📚 Averages and Range: Mean, Median, Mode | 平均数与极差：均值、中位数、众数
This article supports Cambridge KS3 Practice 10.2, covering how to calculate and interpret averages and the range. You will learn the differences between mode, median and mean, how the range measures spread, how to work with frequency tables and missing values, and how to choose the best average for a data set.
本文支持剑桥 KS3 练习 10.2，涵盖如何计算和解释平均数与极差。你将学习众数、中位数和均值之间的差异，极差如何衡量数据的分散程度，如何应对频数表与缺失值问题，以及如何为一组数据选择最合适的平均数。
1. What Are Averages? | 什么是平均数？
An average is a single value that represents the centre of a data set. At KS3 the three main averages are the mode, the median, and the mean. All three are called measures of central tendency because they try to describe the 'typical' or 'middle' value of the data.
平均数是用来表示一组数据中心趋势的单个数值。在 KS3 阶段，三种主要平均数是众数、中位数和均值。它们都被称为集中趋势的度量，因为它们试图描述数据中最"典型"或"中间"的那个值。
Each average is calculated in a different way, so they can give different results for the same data. The range is not an average; it measures how spread out the data are. Together, one average plus the range gives a short but useful summary of a whole data set.
每种平均数的计算方法不同，因此同一组数据可能得到不同结果。极差不是平均数，它衡量数据的分散程度。一个平均数加上极差，就能对整组数据给出简短而有用的概括。
| Average | 平均数 | Definition | 定义 | Works for non-numerical data? | 可用于非数值数据？ |
|---|---|---|
| Mode | 众数 | Most frequent value | 出现次数最多的值 | Yes | 可以 |
| Median | 中位数 | Middle value in order | 排序后的中间值 | No | 不可以 |
| Mean | 均值 | Total divided by how many | 总和除以个数 | No | 不可以 |
| Range | 极差 | Largest minus smallest | 最大值减最小值 | Not an average | 不是平均数 |
2. The Mode: The Most Common Value | 众数：出现次数最多的值
The mode is the value that appears most often. It is the only average that can be used for non-numerical data, such as colours or types of transport. If you are asking 'which one is the most popular?', the mode is the answer.
众数是出现次数最多的数值。它是唯一可用于非数值数据的平均数，例如颜色或交通方式。如果你问的是"哪一个最受欢迎"，答案就是众数。
Example: For 3, 5, 5, 7, 9, the mode is 5 because it occurs twice while every other value occurs once.
示例：对于 3、5、5、7、9，众数是 5，因为它出现了两次，而其他每个数值都只出现一次。
A data set may have one mode, more than one mode, or no mode at all if every value appears the same number of times. When two values tie for the highest frequency, the data set is often described as bimodal and you should give both values as the mode.
一组数据可能有一个众数、多个众数，或者当每个数值出现次数相同时没有众数。当两个数值并列出现次数最多时，这组数据常被称为双众数，你应当把两个数值都写出来。
Watch the wording carefully: the mode is the value itself, not the number of times it appears. In 3, 5, 5, 7, 9 the mode is 5, not 2. The frequency of the mode is 2, but the mode is 5.
要特别留意措辞：众数是那个数值本身，而不是它出现的次数。在 3、5、5、7、9 中，众数是 5，不是 2。众数的频数是 2，但众数是 5。
- Mode | 众数: the most common value | 出现次数最多的数值
- Bimodal | 双众数: two values share the highest frequency | 两个数值并列最高频数
- No mode | 无众数: every value appears equally often | 每个数值出现次数相同
3. The Median: The Middle Value | 中位数：中间值
The median is the middle number when the data are arranged in order. If there are two middle numbers, the median is the mean of those two middle values. Ordering the list first is not optional - it is the step that most students lose marks for skipping.
中位数是把数据按顺序排列后位于中间的那个数。如果中间有两个数，中位数就是这两个中间值的平均数。先排序不是可选项——这正是大多数学生因忽略而丢分的一步。
7, 9 there are 5 values, and (5 + 1) ÷ 2 = 3, so the median is the 3rd value, which is 5.
当数值个数为奇数时，中位数的位置是"数值个数加 1 再除以 2"。对于 3、5、5、7、9，共有 5 个数值，(5 + 1) ÷ 2 = 3，所以中位数是第 3 个数值，即 5。
value. For 2, 4, 6, 8 the two middle values are 4 and 6, so the median is (4 + 6) ÷ 2 = 5.
当数值个数为偶数时，不存在唯一居中的数值。对于 2、4、6、8，中间两个值是 4 和 6，所以中位数是 (4 + 6) ÷ 2 = 5。
Median position (odd n) = (n + 1) ÷ 2
Median (even n) = (valueₙ÷₂ + valueₙ÷₂₊₁) ÷ 2
4. The Mean: The Balance Point | 均值：平衡点
The mean is found by adding all the values and dividing by how many values there are. It uses every piece of data, which is why it is the most informative average - and also why a single extreme value can pull it a long way.
均值是把所有数值相加，再除以数值的个数。它用到了每一个数据，所以信息量最大——也正因如此，一个极端的数值就能把它拉得很远。
Mean = (Σx) ÷ n
Here Σx means 'the sum of all the values' and n means 'how many values there are'. In statistics the Greek letter Σ (sigma) always means 'add them all up'.
这里的 Σx 表示"所有数值之和"，n 表示"数值的个数"。在统计学中，希腊字母 Σ（sigma）永远表示"全部相加"。
3, 5, 5, 7, 9 the total is 3 + 5 + 5 + 7 + 9 = 29 and n = 5, so the mean is 29 ÷ 5 = 5.8.
示例：对于 3、5、5、7、9，总和为 3 + 5 + 5 + 7 + 9 = 29，n = 5，所以均值是 29 ÷ 5 = 5.8。
The mean does not have to be one of the values in the data set, and it can be a decimal even when all the data are whole numbers. Do not 'round it back' to a whole number unless the question asks you to.
均值不必是数据中的某个数值，即使所有数据都是整数，均值也可以是小数。除非题目要求，否则不要把它"四舍五入回"整数。
5. The Range: Measuring Spread | 极差：衡量分散程度
The range is the largest value minus the smallest value. It is not an average: it tells you nothing about the centre of the data, only how far apart the extremes are.
极差是最大值减去最小值。它不是平均数：它不告诉你数据的中心在哪里，只说明两端相距多远。
Range = largest value − smallest value
For 3, 5, 5, 7, 9 the largest value is 9 and the smallest is 3, so the range is 9 − 3 = 6.
示例：对于 3、5、5、7、9，最大值是 9，最小值是 3，所以极差是 9 − 3 = 6。
A small range means the data are clustered together and consistent. A large range means the data are spread out and less predictable. A common mistake is to count how many values are in the list instead of subtracting the smallest from the largest.
极差小说明数据集中、一致；极差大说明数据分散、较难预测。常见错误是去数列表中有多少个数，而不是用最大值减最小值。
Two data sets can share the same mean but have very different ranges. Saying 'both groups scored about the same on average, but group B is more consistent because its range is smaller' is exactly the kind of comparison examiners reward.
两组数据可以有相同的均值，但极差差别很大。说"两组平均得分差不多，但 B 组更稳定，因为它的极差更小"，这正是考官愿意给分的比较方式。
6. Worked Example: One Data Set, Four Answers | 综合例题：一组数据，四个答案
Take the data set 3, 5, 5, 7, 9 and calculate the mode, median, mean and range. This single example shows how all four statistics are produced from the same five numbers.
以数据组 3、5、5、7、9 为例，计算众数、中位数、均值和极差。这一个例子展示了四个统计量如何由同样的五个数字得出。
- Mode | 众数: 5 appears twice, more than any other value, so the mode is 5. | 5 出现两次，多于其他任何数值，所以众数是 5。
- Median | 中位数: the list is already in order; (5 + 1) ÷ 2 = 3, so the median is the 3rd value, 5. | 列表已排好序；(5 + 1) ÷ 2 = 3，所以中位数是第 3 个数值，即 5。
- Mean | 均值: (3 + 5 + 5 + 7 + 9) ÷ 5 = 29 ÷ 5 = 5.8. | 均值：(3 + 5 + 5 + 7 + 9) ÷ 5 = 29 ÷ 5 = 5.8。
- Range | 极差: 9 − 3 = 6. | 极差：9 − 3 = 6。
| Statistic | 统计量 | Working | 计算过程 | Answer | 答案 |
|---|---|---|
| Mode | 众数 | 5 occurs twice | 5 出现两次 | 5 |
| Median | 中位数 | 3rd of 5 values | 5 个值中的第 3 个 | 5 |
| Mean | 均值 | 29 ÷ 5 | 29 ÷ 5 | 5.8 |
| Range | 极差 | 9 − 3 | 9 − 3 | 6 |
Notice that the mode and the median are both 5 while the mean is 5.8. There is no rule that the three averages must agree, and a question that asks you to compare them is really asking you to explain why they differ.
注意众数和中位数都是 5，而均值是 5.8。三种平均数并不必须一致，当题目要求你比较它们时，实际上是在要你解释它们为何不同。
7. Choosing the Best Average | 如何选择最合适的平均数
The 'best' average depends on the type of data and on what you want to show. Choosing the mode, median or mean is a reasoning question, not a guessing game, so always justify your choice with a reason.
"最合适"的平均数取决于数据类型以及你想说明什么。选择众数、中位数或均值是一道说理题，不是猜谜，所以一定要给出理由。
- Use the mode for non-numerical data, or when the most popular item matters - for example the most common shoe size to stock in a shop. | 非数值数据，或最流行的项目更重要时用众数，例如商店应备货的最常见鞋码。
- Use the median when the data contain outliers or extreme values, such as house prices or waiting times, because it is not dragged away by a few very large or very small values. | 数据含异常值或极端值时用中位数，例如房价或等待时间，因为它不会被少数极大或极小的值拖走。
- Use the mean when the data are fairly evenly spread and you want an average that uses every value, such as the average score of a class in a test. | 数据分布较均匀、且希望平均数用到每个数值时用均值，例如全班测验的平均分。
A useful sentence pattern is: 'The median is the most suitable average because there is an extreme value of 100, which would make the mean misleading.' Examiners look for the comparison, not just the name of the average.
一个实用的句式是："中位数最合适，因为存在一个极端值 100，会使均值产生误导。"考官看的是这种比较说明，而不只是你选了哪个名字。
8. Outliers, Skew and Why Averages Disagree | 异常值、偏斜与平均数为何不同
An outlier is a value that is very different from the rest of the data. Outliers pull the mean towards them but leave the median almost unchanged, which is why the two averages can tell very different stories about the same data set.
异常值是与其他数据相差很大的数值。异常值会把均值拉向自己，却几乎不影响中位数，这就是为什么两种平均数会对同一组数据给出截然不同的说法。
the mean is 130 ÷ 6 ≈ 21.7, the median is (6 + 7) ÷ 2 = 6.5 and the range is 100 − 4 = 96.
比较 4、5、6、7、8 与 4、5、6、7、8、100。第一组均值是 30 ÷ 5 = 6，中位数是 6，极差是 4。第二组均值是 130 ÷ 6 ≈ 21.7，中位数是 (6 + 7) ÷ 2 = 6.5，极差是 100 − 4 = 96。
The single value 100 has almost doubled the mean and multiplied the range by 24, yet the median barely moved. When a data set is skewed by an outlier, always quote the median as the typical value and explain that the mean is not representative.
仅仅一个 100 就让均值几乎翻倍、让极差变成原来的 24 倍，而中位数几乎没有变化。当一组数据被异常值带偏时，应引用中位数作为典型值，并说明均值不具代表性。
9. Averages from Frequency Tables | 频数表中的平均数
t frequency, the median is found by counting through the cumulative frequencies, and the mean needs an extra column for value × frequency.
当数据以频数表给出时，你并不知道每一个具体数值，因此要借助频数来处理。众数是频数最大的那个值，中位数要通过累计频数逐个数出位置，而均值需要额外一列"数值 × 频数"。
Mean from a frequency table = (Σfx) ÷ (Σf)
15 and Σfx = 2 + 6 + 15 + 16 + 5 = 44, so the mean is 44 ÷ 15 = 2.93 to 2 decimal places.
对于下表，Σf = 2 + 3 + 5 + 4 + 1 = 15，Σfx = 2 + 6 + 15 + 16 + 5 = 44，所以均值是 44 ÷ 15 = 2.93（保留两位小数）。
| Value x | 数值 x | Frequency f | 频数 f | fx | Cumulative f | 累计频数 |
|---|---|---|---|
| 1 | 2 | 2 | 2 |
| 2 | 3 | 6 | 5 |
| 3 | 5 | 15 | 10 |
| 4 | 4 | 16 | 14 |
| 5 | 1 | 5 | 15 |
equencies reach 10 at x = 3, so the 8th value is 3. Here the mean 2.93, the median 3 and the mode 3 are all close, which suggests the data are fairly evenly spread.
众数是 3，因为它 5 的频数最大。中位数方面，共有 15 个数值，中位数是第 (15 + 1) ÷ 2 = 8 个值；累计频数在 x = 3 时达到 10，所以第 8 个值是 3。这里均值 2.93、中位数 3 和众数 3 都很接近，说明数据分布比较均匀。
10. Working Backwards: Missing Values | 反向求解：缺失值问题
Reverse problems give you the average and ask for a missing number. The key idea is that the mean hides a total: if you know the mean and the number of values, you can always recover the sum.
反向题目给出平均数，要求你求缺失的数。关键思路是：均值背后隐藏着一个总和——只要知道均值和数值个数，你总能还原出总和。
Total = mean × number of values
of them are 6, 7, 9 and 10, which add to 32. The missing number is therefore 40 − 32 = 8.
示例：五个数的均值是 8，所以总和是 8 × 5 = 40。其中四个数是 6、7、9、10，加起来是 32。因此缺失的数是 40 − 32 = 8。
The same idea works for the median and the range. If the range is 12 and the smallest value is 5, the largest must be 17. If the median of five ordered values is 20, then the third value is 20 and two values must lie on each side of it.
同样的思路也适用于中位数和极差。如果极差是 12、最小值是 5，那么最大值必定是 17。如果五个按序排列的数值中位数是 20，那么第三个值就是 20，且它的两侧各有两个数值。
11. Comparing Two Data Sets | 比较两组数据
Comparison questions are common in KS3 and GCSE papers. The reliable method is to compare a measure of average and a measure of spread, then write a conclusion in the context of the question rather than just listing numbers.
比较类题目在 KS3 和 GCSE 试卷中很常见。可靠的方法是：比较一个"平均数类"的统计量和一个
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