📚 A-Level Mathematics: Measures of Central Tendency | A-Level 数学：集中趋势度量指标解析
Measures of central tendency are single values that describe the centre of a data set. In A-Level Mathematics, the three main measures are the mean, median and mode. Choosing the correct measure and interpreting it accurately is essential for data analysis, probability and statistics questions.
集中趋势度量指标是用一个“代表值”来描述整组数据的中心位置。在A-Level数学中，最常用的三个指标是均值（mean）、中位数（median）和众数（mode）。能否正确选择并合理解释这些指标，是数据分析、概率与统计部分的关键考点。
1. The Three Core Measures | 三个核心指标
For any data set, we usually summarise the data using: the mean, the median and the mode. Each measure answers a slightly different question: What is the average? What is the middle value? Which value occurs most often?
对于任意一组数据，我们通常使用均值、中位数和众数进行概括。这三个指标分别回答不同的问题：平均水平是多少？中间位置在哪里？哪个数值出现得最频繁？
- Mean (x̄): the sum of all values divided by the number of values.
- Median: the middle value when the data are arranged in ascending order.
- Mode: the value that appears most frequently.
- 均值 (x̄)：所有数值之和除以数值个数。
- 中位数：将数据按升序排列后处于正中间的数。
- 众数：一组数据中出现次数最多的数。
Data set example: 4, 7, 7, 9, 12
数据示例：4, 7, 7, 9, 12
median is 7 because it is the third value in the ordered list. The mode is 7 because it appears twice.
均值为 (4 + 7 + 7 + 9 + 12) ÷ 5 = 7.8。中位数是7，因为它在排序后的列表中位于第三位。众数也是7，因为它出现了两次。
2. The Mean: Formula and Weighted Version | 均值：公式与加权形式
For raw data, the mean is given by the formula:
对于原始数据，均值的计算公式为：
x̄ = Σx ÷ n
where Σx is the sum of all observations and n is the number of observations. In a frequency table, the mean is calculated using:
其中Σx表示所有观测值的总和，n表示观测值的个数。在频数表中，均值使用如下公式：
x̄ = Σfx ÷ Σf
Here f represents the frequency and x represents the value (or mid-point of a class). This is also called the weighted mean, because values with higher frequency contribute more to the total.
其中f代表频数，x代表取值（如果是分组数据则取组中值）。这个公式也称为加权均值，因为频数越高的取值对总和的贡献越大。
Worked example: The table below shows the number of books read by a group of students.
例题：下表显示了一组学生阅读书本的数量。
| Books (x) | 0 | 1 | 2 | 3 |
| Frequency (f) | 3 | 5 | 4 | 2 |
Σf = 3 + 5 + 4 + 2 = 14. Σfx = (0×3) + (1×5) + (2×4) + (3×2) = 19. Therefore x̄ = 19 ÷ 14 ≈ 1.36.
Σf = 3 + 5 + 4 + 2 = 14。Σfx = (0×3) + (1×5) + (2×4) + (3×2) = 19。因此 x̄ = 19 ÷ 14 ≈ 1.36。
3. The Median: Position and Interpretation | 中位数：位置与解释
The median is the middle value of an ordered data set. For n values, the median is located at position (n + 1) ÷ 2. When n is even, the median is the average of the two middle values.
中位数是有序数据中位于正中间的值。对于n个数据，中位数的位置是 (n + 1) ÷ 2。当n为偶数时，中位数取中间两个数的平均值。
Example: : For the data set 2, 3, 5, 8, 11, n = 5, so the median is the (5 + 1) ÷ 2 = 3rd value, which is 5.
示例：对于数据2, 3, 5, 8, 11，n = 5，因此中位数是第 (5 + 1) ÷ 2 = 3 个数值，即5。
For data presented in a frequency table, the median is found by locating the cumulative frequency which first reaches or exceeds n ÷ 2.
当数据以频数表形式呈现时，中位数通过观察累积频数首次达到或超过 n ÷ 2 的位置来确定。
x̄ is sensitive to outliers; the median is not.
均值对极端值敏感；中位数则不受极端值影响。
This means that when the data include extreme values, the median often gives a more representative “centre” than the mean.
这意味着当数据中存在极端值时，中位数往往比均值更能代表数据的“中心”。
4. The Mode and Modal Class | 众数与模态组
The mode is the value with the highest frequency. A data set may have one mode, more than one mode, or no mode at all if all values occur with equal frequency.
众数是频数最高的数值。一组数据可以有一个众数、多个众数，或者当所有数值出现频数相同时，也可以没有众数。
For grouped data, we cannot identify the exact mode from a frequency table alone. Instead, we identify the modal class, which is the class interval with the highest frequency.
对于分组数据，仅凭频数表无法确定精确的众数。我们通常识别“模态组”（modal class），即频数最高的组区间。
Example: In the frequency table below, the modal class is 20 – 29.
示例：在下表的频数分布中，模态组是20 – 29。
| Class | 0 – 9 | 10 – 19 | 20 – 29 | 30 – 39 |
| Frequency | 4 | 6 | 10 | 5 |
The mode is especially useful for categorical data, such as shoe size or favourite colour.
众数特别适用于分类数据，例如鞋码或最喜欢的颜色这类数据。
5. Estimating Measures from Grouped Data | 从分组数据估计集中趋势
When data are grouped into class intervals, we lose the original raw values. We can still estimate the mean by using class mid-points. The sum of the mid-point multiplied by frequency is divided by the total frequency.
当数据被分组到区间后，原始值便失去了。我们仍然可以通过组中值来估计均值。具体方法是：将每组组中值乘以频数后求和，再除以总频数。
x̄ ≈ Σ(mid-point × frequency) ÷ Σ frequency
x̄ ≈ Σ(组中值 × 频数) ÷ Σ频数
For the median in grouped data, a common A-Level method is linear interpolation. The formula is:
对于分组数据中的中位数，A-Level常用的方法是线性插值。公式为：
Median = L + (n/2 − F) ÷ f × c
where L is the lower class boundary of the median class, F is the cumulative frequency before the median class, f is the frequency of the median class, and c is the class width.
其中L是中位数所在组的下边界，F是中位数所在组之前的累积频数，f是中位数所在组的频数，c是组宽。
Example: Find the median for the following grouped data:
示例：计算以下分组数据的中位数：
| Class | 0 – 9 | 10 – 19 | 20 – 29 | 30 – 39 |
| Frequency | 5 | 8 | 12 | 5 |
otal n = 30, so n/2 = 15. Cumulative frequencies are 5, 13, 25, 30. The median class is 20 – 29. Thus L = 20, F = 13, f = 12 and c = 10.
总频数 n = 30，因此 n/2 = 15。累积频数分别为5、13、25、30。中位数所在组为20 – 29。于是 L = 20, F = 13, f = 12, c = 10。
Median = 20 + (15 − 13) ÷ 12 × 10 = 20 + 1.67 ≈ 21.7
中位数 = 20 + (15 − 13) ÷ 12 × 10 = 20 + 1.67 ≈ 21.7
6. Estimating Measures from Histograms | 从直方图估计集中趋势
Histograms are commonly used in A-Level statistics questions. The mean can be estimated using the same mid-point method. The median can be estimated by finding the value that splits the total area of the histogram into two equal halves, because area is proportional to frequency.
直方图是A-Level统计题中常见的图形。均值仍可通过组中值法估计。中位数则可以通过寻找将直方图总面积等分为两半的数值来估计，因为面积与频数成正比。
To find the median class from a histogram, first calculate the total frequency, then find n/2. Use the cumulative frequencies to locate the class containing this value. Apply linear interpolation using the class boundaries read from the horizontal axis.
要从直方图中找到中位数所在组，首先计算总频数，然后找到n/2。通过累积频数确定该值所在的组。最后用横轴上读取的边界值进行线性插值。
The mode of a histogram can be estimated using the modal class together with the frequencies of neighbouring classes. The formula is:
直方图中众数的估计需要结合模态组以及相邻组的频数。公式为：
Mode ≈ L + d₁ ÷ (d₁ + d₂) × c
quency of previous class, and d₂ = frequency of modal class minus frequency of next class.
其中d₁ = 模态组频数减去前一组的频数，d₂ = 模态组频数减去后一组的频数。
7. Choosing the Best Measure | 如何选择最合适的指标
In exam questions, you are often asked to select the most appropriate measure of central tendency. The answer depends on the distribution of data and the presence of outliers.
在考试中，经常要求你选择最合适的集中趋势指标。答案取决于数据的分布形态以及是否存在异常值。
- Use the mean when data are symmetrically distributed and contain no extreme outliers.
- Use the median when the data are skewed or contain outliers.
- Use the mode for categorical data or when identifying the most common value is more meaningful than computing an average.
- 当数据对称分布且无极端异常值时，使用均值。
- 当数据偏斜或存在异常值时，使用中位数。
- 对于分类数据，或当“最常见的值”比“平均值”更有意义时，使用众数。
For example, consider house prices in a city where one mansion is worth £10,000,000 while most homes are worth around £200,000. The mean will be dragged upwards by the mansion, so the median is a better summary of typical house prices.
例如，考虑一座城市中的房价：一栋豪宅价值1,000万英镑，而大多数房屋价值约20万英镑。均值会被豪宅大幅拉高，因此中位数更能代表典型房价。
8. Common Exam Pitfalls | 常见考试陷阱
Students often lose marks on central tendency questions due to small but avoidable errors. Here are the most common pitfalls:
学生在集中趋势题目中常因一些细小但可以避免的错误而失分。以下是最常见的陷阱：
- Forgetting to order the data before finding the median.
- Using the raw mid-point value instead of the class boundary when applying linear interpolation.
- Confusing frequency with cumulative frequency.
- Calculating the mean of grouped data without multiplying by frequency.
- Choosing the median class as the class with the largest frequency instead of the class containing n/2.
- 求中位数前忘记将数据排序。
- 在线性插值时误用组中值而不是组边界。
- 将频数与累积频数混淆。
- 计算分组数据均值时没有乘以频数。
- 将模态组误认为中位数所在组，而不是包含 n/2 的那一组。
data are given as a histogram, remember that frequency = frequency density × class width.
始终检查你的单位并仔细审题：如果题目要求中位数，不要给出均值。当数据以直方图形式给出时，请记住频数 = 频率密度 × 组宽。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图