Measures of Central Tendency | 集中趋势的度量

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

In A-Level Statistics, a measure of central tendency is a single value that summarises the centre of a data set. The three most common measures are the arithmetic mean, the median and the mode, and each has its own strengths and weaknesses in different contexts.

在 A-Level 统计学中，集中趋势度量是用一个数值来概括数据集中心位置的方法。最常见的三种度量是算术平均数、中位数和众数，它们在不同情境下各有优缺点。

1. What Is a Measure of Central Tendency? | 什么是集中趋势度量？

A measure of central tendency tells us where the bulk of the data values lie. It provides a representative or typical value that can be used to summarise a whole distribution quickly.

集中趋势度量告诉我们大部分数据值位于何处。它提供了一个具有代表性的典型值，可用于快速概括整个分布。

At A-Level, you are expected to calculate these measures for raw data, frequency tables and grouped data, and to understand when each measure is most appropriate.

在 A-Level 阶段，你需要掌握对原始数据、频数分布表和分组数据计算这些度量，并理解每种度量在何种情况下最合适。

2. The Arithmetic Mean | 算术平均数

The arithmetic mean, usually called the mean, is found by adding all data values and dividing by the number of values. For raw data, the formula is:

算术平均数通常简称为平均数，是将所有数据值相加后除以数据个数得到的。对于原始数据，公式为：

x̄ = Σx / n

Here, Σx is the sum of all observations and n is the number of observations. For example, for the data set 4, 7, 8, 10, 12, the mean is (4 + 7 + 8 + 10 + 12) / 5 = 41 / 5 = 8.2.

其中 Σx 表示所有观测值的总和，n 表示观测值的个数。例如，对于数据集 4、7、8、10、12，平均数为 (4 + 7 + 8 + 10 + 12) / 5 = 41 / 5 = 8.2。

The mean uses every value in the data set, so it is widely used in further statistical calculations such as variance and standard deviation.

平均数使用了数据集中的每一个数值，因此它被广泛用于方差、标准差等进一步的统计计算中。

3. The Median | 中位数

The median is the middle value when the data are arranged in ascending order. If n is odd, the median is the middle observation; if n is even, it is the mean of the two middle observations.

中位数是将数据按升序排列后位于中间位置的数值。如果 n 为奇数，中位数就是正中间的观测值；如果 n 为偶数，中位数则是中间两个观测值的平均数。

To locate the median position, use the formula:

要确定中位数的位置，可使用公式：

Median position = (n + 1) / 2

For example, in the ordered data set 3, 5, 8, 11, 15, 20, the median position is (6 + 1) / 2 = 3.5, so the median is (8 + 11) / 2 = 9.5.

例如，在升序排列的数据集 3、5、8、11、15、20 中，中位数位置为 (6 + 1) / 2 = 3.5，因此中位数为 (8 + 11) / 2 = 9.5。

The median is particularly useful when a distribution is skewed or contains outliers, because it is not influenced by extreme values.

当分布偏斜或包含异常值时，中位数尤其有用，因为它不受极端值的影响。

4. The Mode | 众数

The mode is the value that occurs most frequently in a data set. A data set may have one mode, more than one mode, or no mode at all.

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、多个众数，或者没有众数。

For example, in the data set 2, 4, 4, 6, 7, 7, 7, 9, the mode is 7 because it appears three times. If two values have the same highest frequency, the data are bimodal.

例如，在数据集 2、4、4、6、7、7、7、9 中，众数为 7，因为它出现了三次。如果两个数值具有相同的最高频率，则数据为双峰。

The mode is the only measure of central tendency that can be used for categorical data, such as favourite colour or type of transport.

众数是唯一可用于类别数据的集中趋势度量，例如最喜欢的颜色或交通方式。

5. Mean from a Frequency Table | 频数分布表中的平均数

When data are presented in a frequency table, the mean is calculated using the formula:

当数据以频数分布表的形式呈现时，平均数可使用以下公式计算：

x̄ = Σfx / Σf

Here, x is the data value, f is its frequency, and Σfx is the sum of each value multiplied by its frequency. For example:

其中 x 是数据值，f 是其频数，Σfx 是每个数据值乘以其频数后的总和。例如：

x	f	fx
1	3	3
2	7	14
3	5	15
4	2	8

Here Σf = 17 and Σfx = 40, so the mean is 40 / 17 ≈ 2.35.

此处 Σf = 17，Σfx = 40，因此平均数约为 40 / 17 ≈ 2.35。

6. Estimating the Mean from Grouped Data | 分组数据中平均数的估算

For grouped continuous data, we cannot calculate the exact mean because we only know class intervals, not the individual values. We therefore estimate the mean using class midpoints.

对于分组连续数据，我们无法计算精确的平均数，因为我们只知道组距，而不知道每个具体数值。因此，我们使用组中点来估算平均数。

The estimated mean is given by:

平均数的估算公式为：

x̄ ≈ Σfm / Σf

Here, m is the midpoint of each class, calculated as (lower class boundary + upper class boundary) / 2, and f is the class frequency.

其中 m 是每组的组中点，计算方法为 (组下限 + 组上限) / 2，f 是该组的频数。

For example, for the class 10 ≤ x < 20, the midpoint is 15. Every value in the class is assumed to be at the midpoint, which introduces the approximation.

例如，对于组 10 ≤ x < 20，组中点为 15。假设该组中的所有数值都位于中点，这就引入了近似。

7. Interpolating the Median from Grouped Data | 分组数据中位数的插值

In a grouped frequency table, the median is estimated by linear interpolation within the median class. Use the cumulative frequency to identify the class containing the median.

在分组频数表中，中位数通过在中位数所在组内进行线性插值来估算。使用累积频数来确定包含中位数的组。

The interpolation formula is:

插值公式为：

Median = L + [(n/2 – F) / f] × w

Here, L is the lower class boundary of the median class, n is the total frequency, F is the cumulative frequency before the median class, f is the frequency of the median class, and w is the class width.

其中 L 是中位数所在组的下边界，n 是总频数，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，w 是组距。

For example, with classes 0-10, 10-20, 20-30, 30-40 and frequencies 3, 7, 5, 2, the total n = 17, so n/2 = 8.5. The cumulative frequencies are 3, 10, 15, 17, so the median class is 10-20. Then L = 10, F = 3, f = 7, w = 10, giving:

例如，组为 0-10、10-20、20-30、30-40，频数分别为 3、7、5、2，总频数 n = 17，因此 n/2 = 8.5。累积频数为 3、10、15、17，所以中位数所在组为 10-20。于是 L = 10，F = 3，f = 7，w = 10，得到：

Median = 10 + [(8.5 – 3) / 7] × 10 ≈ 17.86

Remember that for grouped continuous data, the median position is usually taken as n/2 rather than (n + 1)/2.

请记住，对于分组连续数据，中位数的位置通常取 n/2，而不是 (n + 1)/2。

8. Choosing an Appropriate Measure | 选择适当的集中趋势度量

Choosing the best measure depends on the type of data and the shape of the distribution. The table below summarises the main considerations.

选择最佳度量取决于数据的类型和分布的形状。下表总结了主要的考虑因素。

Measure	Advantages	Disadvantages
Mean	Uses all values; useful in further statistics	Affected by outliers and skewness
Median	Resistant to outliers; good for skewed data	Ignores actual sizes of most values
Mode	Works for categorical data; easy to identify	May not exist or may not be unique

In general, use the mean for symmetric distributions without outliers, the median for skewed distributions or data with outliers, and the mode for categorical data.

一般来说，对于无异常值的对称分布使用平均数，对于偏斜分布或含异常值的数据使用中位数，对于类别数据使用众数。

9. Effect of Outliers and Skewness | 异常值与偏态的影响

An outlier is an extreme value that lies far away from the rest of the data. The mean is very sensitive to outliers because it includes every value in the calculation.

异常值是指远离其余数据的极端值。平均数对异常值非常敏感，因为它在计算中包含了每一个数值。

For example, in the data set 5, 6, 7, 8, 100, the mean is 25.2, which does not represent a typical value well. The median is 7, which is much more representative.

例如，在数据集 5、6、7、8、100 中，平均数为 25.2，不能很好地代表典型值。中位数为 7，更具代表性。

In a positively skewed distribution, the mean is usually greater than the median, which is usually greater than the mode. In a negatively skewed distribution, the order is reversed.

在正偏态分布中，平均数通常大于中位数，中位数通常大于众数。在负偏态分布中，顺序则相反。

Positive skew: mean > median > mode

Negative skew: mean < median < mode

10. Coding and Linear Transformations | 编码与线性变换

Data are often coded to make calculations easier, especially when values are large or close together. If a new variable y is defined by:

数据通常经过编码以简化计算，尤其是当数值较大或彼此接近时。如果定义新变量 y 为：

y = (x – a) / b

then the original x values can be recovered by x = a + by. The mean of the coded data satisfies the reverse transformation:

那么原始 x 值可以通过 x = a + by 还原。编码数据的平均数满足反向变换：

x̄ = a + b ȳ

For example, to find the mean of 102, 106, 110, 114, code using y = x – 100. The coded values are 2, 6, 10, 14, whose mean is 8. Therefore x̄ = 100 + 8 = 108.

例如，要求 102、106、110、114 的平均数，可使用 y = x – 100 进行编码。编码后的值为 2、6、10、14，其平均数为 8。因此 x̄ = 100 + 8 = 108。

The median and mode also transform in the same linear way, so linear coding does not change the underlying shape of the distribution.

中位数和众数也以同样的线性方式变换，因此线性编码不会改变分布的潜在形状。

11. Comparing Data Sets Using Averages | 使用平均数比较数据集

When comparing two or more data sets, a measure of central tendency gives a starting point. You should compare the means or medians and state which data set is on average higher or lower.

在比较两个或多个数据集时，集中趋势度量是一个起点。你应比较平均数或中位数，并说明哪个数据集平均而言更高或更低。

For example, if class A has a mean test score of 68 and class B has a mean of 74, class B performed better on average, assuming both distributions have similar shapes and no extreme outliers.

例如，如果 A 班的平均测验成绩为 68，B 班为 74，那么 B 班平均表现更好，前提是两个分布形状相似且没有极端异常值。

Central tendency alone does not describe spread. For a complete comparison, you should also consider standard deviation or interquartile range, but central measures remain essential.

仅靠集中趋势本身无法描述离散程度。要进行完整比较，还应考虑标准差或四分位距，但集中趋势度量仍然至关重要。

12. Exam Tips and Common Mistakes | 考试技巧与常见错误

Edexcel exam questions often combine calculation with interpretation. Keep the following points in mind:

Edexcel 考试题目通常将计算与解释结合起来。请牢记以下几点：

Use n/2 for grouped median interpolation, not (n + 1)/2 – 在分组数据中位数插值时使用 n/2，而不是 (n + 1)/2。
Use class midpoints correctly when estimating the mean from grouped data – 在分组数据中估算平均数时正确使用组中点。
Check that your answer lies inside the median class or within the data range – 检查你的答案是否位于中位数所在组内或数据范围内。
Do not confuse the median class with the modal class – 不要将中位数所在组与众数所在组混淆。
Always include units if the data measure physical quantities – 如果数据测量的是物理量，请始终写明单位。

By practising these calculations in frequency tables and grouped data contexts, you will be well prepared for Edexcel A-Level statistics questions on central tendency.

通过在频数表和分组数据情境中练习这些计算，你将为 Edexcel A-Level 统计学中关于集中趋势的题目做好充分准备。

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Measures of Central Tendency | 集中趋势的度量