Measures of Central Tendency | 集中趋势的度量

📚 Measures of Central Tendency | 集中趋势的度量

In statistics, a measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that data set. It is often called an ‘average’ and is one of the fundamental concepts in data analysis. At A-Level, particularly under the Edexcel specification, you are expected to understand three main measures: the mean, the median, and the mode. You must also be able to calculate them for raw data, frequency tables, and grouped data, using techniques such as linear interpolation. This article will walk you through all the essential methods, offer worked examples, and explain how transformations affect these measures.

在统计学中，集中趋势的度量是一个用于描述数据集中间位置的单一数值，常被称为“平均数”，是数据分析中的基本概念之一。在 A-Level 阶段，尤其是 Edexcel 考试大纲中，你需要掌握三种主要的集中趋势度量：均值、中位数和众数。你还必须能够在原始数据、频数表和分组数据中计算它们，并使用诸如线性插值等技巧。本文将带你梳理所有核心方法，提供详细示例，并解释数据变换如何影响这些统计量。

1. Introduction to Central Tendency | 集中趋势简介

Central tendency measures summarise a dataset with one representative number, providing a quick snapshot of the data’s typical value. In Edexcel A-Level Mathematics, the ‘Location and Measures’ topic appears both in the applied statistics units and across the data handling sections. The mean, median, and mode each capture this in a different way, and understanding their behaviour is crucial for choosing the right summary in a given context.

集中趋势的度量用一个代表性数字来概括数据集，能够快速呈现数据的典型值。在 Edexcel A-Level 数学中，“位置与度量”这一主题不仅出现在应用统计学单元中，也贯穿于数据处理部分的各个章节。均值、中位数和众数各以不同的方式捕捉数据的中心，理解它们的特点对于在特定情境下选择合适的概括量至关重要。

In addition to the basic calculations, the specification requires you to estimate these measures from grouped frequency distributions, appreciate the effect of linear transformations, and interpret the location of a distribution in terms of skewness. All these skills will be covered in the sections that follow.

除了基本的计算之外，大纲还要求你从分组频数分布中估计这些度量、理解线性变换的影响，并根据偏斜程度解读分布的位置。所有这些技能都将在后续小节中逐一呈现。

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2. The Arithmetic Mean | 算术平均

The arithmetic mean, commonly called the mean, is the sum of all observed values divided by the number of observations. For a set of n data values x₁, x₂, …, xₙ the mean, denoted by x̄, is given by:

算术平均，通常简称为均值，是所有观测值的总和除以观测值的个数。对于包含 n 个数据值 x₁, x₂, …, xₙ 的集合，用 x̄ 表示的均值计算公式为：

x̄ = Σxᵢ / n

Here Σxᵢ represents the sum of all the x-values. The mean takes every piece of data into account, which makes it very sensitive to extreme values or outliers. For example, the mean of 5, 7, 8, 9, 11 is (5+7+8+9+11) ÷ 5 = 40 ÷ 5 = 8.

其中 Σxᵢ 表示所有 x 值的总和。均值考虑了每一个数据点，这使得它对极端值或异常值非常敏感。例如，数据 5, 7, 8, 9, 11 的均值为 (5+7+8+9+11) ÷ 5 = 40 ÷ 5 = 8。

The mean is widely used because of its mathematical convenience and its role in further statistical work, such as variance and regression. However, when a dataset contains a small number of very large or very small values, the mean can be pulled away from the centre of the bulk of the data and may no longer be representative.

均值因其数学上的便利性以及在方差和回归等后续统计工作中的重要作用而被广泛使用。然而，当数据集包含少量极大或极小的数值时，均值可能会被拉向这些值，从而偏离大部分数据的中心，可能不再具有代表性。

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3. The Median | 中位数

The median is the middle value when the data have been arranged in order of size. If there are n observations, the position of the median is found using (n+1)/2. For an odd number of values, the median is the exact middle data point. For an even number, it is the mean of the two middle values. This measure effectively splits the dataset into two equal halves.

中位数是将数据按大小顺序排列后位于中间的数值。如果观测值有 n 个，中位数的位置由 (n+1)/2 确定。当数据个数为奇数时，中位数就是正中间的那个数据点；当个数为偶数时，它是中间两个值的平均。这种度量可以将数据集有效地分成两个相等的部分。

For example, in the ordered dataset 2, 4, 7, 9, 12, the median is 7 because there are 5 values and the 3rd value is 7. In the dataset 2, 4, 7, 9, the median is (4+7)/2 = 5.5. The key advantage of the median is its resistance to outliers; it only depends on the central order statistics and ignores the actual magnitudes of extreme observations.

例如，在已排序的数据集 2, 4, 7, 9, 12 中，中位数为 7，因为共有 5 个值，第 3 个值是 7。在数据集 2, 4, 7, 9 中，中位数为 (4+7)/2 = 5.5。中位数的主要优点是对异常值具有抵抗力；它只依赖于中间的顺序统计量，而忽略极端观测值的实际大小。

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4. The Mode | 众数

The mode is the value that occurs most frequently in a dataset. It is possible for a dataset to have one mode (unimodal), two modes (bimodal), or several modes (multimodal). If all values appear with the same frequency, the dataset has no mode. Because the mode relies only on frequency counts, it is the easiest measure to identify but the least used in deeper statistical analysis.

众数是数据集中出现频率最高的数值。一个数据集可能有一个众数（单峰）、两个众数（双峰）或多个众数（多峰）。如果所有值出现的频数都相同，那么该数据集就没有众数。由于众数只依赖于频数计数，因此它是最容易识别的度量，但在深入的统计分析中使用得也最少。

The mode can be applied to qualitative as well as quantitative data. For instance, in a survey of favourite colours, the mode would indicate the most popular colour. In grouped data, we refer to the modal class rather than a precise mode, though an estimate can be obtained using a standard formula.

众数既可应用于定量数据，也可应用于定性数据。例如，在一项关于最喜欢的颜色的调查中，众数将表示最受欢迎的颜色。在分组数据中，我们通常提及众数所在的区间而非精确的众数，但也可以通过标准公式得到估计值。

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5. Mean from a Frequency Table | 从频率表计算均值

When data are presented in an ungrouped frequency table, the mean is calculated by multiplying each distinct data value x by its frequency f, summing all these products, and dividing by the total frequency Σf. The formula becomes:

当数据以未分组的频数表形式呈现时，均值的计算方法是：将每个不同的数据值 x 乘以其频数 f，将所有这些乘积相加，再除以总频数 Σf。公式变为：

x̄ = Σ(f × x) / Σf

This approach saves time when many values are repeated and ensures the same result as listing every individual data point. In an exam, constructing an additional column for f × x in the table is recommended. For example, a dice rolled 20 times with frequency distribution: Score 1 (f=3), Score 2 (f=4), Score 3 (f=3), Score 4 (f=5), Score 5 (f=3), Score 6 (f=2) yields Σfx = 1×3 + 2×4 + 3×3 + 4×5 + 5×3 + 6×2 = 3+8+9+20+15+12 = 67. Mean = 67/20 = 3.35.

当许多数值重复出现时，这种方法可以节省时间，并且可以确保与列出每个数据点得到相同的结果。在考试中，建议在表格中添加一列 f × x。例如，掷骰子 20 次的频数分布为：点数 1（f=3），点数 2（f=4），点数 3（f=3），点数 4（f=5），点数 5（f=3），点数 6（f=2）。Σfx = 1×3+2×4+3×3+4×5+5×3+6×2 = 3+8+9+20+15+12 = 67。均值 = 67/20 = 3.35。

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6. Estimating the Mean for Grouped Data | 分组数据的均值估计

For continuous data organised into class intervals, we cannot calculate the exact mean unless the original raw data are known. Instead, we estimate the mean using the midpoint of each interval. The midpoint xₘ is taken as the representative value for all observations in that class. The formula is:

对于被组织成组距区间的连续数据，除非已知原始数据，否则我们无法计算精确的均值。此时，我们使用每个区间的中点来估计均值。中点 xₘ 被视为该组内所有观测值的代表值。计算公式为：

x̄ ≈ Σ(f × xₘ) / Σf

It is crucial to correctly identify the class boundaries and midpoints. If a class is given as ’10–19′, assuming continuous data, the class boundaries are 9.5 to 19.5 and the midpoint is 14.5. The estimation works well when the data are roughly symmetric within each interval, but it remains an approximation that can be influenced by interval width and skewness.

正确识别组的边界和中点至关重要。如果一组给出为“10–19”，假设数据连续，组边界为 9.5 至 19.5，中点为 14.5。当每个区间内的数据大致对称时，这种估计效果不错，但这仍然是一种近似，会受到组距宽度和偏斜程度的影响。

In exam questions, you will often need to fill in the midpoint column and complete the f × xₘ column, sum them, and divide by the total frequency. Edexcel may also ask you to interpret the result in context or compare with other measures.

在考试题中，你通常需要填写中点列，完成 f × xₘ 列，求和，再除以总频数。Edexcel 可能还会要求你根据情境解释结果，或与其他度量进行比较。

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7. Linear Interpolation for the Median with Grouped Data | 分组数据中位数的线性插值

When data are grouped, the exact median cannot be located. We approximate the median by assuming that the data values within the median class are evenly spread. Linear interpolation uses cumulative frequency to pinpoint the position. The position of the median is n/2, where n = Σf.

当数据被分组后，无法找到精确的中位数。我们假设中位数所在组内的数据值是均匀分布的，从而对中位数进行近似。线性插值使用累积频数来确定位置。中位数的位置为 n/2，其中 n = Σf。

The median is estimated using the formula:

中位数的估计值为：

Median = L + ((n/2 – F) / fₘ) × w

where L is the lower class boundary of the median class, F is the cumulative frequency before the median class, fₘ is the frequency of the median class, and w is the class width. The median class is the first class for which the cumulative frequency reaches or exceeds n/2.

其中 L 为中位数所在组的下边界，F 为低于中位数所在组的累积频数，fₘ 为中位数所在组的频数，w 为组距宽度。中位数所在组是第一个累积频数达到或超过 n/2 的那一组。

Worked example: In a grouped table of heights, Σf = 50, n/2 = 25. The cumulative frequencies are 8, 20, 35, 45, 50. The median class is the third class (cumulative 35). If the class is ‘160–164 cm’, boundaries are 159.5–164.5, w = 5, L = 159.5, F = 20, fₘ = 15. Then Median = 159.5 + ((25 – 20)/15) × 5 = 159.5 + (5/15)×5 = 159.5 + 1.666… ≈ 161.17 cm.

计算示例：在一个身高的分组表中，Σf = 50，n/2 = 25。累积频数依次为 8, 20, 35, 45, 50。中位数所在组是第三组（累积 35）。如果该组为“160–164 cm”，边界为 159.5–164.5，w = 5，L = 159.5，F = 20，fₘ = 15。那么中位数 = 159.5 + ((25 – 20)/15) × 5 = 159.5 + (5/15)×5 = 159.5 + 1.666… ≈ 161.17 cm。

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8. Estimating the Mode for Grouped Data | 分组数据众数的估计

For grouped data, the mode is not simply the midpoint of the tallest frequency class. Instead, we identify the modal class (the class with the highest frequency) and apply a formula that adjusts for neighbouring frequencies. The estimated mode is:

对于分组数据，众数并不是简单地位于频数最高组的中点。我们需要确定众数所在的组（即频数最高的组），并应用一个考虑了相邻组频数的公式。众数的估计值为：

Mode ≈ L + ( (fₘ – fₚ) / (2fₘ – fₚ – fₛ) ) × w

Here L is the lower boundary of the modal class, fₘ is the frequency of the modal class, fₚ is the frequency of the class preceding it, fₛ is the frequency of the class following it, and w is the class width. This formula works best for continuous data with unimodal distributions.

其中 L 是众数所在组的下边界，fₘ 是众数所在组的频数，fₚ 是前一组的频数，fₛ 是后一组的频数，w 是组距。该公式最适用于单峰分布的连续数据。

When the specification does not require a precise estimate, simply stating the modal class is acceptable. However, Edexcel S1 (and newer specification) may ask for the modal class or this estimate. Be sure to check the question’s wording.

当考试大纲不要求精确估计时，只需给出众数所在组即可。但 Edexcel S1（以及新大纲）可能会要求提供该组或此估计值。请务必仔细审题。

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9. Comparing Measures of Central Tendency | 集中趋势度量的比较

Each measure has strengths and weaknesses. The mean uses all data and is essential for advanced statistics, but it is sensitive to outliers. The median is robust and better for skewed distributions, but it ignores the actual values of data outside the centre. The mode is simple and can be used for categorical data, but it may not exist or may not be unique, and it discards most information about the dataset.

每种度量都有其优缺点。均值使用了所有数据，对高级统计学至关重要，但它对异常值敏感。中位数具有较强的稳健性，更适用于偏斜分布，但它忽略了中心以外数据的实际数值。众数简单易懂，可用于分类数据，但它可能不存在或不唯一，并且舍弃了数据集中大部分的信息。

In a perfectly symmetric distribution, the mean, median, and mode coincide. In a right-skewed (positively skewed) distribution, the mean is typically greater than the median, and the median is greater than the mode. In a left-skewed (negatively skewed) distribution, the order is reversed. Understanding these relationships helps you select the most representative measure.

在完全对称的分布中，均值、中位数和众数三者重合。在右偏（正偏）分布中，均值通常大于中位数，而中位数又大于众数。在左偏（负偏）分布中，顺序则相反。理解这些关系有助于你选择最具代表性的度量。

In real-life contexts such as income or house prices, the median is often preferred over the mean because a few very high earners or expensive properties would inflate the mean significantly. The mode is rarely used alone for quantitative decision-making but can describe the most typical case.

在现实情境中，比如收入或房价，中位数通常比均值更受青睐，因为少数极高的收入或昂贵的房产会显著抬高均值。众数很少单独用于定量决策，但可以用来描述最典型的情况。

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10. Effect of Transformations on Central Tendency | 数据变换对集中趋势的影响

Linear transformations of the form y = a + bx affect the measures of central tendency in predictable ways, as required by the Edexcel specification. Adding a constant a to every observation shifts the mean, median, and mode by exactly a units. Multiplying every observation by a constant b multiplies the mean, median, and mode by b.

形如 y = a + bx 的线性变换会按照可预测的方式影响集中趋势的度量，这也是 Edexcel 大纲所要求的内容。给每个观测值加上常数 a，会使均值、中位数和众数都恰好平移 a 个单位。将每个观测值乘以常数 b，则会使均值、中位数和众数都乘以 b。

Specifically, if new variable y = a + bx where x is the original variable, then ȳ = a + b x̄, Median_y = a + b × Median_x, and Mode_y = a + b × Mode_x. This property is extremely useful when coding data to simplify calculations. After working out the mean of coded data, you can back-transform to find the original mean.

具体来说，若新变量 y = a + bx，x 为原始变量，则 ȳ = a + b x̄，y 的中位数 = a + b × x 的中位数，y 的众数 = a + b × x 的众数。当你对数据进行编码以简化计算时，这一性质极为有用。在算出编码数据的均值后，你可以反向变换得到原始均值。

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11. Skewness and the Measures’ Relationship | 偏斜与各度量的关系

The relative positions of the mean, median, and mode provide a quick indication of skewness. For a positive skew (tail on the right), the order typically is: mode < median < mean. For a negative skew (tail on the left), mean < median < mode. In exam questions, you may be given a histogram or box plot and asked to deduce skewness from these relationships.

均值、中位数和众数的相对位置可以快速指示分布偏斜的方向。对于正偏斜（尾部在右侧），顺序通常是：众数 < 中位数 < 均值。对于负偏斜（尾部在左侧），均值 < 中位数 < 众数。在考试题目中，你可能会看到一个直方图或箱线图，并被要求根据这些关系推断偏斜情况。

Knowledge of skewness is important for choosing between the mean and median. When the data are skewed, the median is usually the better measure of central tendency because it resists the pull of the long tail. The mean, however, remains the basis for many inferential statistics tests that assume approximate symmetry.

了解偏斜对于在均值和中位数之间做出选择非常重要。当数据偏斜时，中位数通常是集中趋势的更优度量，因为它能抵抗长尾的拉扯。然而，均值为许多基于近似对称假定的推断统计检验奠定了基础。

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12. Weighted Mean | 加权平均

Sometimes different values in a dataset carry different levels of importance. The weighted mean accounts for this by assigning a weight wᵢ to each value xᵢ. The formula is:

有时数据集中的不同数值具有不同的重要程度。加权平均通过为每个数值 xᵢ 分配一个权数 wᵢ 来考虑这一点。其计算公式为：

x̄_w = Σ(wᵢ × xᵢ) / Σwᵢ

This concept appears in index numbers, expected values, and in situations like calculating a student’s grade when coursework and exams have different weightings. The unweighted mean is a special case where all weights equal 1.

这一概念出现在指数、期望值以及像计算学生成绩这种情况下（当作业和考试具有不同权重时）。未加权平均是所有权数都等于 1 时的特例。

For example, if two assessments: score 70 with weight 2 and score 85 with weight 3, then weighted mean = (70×2 + 85×3) / (2+3) = (140+255)/5 = 79. The ordinary mean would have been (70+85)/2 = 77.5. The weighting pulls the result towards the higher-scoring assessment.

例如，两项评估：分数 70 权重为 2，分数 85 权重为 3，那么加权平均 = (70×2 + 85×3) / (2+3) = (140+255)/5 = 79。而普通均值则为 (70+85)/2 = 77.5。权重的拉动使得结果向分数较高的评估倾斜。

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