📚 Measuring the Centre of Data in IB Mathematics | IB数学：数据中心度量
Measures of central tendency - the mean, the median and the mode - form the first block of the descriptive statistics unit in IB Mathematics: Applications and Interpretation (topic 4.1), and they resurface in Mathematics: Analysis and Approaches whenever a data set has to be summarised. This revision guide works through every formula you are expected to know, shows fully worked examples for both ungrouped and grouped data, explains quartiles, the five-number summary and outlier tests, and finishes with the calculator routines and exam traps that most often cost students marks.
数据中心度量——平均数、中位数与众数——是 IB 数学：应用与解释（AI）大纲描述性统计部分的第一个知识板块（4.1），在数学：分析与方法（AA）中只要涉及数据汇总也一定会再次出现。本复习指南系统梳理你必须掌握的全部公式，给出未分组数据与分组数据的完整例题，讲解四分位数、五数概括与异常值判定，并在最后总结图形计算器的操作流程和最常导致失分的考试陷阱。
1. What Are Measures of Central Tendency? | 什么是集中趋势的度量？
A measure of central tendency is a single number that is used to represent the "middle" or "typical" value of a data set. The IB syllabus names three: the mean (the arithmetic average), the median (the middle value once the data is ordered) and the mode (the most frequent value). Each one answers a slightly different question, and the correct choice depends on the shape of the distribution and on whether extreme values are present.
集中趋势的度量是用一个数值来代表整组数据的“中间值”或“典型值”。IB 大纲明确列出三种：平均数（算术平均）、中位数（数据排序后位于中间的值）和众数（出现次数最多的值）。它们各自回答的问题略有不同，选用哪一种取决于数据的分布形态以及是否存在极端值。
Every measure of centre must be reported with its unit, and it should always be quoted together with a measure of spread - usually the range, the interquartile range or the standard deviation - because a centre with no spread tells only half the story.
任何中心度量都必须带上单位，并且应当与离散程度的度量一起报告——通常是全距、四分位距或标准差——因为只有中心而没有离散程度，等于只讲了一半的事实。
- Mean / 平均数: uses every value, sensitive to outliers. / 用上全部数据，对异常值敏感。
- Median / 中位数: positional, resistant to outliers. / 基于位置，抗异常值。
- Mode / 众数: frequency-based, works for categorical data too. / 基于频数，也适用于分类数据。
2. The Arithmetic Mean | 算术平均数
For a sample of n values, the mean is the sum of all observations divided by the number of observations. The Greek capital letter sigma (Σ) means "add up all of them", so the formula is compact and easy to apply.
对于含有 n 个观测值的样本，平均数等于所有观测值之和除以观测值的个数。希腊大写字母 Σ 表示“全部相加”，因此公式非常简洁、便于使用。
x̄ = Σx ÷ n
where Σx is the sum of the data values and n is the number of values. Note the notation convention: x̄ (x-bar) is used for a sample mean, while the Greek letter μ (mu) is reserved for a population mean.
其中 Σx 是全部数据值之和，n 是数据个数。注意符号约定：x̄（读作 x-bar）表示样本平均数，而希腊字母 μ（mu）专门表示总体平均数。
t-answer questions. First, the sum of the deviations from the mean is always zero: Σ(x − x̄ et is multiplied by a constant k and then increased by a constant c, the new mean is k × x̄ + c. These properties make the mean the natural measure to use in algebra-based questions.
有两个性质值得牢记，因为它们经常出现在简答题中。第一，各观测值与平均数之差的代数和恒为零：Σ(x − x̄) = 0。第二，若数据中每个值都乘以常数 k 再加上常数 c，则新的平均数为 k × x̄ + c。这些性质使平均数在代数类题目中成为最自然的度量。
3. The Mean from a Frequency Table | 由频数表求平均数
When the data is presented as a frequency table, the mean is found by multiplying each value by its frequency, adding the products, and dividing by the total frequency. Each value x therefore contributes with a weight f, which is exactly why the formula below is sometimes called a weighted mean.
当数据以频数表形式给出时，求平均数的方法是：把每个数值乘以其频数，将所有乘积相加，再除以总频数。于是每个数值 x 都以频数 f 为权重参与计算，这也是下面这个公式有时被称为加权平均数的原因。
x̄ = Σfx ÷ Σf
Worked example: the table below records the number of siblings reported by 25 students.
例题：下表记录了 25 名学生所填写的兄弟姐妹人数。
| x | 1 | 2 | 3 | 4 | 5 |
| f | 3 | 7 | 8 | 5 | 2 |
| fx | 3 | 14 | 24 | 20 | 10 |
Here Σf = 25 and Σfx = 3 + 14 + 24 + 20 + 10 = 71, so x̄ in the group x = 3, so the median is 3 and the mode is also 3 (the highest frequency, 8).
这里 Σf = 25，Σfx = 3 + 14 + 24 + 20 + 10 = 71，所以 x̄ = 71 ÷ 25 = 2.84。由于累计频数依次为 3、10、18、23、25，排序后的第 13 个值落在 x = 3 这一组中，因此中位数为 3，众数同样是 3（频数最高，为 8）。
A common error is to average the x-values themselves and ignore the frequencies; that would give 3 here instead of 2.84. Always show the Σf and Σfx rows - IB mark schemes usually award a method mark for them.
常见错误是直接对 x 求平均而忽略频数，那样会得到 3 而不是 2.84。务必写出 Σf 与 Σfx 两栏——IB 评分标准通常会为此给方法分。
4. Grouped Data and the Midpoint | 分组数据与组中值
When data is grouped into class intervals, the individual values are no longer visible, so each interval is represented by its midpoint (also called the class mark). The midpoint is simply the average of the lower and upper class boundaries, and the mean is then calculated exactly as for a frequency table.
当数据被合并为组区间后，个体数值已经不可见，于是用每组的组中值（也叫组标）来代表该组。组中值就是组下限与组上限的平均数，然后按照频数表的方法求平均数即可。
midpoint = (lower boundary + upper boundary) ÷ 2 x̄ ≈ Σf xm ÷ Σf
Worked example: the times (in minutes) taken by 40 students to complete a puzzle.
例题：40 名学生完成一道拼图所用的时间（分钟）。
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| f | 4 | 9 | 15 | 8 | 4 |
| xm | 5 | 15 | 25 | 35 | 45 |
| f xm | 20 | 135 | 375 | 280 | 180 |
Here Σf = 40 and Σf xm = 20 + 135 + 375 + 280 + 180 = 990, so x̄ an estimate, not an exact value - a point that IB examiners like to see stated explicitly.
这里 Σf = 40，Σf xm = 20 + 135 + 375 + 280 + 180 = 990，所以 x̄ = 990 ÷ 40 = 24.75 分钟。由于用组中值代替了原始数据，这个结果只是估计值而非精确值——IB 阅卷官很喜欢看到考生明确指出这一点。
an interval is written as 40 < x ≤ 50 there is no ambiguity and the midpoint is still 45.
注意组界：写成 0-10、10-20 的区间通常视为连续，因此 10 属于第二组而非第一组。如果区间写作 40 < x ≤ 50，则不存在歧义，组中值仍为 45。
5. The Median for Ungrouped Data | 未分组数据的中位数
The median is the middle value of an ordered data set. The first step is always to sort the data into ascending order; skipping this step is the single most common reason for losing marks on median questions. If n is odd, the median is the value at position (n + 1) ÷ 2. If n is even, the median is the mean of the two middle values.
中位数是数据排序后位于中间的那个值。第一步永远是先把数据按从小到大排列；跳过这一步是解答中位数题目时最常见的失分原因。若 n 为奇数，中位数位于第 (n + 1) ÷ 2 个位置；若 n 为偶数，中位数是中间两个数值的平均数。
value, which is 5. For 1, 3, 4, 5, 7, 9 we have n = 6, so the median is (4 + 5) ÷ 2 = 4.5.
例如数据 4, 7, 1, 9, 5, 3, 11，先排序为 1, 3, 4, 5, 7, 9, 11。因为 n = 7，中位数是第 (7 + 1) ÷ 2 = 4 个值，即 5。而对 1, 3, 4, 5, 7, 9，n = 6，中位数为 (4 + 5) ÷ 2 = 4.5。
The median is a positional measure: it depends only on where a value sits in the ordered list, not on how large it is. That is precisely why it resists outliers, and why the median is the preferred measure of centre for income, house prices and waiting times.
中位数是一种基于位置的度量：它只取决于数值在排序中的位置，而与数值本身有多大无关。正因如此它不受异常值影响，也正因如此，收入、房价与等待时间这类数据更偏好用中位数来刻画中心。
6. The Median and Quartiles for Grouped Data | 分组数据的中位数与四分位数
For grouped data the exact middle value cannot be identified, so the median is estimated by interpolation inside the median class. The term n ÷ 2 locates the position, F is the cumulative frequency of all classes before the median class, f is the frequency of the median class and w is the class width.
对于分组数据，无法确定精确的中间数值，因此要在中位数组内通过插值来估计中位数。其中 n ÷ 2 确定位置，F 是中位数组之前所有组的累计频数，f 是中位数组的频数，w 是组宽。
median = L + ((n ÷ 2 − F) ÷ f) × w
10. Therefore median = 20 + ((20 − 13) ÷ 15) × 10 = 20 + 4.67 = 24.67 minutes, which agrees well with the estimated mean of 24.75 and confirms the distribution is close to symmetric.
仍以时间例题为例：n = 40，故 n ÷ 2 = 20。累计频数为 4、13、28、36、40，所以中位数组是 20-30，其中 L = 20，F = 13，f = 15，w = 10。因此中位数 = 20 + ((20 − 13) ÷ 15) × 10 = 20 + 4.67 = 24.67 分钟。它与估计平均数 24.75 相当接近，说明该分布接近对称。
+ ((30 − 28) ÷ 8) × 10 = 32.5, so the interquartile range is 32.5 − 16.67 = 15.83 minutes.
把 n ÷ 2 换成 n ÷ 4 和 3n ÷ 4，同一套插值方法就给出四分位数。本题中 Q1 = 10 + ((10 − 4) ÷ 9) × 10 = 16.67，Q3 = 30 + ((30 − 28) ÷ 8) × 10 = 32.5，因此四分位距为 32.5 − 16.67 = 15.83 分钟。
7. The Mode and Bimodal Data | 众数与双峰数据
The mode is the value that occurs most often. It is the only measure of centre that can be used for qualitative (categorical) data, such as favourite colour or blood group, because it does not require any arithmetic. A data set may have one mode (unimodal), two modes (bimodal), several modes, or no mode at all if every value occurs equally often.
众数是出现次数最多的数值。它是唯一可用于定性（分类）数据的中心度量，例如最喜欢的颜色或血型，因为它完全不需要算术运算。一组数据可能只有一个众数（单峰）、两个众数（双峰）、多个众数，也可能所有数值出现次数相同而无众数。
l class is 20-30 minutes, because f = 15 is the largest frequency. Exam answers that state "the mode is the 20-30 group" are fully acceptable and are often worth a single mark.
对于分组数据，我们并不报告单一的众数值，而是报告众数组——频数最大的那一组。在时间例题中，众数组是 20-30 分钟，因为 f = 15 为最大频数。答题时写“众数组为 20-30 组”完全可以接受，通常对应 1 分。
If your syllabus requires a numerical estimate for grouped data, the modal value can be interpolated as L + ((fm − f1) ÷ ((fm − f1) + (fm − f2))) × w, where f1 and f2 are the frequencies of the classes immediately before and after the
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