📚 Mode, Median, Percentiles and Skewness | 众数、中位数、百分位数与偏度
In A-Level Statistics, understanding how data is centred and spread is fundamental. Beyond the mean, the mode, median and percentiles provide robust ways to summarise distributions. Equally important is the concept of skewness, which describes the asymmetry of data and influences the relationship between these averages. This article covers the calculation, interpretation and application of these measures, focusing on both raw and grouped data, following the Edexcel specification.
在 A-Level 统计中，理解数据的集中趋势和离散程度是基础。除了均值，众数、中位数和百分位数提供了概括分布的稳健方法。同样重要的是偏度概念，它描述数据的不对称性，并影响这些平均指标之间的关系。本文涵盖这些度量的计算、解释与应用，兼顾原始数据和分组数据，严格遵循 Edexcel 考试大纲。
1. Understanding Mode | 理解众数
The mode is the value or category that occurs most frequently in a data set. A set may have one mode (unimodal), two modes (bimodal) or more (multimodal). If all values appear equally often, there is no mode. For qualitative data, the mode is the only meaningful average.
众数是数据集中出现频率最高的数值或类别。数据集可以只有一个众数（单峰）、两个众数（双峰）或更多（多峰）。如果所有值出现次数相同，则不存在众数。对于定性数据，众数是唯一有意义的平均指标。
. Instead, we estimate the modal class and interpolate using the formula: Mode ≈ L + [ (fₘ – fₚ) / ( (fₘ – fₚ) + (fₘ – y of the preceding class, fₙ is the frequency of the next class, and w is the class width.
对于分组频数分布，仅凭表格无法确定精确众数。此时我们估计众数所在组，并用公式插值：众数 ≈ L + [ (fₘ – fₚ) / ( (fₘ – fₚ) + (fₘ – fₙ) ) ] × w，其中 L 为众数组的下限，fₘ 为其频数，fₚ 为前一组的频数，fₙ 为后一组的频数，w 为组距。
The mode is not affected by extreme values (outliers), making it useful in skewed distributions where the mean might be misleading.
众数不受极端值（离群值）影响，因此在偏态分布中当均值可能产生误导时，众数很有用。
2. The Median – The Middle Value | 中位数——中间值
The median is the value that divides an ordered data set into two equal halves. For ungrouped data with n observations, the position of the median is (n + 1)/2. If this position is not an integer, we take the average of the two adjacent values. The median is robust to outliers and skewed distributions.
中位数是将有序数据集分成两半的值。对于具有 n 个观测值的未分组数据，中位数的位置是 (n + 1)/2。如果该位置不是整数，我们取相邻两个值的平均数。中位数对离群值和偏态分布具有稳健性。
n uses n/2 for grouped data, but it is safer to check marking schemes; the linear interpolation principle remains the same). The median is then estimated using: Median = L + [ ( (n/2) – before the median class, f is the frequency of the median class, and w is the class width.
对于分组连续数据，我们先通过累计频数找到中位数所在组，中位数位置通常取 n/2（Edexcel 分组数据中常用 n/2，但原理一致）。然后用线性插值估计中位数：中位数 = L + [ ( (n/2) – F ) / f ] × w，其中 L 为中位数组下限，F 为低于中位数组的累计频数，f 为该组频数，w 为组距。
3. Percentiles and Quartiles | 百分位数与四分位数
les divide a data set into 100 equal parts. The k-th percentile is the value below which k% ile (upper quartile). They provide a picture of spread and are essential in constructing box plots.
百分位数将数据集分成 100 等份。第 k 百分位数是有 k% 观测值低于该值的数值。四分位数是特定的百分位数：Q₁ = 第 25 百分位数（下四分位数），Q₂ = 第 50 百分位数（中位数），Q₃ = 第 75 百分位数（上四分位数）。它们描绘数据分布范围，是构建箱线图的关键。
For raw data, the position of Q₁ is (n+1)/4, and Q₃ is 3(n+1)/4. When these positions are not whole numbers, use linear interpolation between adjacent observations. Some exam boards accept different conventions, but Edexcel typically uses the (n+1) approach for discrete data.
对于原始数据，Q₁ 的位置是 (n+1)/4，Q₃ 的位置是 3(n+1)/4。当这些位置不是整数时，在相邻观测值间进行线性插值。虽然不同考试局可能有差异，但 Edexcel 通常对离散数据采用 (n+1) 方法。
The interquartile range (IQR = Q₃ – Q₁) measures the spread of the middle 50% of data. It is resistant to outliers and is preferred over range in skewed contexts.
四分位距（IQR = Q₃ – Q₁）衡量中间 50% 数据的离散程度。它不受离群值影响，在偏态情形中比极差更可取。
4. Calculating Percentiles for Grouped Data | 分组数据百分位数的计算
te the position P = (k/100) × n. Locate the class where the cumulative frequency first exceeds P. Then apply: Pₖ = L + [ (P – F) / f ] × w, where symbols retain their earlier meanings.
对于分组数据，百分位数通过线性插值求得。公式与中位数相似。对于第 k 百分位数，先计算位置 P = (k/100) × n。找到累计频数首次超过 P 的组，然后应用公式：Pₖ = L + [ (P – F) / f ] × w，符号含义同前。
e, to find Q₃ (75th percentile) with n = 80, calculate P = 0.75 × 80 = 60. If the class 50–60 has boundaries 50–60, frequency 15, and cumulative frequency before the class is 55, then Q₃ = 50 + [ (60 – 55) / 15 ] × 10 = 53.33 (approximately). Always check that boundaries are used correctly.
例如，求 Q₃（第 75 百分位数），n = 80，P = 0.75 × 80 = 60。若数据组 50–60 有边界 50–60，频数 15，该组前累计频数为 55，则 Q₃ = 50 + [ (60 – 55) / 15 ] × 10 = 53.33（约）。务必正确使用组界。
This method applies to any percentile and is regularly tested in Edexcel S1.
该方法适用于任何百分位数，是 Edexcel S1 常见考点。
5. Introduction to Skewness | 偏度概述
Skewness quantifies the asymmetry of a probability distribution. A symmetric distribution has no skew; the left and right sides mirror each other. Positive skew (right-skewed) occurs when the right tail is longer, and most values concentrate on the left. Negative skew (left-skewed) has a longer left tail.
偏度量化概率分布的不对称性。对称分布无偏斜，左右两侧互为镜像。正偏态（右偏）指右尾更长，大多数值聚集在左侧。负偏态（左偏）则左尾更长。
In real data, income distributions are often positively skewed, while exam scores where a test is very easy may be negatively skewed. Skewness affects the choice of appropriate average: the median is usually preferred for skewed data, and the mode can indicate the peak.
现实数据中，收入分布常呈正偏态，而一场非常简单的考试分数可能呈负偏态。偏度影响恰当平均指标的选择：偏态数据通常优选中位数，众数则可指示峰值位置。
6. Quartile-Based Skewness | 基于四分位数的偏度
A simple way to assess skewness is to compare the distances from the median to the quartiles. If (Q₃ – Q₂) > (Q₂ – Q₁), the upper half is more spread, indicating positive skew. If (Q₃ – Q₂) < (Q₂ – Q₁), the data are negatively skewed. When the two distances are approximately equal, the distribution is symmetric.
评估偏度的一个简单方法是比较中位数到四分位数的距离。如果 (Q₃ – Q₂) > (Q₂ – Q₁)，上半部分更分散，表明正偏态。如果 (Q₃ – Q₂) < (Q₂ – Q₁)，数据为负偏态。两距离大致相等时，分布对称。
Edexcel explicitly mentions this quartile method: examine the box plot — if the right whisker and right box segment are longer, positive skew is likely. A quartile skewness coefficient can be defined as (Q₃ + Q₁ – 2Q₂) / (Q₃ – Q₁), producing values between –1 and 1; zero indicates symmetry.
Edexcel 明确提及此四分位数方法：观察箱线图——若右须和右箱段更长，很可能呈正偏态。可定义四分位偏度系数为 (Q₃ + Q₁ – 2Q₂) / (Q₃ – Q₁)，取值在 –1 到 1 之间；零表示对称。
7. Using Mean, Median and Mode to Infer Skewness | 利用均值、中位数和众数推断偏度
positively skewed distribution, the order is typically: mode < median < mean, because the long right tail pulls the mean upwards. In a negatively skewed distribution, the order reverses: mean < median < mode.
在完美的对称单峰分布中，均值 = 中位数 = 众数。在正偏态分布中，大小顺序通常为：众数 < 中位数 < 均值，因为长右尾将均值上拉。在负偏态分布中，顺序相反：均值 < 中位数 < 众数。
| Skewness Type 偏度类型 | Order of Averages 平均指标顺序 | Tail Direction 尾部方向 |
|---|---|---|
| Positive (right) skew 正偏态 | mode < median < mean | Long right tail 长右尾 |
| Symmetric 对称 | mean = median = mode | None 无 |
| Negative (left) skew 负偏态 | mean < median < mode | Long left tail 长左尾 |
tion of the tail. Another formal measure is Pearson’s coefficient of skewness: Sk = 3(mean – median) / standard deviation. If Sk > 0, positive skew; Sk < 0, negative skew.
可通过均值被拉向尾部方向来记忆这一关系。另一个正式的度量是皮尔逊偏度系数：Sk = 3(均值 – 中位数) / 标准差。若 Sk > 0 为正偏态；Sk < 0 为负偏态。
8. Worked Example: Grouped Data and Skewness | 实例分析：分组数据与偏度
Consider the grouped frequency table of exam marks (n = 100). Class intervals: 0–20 (f=8), 20–40 (12), 40–60 (35), 60–80 (30), 80–100 (15). Boundaries are used, so classes are 0–20 etc. with width 20.
考虑考试成绩的分组频数表（n=100）。组区间：0–20（f=8），20–40（12），40–60（35），60–80（30），80–100（15）。使用组界，因此组距为 20。
Estimate the mode: modal class 40–60 (fₘ=35). fₚ=12, fₙ=30. L=40. Mode ≈ 40 + [ (35–12) / ((35–12)+(35–30)) ] × 20 = 40 + [23 / (23+5)] × 20 = 40 + (23/28)×20 ≈ 56.43.
估计众数：众数组 40–60（fₘ=35），fₚ=12，fₙ=30，L=40。众数 ≈ 40 + [ (35–12) / ((35–12)+(35–30)) ] × 20 = 40 + (23/28)×20 ≈ 56.43。
quencies: <20:8, <40:20, <60:55, <80:85, <100:100. Median position n/2=50. Median class 40–60. Median = 40 + [ (50–20)/35 ] × 20 = 40 + (30/35)×20 ≈ 57.14.
累计频数：<20:8, <40:20, <60:55, <80:85, <100:100。中位数位置 50。中位数 = 40 + [ (50–20)/35 ] × 20 ≈ 57.14。
Q₁ position = n/4 = 25, class 40–60: Q₁ = 40 + [ (25–20)/35 ] × 20 ≈ 42.86. Q₃ position = 3n/4 = 75, class 60–80: Q₃ = 60 + [ (75–55)/30 ] × 20 ≈ 73.33. IQR = 30.47.
Q₁ 位置 25：Q₁ ≈ 42.86。Q₃ 位置 75：Q₃ ≈ 73.33。IQR = 30.47。
Now assess skewness: compare (Q₃ – Q₂) = 73.33 – 57.14 = 16.19, and (Q₂ – Q₁) = 57.14 – 42.86 = 14.28. Since (Q₃ – Q₂) > (Q₂ – ×12+50×35+70×30+90×15)/100 = 57.8). Indeed mode < median < mean, confirming positive skew.
评估偏度：比较 (Q₃ – Q₂)=16.19 和 (Q₂ – Q₁)=14.28，由于 (Q₃ – Q₂) > (Q₂ – Q₁)，存在轻微正偏态。众数 (56.43) < 中位数 (57.14) < 均值 (计算均值 ≈ 57.8)，确认正偏态。这一完整流程在考试中可直接使用。
9. Common Pitfalls | 常见失误
Students often confuse the position formulas for ungrouped and grouped data. Remember: for raw data use (n+1)/2, (n+1)/4; for grouped data, use n/2, n/4, etc., and always interpolate. Another mistake is using the wrong class boundaries—ensure you use the true limits (e.g., if classes are 10–19, 20–29, boundaries are 9.5–19.5 etc. unless continuous).
学生常混淆未分组与分组数据的位置公式。记住：原始数据用 (n+1)/2、(n+1)/4；分组数据用 n/2、n/4 等，并始终进行插值。另一个错误是组界使用不当——务必使用真实界限（例如，若分组为 10–19，20–29，连续组界为 9.5–19.5 等，除非题目明确为离散）。
When estimating the mode for grouped data, using the correct adjacent frequencies is crucial. Never just pick the midpoint of the modal class. Also, be careful with cumulative frequency additions—an off-by-one error can affect the median and quartiles.
估计分组数据众数时，正确使用相邻频数至关重要。切勿只取众数组的组中点。同时，累计频数相加要谨慎——相差一位就会影响中位数和四分位数。
Finally, interpreting skewness only via the mean-median relationship can be misleading if outliers distort the mean. Always support with quartile distances or a box plot.
最后，仅通过均值-中位数关系推断偏度可能因离群值扭曲均值而产生误导，应始终用四分位数距离或箱线图加以佐证。
10. Summary and Exam Tips | 总结与考试技巧
To ace questions on mode, median, percentiles and skewness, master the interpolation method for grouped data and be able to switch between raw and grouped formulas. Practice estimating the mode using the correct formula, not just the modal class midpoint. In Edexcel papers, clearly state which measure you are using and show all working.
要攻克众数、中位数、百分位数和偏度的考题，必须熟练掌握分组数据的插值方法，并能区分原始数据与分组数据的公式应用。练习用正确公式估计众数，而非只取组中点。在 Edexcel 试卷中，明确说明所用度量并展示完整演算过程。
When asked to comment on skewness, always reference both the quartile distances and the ordering of mean, median and mode. If standard deviation is provided, calculate Pearson’s coefficient to support your argument. Include a sketched box plot or a sentence describing tail direction. These small touches demonstrate thorough understanding and secure method marks.
当要求对偏度发表评论时，务必同时提及四分位数距离和均值、中位数、众数的顺序。若给出了标准差，可计算皮尔逊系数佐证。附上箱线图草图或描述尾部方向。这些小细节能体现全面理解，确保方法分。
Finally, check your answers for reasonableness: do the computed positions lie within the expected class? Do the quartiles and median fall in the correct order? A quick mental check avoids careless mistakes.
最后，检查答案的合理性：计算出的位置是否在预期的组内？四分位数与中位数的顺序是否正确？快速心算核查可避免粗心失误。
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