 2.09 (to 3 significant figures).
例如：数值为 x = 1、2、3、4，对应频数为 f = 3、5、2、1。则 Σf = 3 + 5 + 2 + 1 = 11，(1 × 3) + (2 × 5) + (3 × 2) + (4 × 1) = 3 + 10 + 6 + 4 = 23。所以平均数 = 23 ÷ 11 = 2.09（保留三位有效数字）。
| x | f | fx |
| 1 | 3 | 3 |
| 2 | 5 | 10 |
| 3 | 2 | 6 |
| 4 | 1 | 4 |
| Total | 合计 | Σf = 11 | Σ(fx) = 23 |
The modal value is 2, because it has the highest frequency (5). Note that the mode from a frequency table is the value with the largest f, not the largest fx.
众数是 2，因为它的频数最大（5）。注意：频数表中的众数是 f 最大的那个 x 值，而不是 fx 最大的那个。
5. Estimating the Mean from Grouped Data | 从分组数据估算平均数
When data are grouped into classes such as 0-9, 10-19, 20-29, you no longer know the exact values, only how many fall into each class. To estimate the mean, use the midpoint of each class as the representative value x.
当数据被分成 0-9、10-19、20-29 这样的组时，你已经不知道确切数值，只知道每组里有多少个数据。要估算平均数，就用每组的组中值作为代表值 x。
Midpoint = (lower bound + upper bound) ÷ 2 Estimated mean = Σ(fx) ÷ Σf
and 34.5. Then Σf = 4 + 7 + 5 + 2 = 18 and Σ(fx) = (4 × 4.5) + (7 × 14.5) + (5 × 24.5) + (2 × 34.5) = 18 + 101.5 + 122.5 + 69 = 311. The estimated mean is 311 ÷ 18 = 17.3 (to 3 significant figures).
例题：0-9 组有 4 个数据，10-19 组有 7 个，20-29 组有 5 个，30-39 组有 2 个。组中值分别为 4.5、14.5、24.5、34.5。则 Σf = 4 + 7 + 5 + 2 = 18，(4 × 4.5) + (7 × 14.5) + (5 × 24.5) + (2 × 34.5) = 18 + 101.5 + 122.5 + 69 = 311。估算平均数为 311 ÷ 18 = 17.3（保留三位有效数字）。
Always write "estimate" or "approximately" when working with grouped data, because the answer depends on the assumption that every value sits at the midpoint. The modal class is simply the class with the highest frequency.
处理分组数据时一定要写 "估算" 或 "约等于"，因为答案依赖于 "每个数据都落在组中值处" 这一假设。众数组就是频数最大的那一组。
6. Finding the Median, Quartiles and IQR | 求中位数、四分位数与四分位距
8th value. If n = 20, the position is 10.5, so take the mean of the 10th and 11th values.
求 n 个已排序数据的中位数时，用位置 (n + 1) ÷ 2。若 n = 15，中位数是第 8 个值。若 n = 20，位置为 10.5，因此取第 10 个与第 11 个值的平均数。
The lower quartile Q₁ is the median of the lower half of the data, and the upper quartile Q₃ is the median of the upper half. The interquartile range is the middle 50% of the data:
下四分位数 Q₁ 是下半部分数据的中位数，上四分位数 Q₃ 是上半部分数据的中位数。四分位距是中间 50% 的数据范围：
IQR = Q₃ − Q₁
7) ÷ 2 = 6, and the upper half is 11, 13, 15, 17, giving Q₃ = (13 + 15) ÷ 2 = 14. So IQR = 14 − 6 = 8, while the range is 17 − 3 = 14.
对已排序数据集 3、5、7、9、11、13、15、17：n = 8，中位数位置为 4.5，中位数 = (9 + 11) ÷ 2 = 10。下半部分为 3、5、7、9，Q₁ = (5 + 7) ÷ 2 = 6；上半部分为 11、13、15、17，Q₃ = (13 + 15) ÷ 2 = 14。因此 IQR = 14 − 6 = 8，而极差 = 17 − 3 = 14。
7. Range, Interquartile Range and Outliers | 极差、四分位距与异常值
The range is quick to calculate and easy to understand, but it only uses two values, so one extreme score can make it very misleading. The interquartile range ignores the bottom 25% and the top 25%, so it is a more reliable measure of spread when outliers are present.
极差计算快、易理解，但它只用到两个数值，因此一个极端分数就可能使它严重失真。四分位距忽略了最低 25% 和最高 25% 的数据，所以在存在异常值的时候，它是更可靠的离散程度度量。
A common rule for identifying an outlier is:
判断异常值的常用规则是：
Outlier if x > Q₃ + 1.5 × IQR or x < Q₁ − 1.5 × IQR
ould be an outlier. Since the largest value is 17, there are no outliers in that data set.
在上面的例子中，Q₁ = 6、Q₃ = 14、IQR = 8，所以 1.5 × IQR = 12。任何小于 6 − 12 = −6 或大于 14 + 12 = 26 的值都算异常值。由于最大值是 17，该数据集没有异常值。
When you compare two sets, it is often best to report both the range and the IQR. If the range is much larger than the IQR, this is a signal that at least one extreme value is present.
比较两组数据时，最好同时给出极差和四分位距。如果极差远大于四分位距，就说明至少存在一个极端值。
8. Writing Comparison Sentences - Exam Technique | 比较句的写法——考试技巧
Many students lose marks not because their arithmetic is wrong, but because their comparison is too vague. Writing "Class A is better" earns nothing. You must quote the values and say what they mean.
很多学生丢分不是因为算错了，而是因为比较写得太笼统。只写 "A 班更好" 是拿不到分的。你必须引用具体数值，并说明它们的含义。
A reliable three-sentence template is:
一个可靠的三句式模板是：
- On average, [group A] is higher than [group B], because the median of A (value) is greater than the median of B (value). | 平均来看，[A 组] 高于 [B 组]，因为 A 的中位数（数值）大于 B 的中位数（数值）。
- The range of [group A] (value) is greater / smaller than the range of [group B] (value), so A is more / less spread out. | [A 组] 的极差（数值）大于／小于 [B 组] 的极差（数值），所以 A 更分散／更集中。
- Therefore [group A] is more consistent / more reliable on average. | 因此 [A 组] 整体上更稳定／更可靠。
Useful comparison vocabulary includes: higher, lower, more consistent, less consistent, more spread out, more variable, more reliable. Avoid the word "better" on its own - always explain better at what.
有用的比较词汇包括：更高、更低、更稳定、较不稳定、更分散、波动更大、更可靠。不要单独使用 "更好"——一定要说明在哪方面更好。
9. Worked Example: Two Classes' Test Scores | 例题：两个班的测验成绩
12, 14. Both sets have n = 10 and both have a total of 80, so both means are 80 ÷ 10 = 8.
A 班成绩：5、6、7、7、8、8、8、9、10、12。B 班成绩：2、4、5、6、7、9、10、11、12、14。两组都有 n = 10，总和都是 80，所以两班的平均数都是 80 ÷ 10 = 8。
9) ÷ 2 = 8, there is no repeated value so there is no mode, and the range is 14 − 2 = 12.
A 班的中位数为 (8 + 8) ÷ 2 = 8，众数为 8，极差为 12 − 5 = 7。B 班的中位数为 (7 + 9) ÷ 2 = 8，没有重复值所以没有众数，极差为 14 − 2 = 12。
| Statistic | 统计量 | Class A | A 班 | Class B | B 班 |
| Mean | 平均数 | 8 | 8 |
| Median | 中位数 | 8 | 8 |
| Mode | 众数 | 8 | none | 无 |
| Range | 极差 | 7 | 12 |
Model answer: "On average, the two classes performed equally well, because both the mean and the median are 8. However, the range for Class B (12) is greater than the range for Class A (7),
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