📚 Mean, Median, Mode and Range | 平均数、中位数、众数和极差
In IGCSE Edexcel Mathematics, the ability to summarise data using averages and the range underpins much of statistics. The three main averages – mean, median and mode – each describe the 'centre' of a data set in a unique way, while the range measures the spread. This guide explains how to calculate each measure from both simple lists and frequency tables, and shows you how to compare data sets effectively using these tools.
在 IGCSE Edexcel 数学中，使用平均数和极差进行数据汇总的能力是统计学的基础。三种主要的平均数——平均数、中位数和众数——分别从独特的角度描述数据集的“中心”，而极差则度量数据的分散程度。本指南将说明如何从简单列表和频数表中计算各个度量，并展示如何借助这些工具有效比较不同数据组。
1. Understanding Averages and Range | 理解平均数与极差
An average is a single value that represents the typical or central point of a data set. The three common averages are the mean, median and mode. In contrast, the range is not an average but a measure of dispersion – it tells us how far apart the smallest and largest values are.
平均数是代表数据集典型值或中心点的一个单一数值。常见的三个平均数是平均数、中位数和众数。相反，极差不是平均数，而是一个离散程度的度量——它告诉我们最小值与最大值之间相差多远。
Each average has its own strengths and is useful in different situations. For example, the mean uses all the data values, making it sensitive to extreme values. The median focuses on the middle and is unaffected by outliers. The mode simply identifies the most frequent value. Together with the range, these measures give a clear picture of both the centre and the spread of data.
每种平均数都有各自的优点，并适用于不同的场景。例如，平均数使用了所有数据值，因此对极端值很敏感。中位数关注中间位置，不受异常值影响。众数只是找出出现最频繁的值。结合极差，这些度量能清晰地描绘数据组的中心和分散情况。
2. The Mean (Arithmetic Average) | 平均数（算术平均值）
The mean, often called the average, is found by adding all the data values together and then dividing by the number of values. It can be written compactly using sigma notation, where Σx denotes the sum of all values and n is the total number of observations.
平均数，通常简称为平均值，是将所有数据值相加，然后除以值的个数而得到的。可以用求和符号简洁地表示，其中 Σx 表示所有数值的总和，n 是观测值的总数。
Mean = Σx / n
平均数 = Σx / n
n the data 5, 7, 8, 6, 9: the sum is 5+7+8+6+9 = 35, and n = 5, so the mean is 35 ÷ 5 = 7.
例如，给定数据 5, 7, 8, 6, 9：总和为 5+7+8+6+9 = 35，n = 5，因此平均数为 35 ÷ 5 = 7。
If any values are repeated, the calculation is the same. The mean gives equal weight to every data point, which is why a single very large or very small value can pull the mean up or down significantly.
如果某些数值重复出现，计算方法相同。平均数赋予每个数据点相等的权重，这就是为什么一个极大或极小的值会显著拉升或拉低平均数。
3. Mean from a Frequency Table | 从频数表求平均数
When data is organised in a frequency table, we multiply each distinct value x by its frequency f, sum these products, and then divide by the total frequency Σf. This approach avoids having to add every single value individually.
当数据以频数表的形式整理时，我们将每个不同的数值 x 乘以其频数 f，再求这些乘积的总和，然后除以总频数 Σf。这种方法避免了对每个单一数值逐一相加。
Mean = Σ(f × x) / Σf
平均数 = Σ(f × x) / Σf
) = (3×2) + (4×5) + (2×8) = 6 + 20 + 16 = 42, Σf = 3+4+2 = 9, so the mean = 42 / 9 ≈ 4.67.
例如，一张表格显示数值 2 出现 3 次，数值 5 出现 4 次，数值 8 出现 2 次：Σ(f×x) = (3×2) + (4×5) + (2×8) = 6 + 20 + 16 = 42，Σf = 3+4+2 = 9，因此平均数 = 42 / 9 ≈ 4.67。
class interval as the value of x. The estimated mean is only an approximation because the raw data is not known.
如果表格包含分组数据（例如 0 ≤ x < 10），则用每个组区间的中点作为 x 值。由于不知道原始数据，估算的平均数只是一个近似值。
4. The Median (Middle Value) | 中位数（中间值）
The median is the middle number when the data is arranged in ascending order. If there is an odd number of values, the median is simply the central one. If there is an even number of values, the median is the average of the two middle numbers.
中位数是将数据升序排列后位于中间位置的数。如果数值个数为奇数，中位数就是正中央的那个值。如果是偶数个，中位数是中间两个数的平均值。
Median position = (n + 1) / 2
中位数位置 = (n + 1) / 2
(4+1)/2 = 2.5th value, meaning we take the mean of the 2nd and 3rd values: (12+14)/2 = 13.
以数据集 12, 15, 14, 10, 20 为例。先排序：10, 12, 14, 15, 20。n = 5，中位数位置 = (5+1)/2 = 第 3 个值，因此中位数 = 14。对于偶数个数据如 10, 12, 14, 18，位置 = (4+1)/2 = 第 2.5 个值，意味着取第 2 与第 3 个值的平均数：(12+14)/2 = 13。
The median is robust: extremely high or low values do not shift it, making it especially useful when data contains outliers or is skewed.
中位数具有稳健性：极大或极小的数值不会使其偏移，因此在数据含有异常值或呈偏态时特别有用。
5. Median from a Frequency Table | 从频数表求中位数
When working with a frequency table of ungrouped discrete data, create a cumulative frequency column to locate the median position. The median is the first data value whose cumulative frequency is equal to or greater than (n+1)/2.
处理未分组离散数据的频数表时，要建立一个累积频数列来定位中位数位置。中位数是累积频数首次达到或超过 (n+1)/2 所对应的数据值。
e group of value 5 because cumulative frequency reaches 7 at that point. Thus, median = 5.
例如，数值 2 出现 3 次（累积频数 3），数值 5 出现 4 次（累积频数 7），数值 8 出现 2 次（累积频数 9）。总数 n = 9，中位数位置 = (9+1)/2 = 第 5 个值。第 5 个值落在数值 5 的组内，因为累积频数在此时达到 7。因此，中位数 = 5。
If data is grouped, the median class is found using cumulative frequency, and linear interpolation is typically used to estimate the median – a skill further developed in cumulative frequency diagrams.
如果数据是分组的，则使用累积频数找出中位数所在组，然后通常采用线性插值来估算中位数——这是在累积频数图学习中需要进一步培养的技能。
6. The Mode (Most Frequent) | 众数（出现频率最高）
The mode is the value that occurs most often. A data set can have one mode (unimodal), more than one mode (bimodal or multimodal), or no mode if all values appear equally often.
众数是出现次数最多的值。一个数据集可以有一个众数（单峰）、多个众数（双峰或多峰），或者如果没有哪个值重复出现就没有众数。
From a frequency table, the mode is simply the value with the highest frequency. For grouped data, the modal class is the class interval with the highest frequency.
从频数表中，众数就是频数最高的那个值。对于分组数据，众数组是频数最高的那个组区间。
The mode is easy to find and is the only average that can be used with non-numerical data (e.g. the most popular colour).
众数容易找到，而且是唯一可用于非数值数据（例如最受欢迎的颜色）的平均数。
7. The Range (Spread of Data) | 极差（数据的分散程度）
The range is a simple measure of spread. It is calculated by subtracting the smallest value from the largest value in the data set.
极差是一个简单的离散度量。它通过从数据集中最大值减去最小值来计算。
Range = Maximum value − Minimum value
极差 = 最大值 − 最小值
, 94, the range = 94 − 45 = 49. A larger range suggests greater variation within the data.
例如，若测试分数为 45, 62, 78, 81, 94，极差 = 94 − 45 = 49。极差越大表明数据内的变异越大。
The range is quick to calculate but can be heavily influenced by outliers because only the two extreme values are considered.
极差计算快速，但容易受异常值的严重影响，因为它只考虑两个极端值。
8. Comparing Data Sets Using Averages and Range | 使用平均数与极差比较数据组
To compare two data sets, look at both a measure of central tendency (mean or median) and a measure of spread (range). For instance, two classes might have the same mean score, but the range reveals which class is more consistent.
比较两个数据组时，既要看集中趋势量数（平均数或中位数），也要看离散量数（极差）。例如，两个班可能有相同的平均分，但极差能揭示哪个班的成绩更稳定。
| Measure | Class A | Class B |
| Mean | 72 | 72 |
| Median | 71 | 73 |
| Range | 30 | 18 |
In the table above, both classes have a mean of 72, but Class B's smaller range (18 compared to 30) indicates their scores are more tightly clustered around the average, implying more consistent performance.
在上表中，两个班的平均数都是 72，但 B 班较小的极差（18 对 30）表明其分数更紧密地围绕在平均值周围，意味着表现更一致。
Always pair a concluding statement with numerical evidence when answering exam questions. For example: 'Class B has a higher median and a smaller range, so on average they performed better and more consistently.'
在回答考题时，始终要将结论与数据证据结合。例如：“B 班的中位数更高，极差更小，因此平均而言他们的表现更好且更稳定。”
9. Choosing the Best Average | 选择最合适的平均数
Not all averages are equally appropriate for every situation. The mean is best when all values are fairly representative, but the median is preferred when there are outliers because it is not distorted by them. The mode is most useful when looking for the most common category or when data is non-numerical.
并非所有平均数都同等适用于每种情况。当所有值都具有代表性时，用平均数最佳；但存在异常值时，中位数更合适，因为它不会被扭曲。众数在寻找最常见的类别或数据为非数值时最有用。
For example, house prices in an area often have a few very expensive properties. The mean house price is pulled up, so estate agents often report the median, which reflects the typical price more accurately. Knowing which average to use is a key skill in data handling.
例如，一个区域的房价常有少数非常昂贵的房产。平均房价会被拉高，因此房地产经纪人常报告中位数，它能更准确地反映典型价格。知道何时使用哪种平均数是数据处理的一项关键技能。
10. Worked Example: Mixed Dataset | 综合例题：混合数据集
A school recorded the number of late days for 10 students: 0, 1, 0, 3, 2, 1, 0, 2, 14, 2. Calculate the mean, median, mode and range, and discuss which average is most representative.
一所学校记录了 10 名学生的迟到天数：0, 1, 0, 3, 2, 1, 0, 2, 14, 2。计算平均数、中位数、众数和极差，并讨论哪个平均数最具代表性。
Step 1 – Mean: Sum = 0+1+0+3+2+1+0+2+14+2 = 25, n = 10, Mean = 25/10 = 2.5. Step 2 – = 10, median position = (10+1)/2 = 5.5th value → mean of 5th (1) and 6th (2) → 1.5. Step 3 – Mode = 0 and 2 (both appear 3 times, so bimodal). Step 4 – Range = 14 − 0 = 14.
步骤 1 – 平均数：总和 = 0+1+0+3+2+1+0+2+14+2 = 25，n = 10，平均数 = 25/10 = 2.5。步骤 2 – 中位数：排序后数据：0, 0, 0, 1, 1, 2, 2, 2, 3, 14。n = 10，中位数位置 = (10+1)/2 = 第 5.5 个值 → 第 5 个 (1) 和第 6 个 (2) 的平均 → 1.5。步骤 3 – 众数 = 0 和 2（两者均
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