KS3 Cambridge Math: Mean, Median, Mode and Range Guide

Introduction | 引言

Statistics is one of the most practical branches of mathematics, and at the KS3 level, the Cambridge curriculum introduces students to the fundamentals of data handling. Among the most essential concepts are the three measures of central tendency — mean, median, and mode — along with the measure of spread known as the range.

统计学是数学中最实用的分支之一，在KS3阶段，剑桥课程向学生介绍了数据处理的基础知识。其中最基本的概念包括三种集中趋势的度量——平均数、中位数和众数——以及衡量数据分散程度的极差。

Understanding these four statistical measures is critical not only for success in the Cambridge Checkpoint tests but also as a foundation for IGCSE Mathematics and even A-Level Statistics. These concepts help us make sense of data in everyday life — from calculating average test scores to interpreting weather data and sports statistics.

理解这四种统计度量不仅对剑桥Checkpoint考试的成功至关重要，也是IGCSE数学甚至A-Level统计学的基础。这些概念帮助我们在日常生活中理解数据——从计算平均考试成绩到解读天气数据和体育统计。

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1. Mean (Average) | 平均数

What is the Mean? | 什么是平均数？

The mean (often called the average) is calculated by adding all the values in a dataset and then dividing by the number of values. It represents the “fair share” — if all the data values were redistributed equally, each would equal the mean.

平均数（通常称为平均值）是通过将数据集中所有值相加，然后除以值的个数来计算的。它代表了”公平份额”——如果所有数据值被平均重新分配，每个值将等于平均数。

Formula | 公式

Mean = (Sum of all values) ÷ (Number of values)

平均数 = 所有值之和 ÷ 值的个数

In mathematical notation: Mean = Σx / n, where Σx represents the sum of all data values and n is the total count.

用数学符号表示：平均数 = Σx / n，其中Σx代表所有数据值的总和，n是总计数。

Worked Example | 示例

Question: Sarah recorded the number of pages she read each day for one week: 12, 15, 8, 20, 14, 18, 11. Calculate the mean number of pages read per day.

问题：Sarah记录了她一周每天阅读的页数：12, 15, 8, 20, 14, 18, 11。计算每天阅读页数的平均数。

Solution: Step 1: Add all values → 12 + 15 + 8 + 20 + 14 + 18 + 11 = 98. Step 2: Count the number of values → 7 values. Step 3: Divide → 98 ÷ 7 = 14. Answer: The mean number of pages read per day is 14.

解答：第一步：将所有值相加 → 12 + 15 + 8 + 20 + 14 + 18 + 11 = 98；第二步：计算值的个数 → 7个值；第三步：除法 → 98 ÷ 7 = 14。答案：每天阅读页数的平均数为14页。

Common Pitfall: The Mean and Outliers | 常见陷阱：平均数与异常值

One important limitation of the mean is that it is sensitive to outliers — extreme values that are much larger or smaller than the rest. For example, if a billionaire walks into a room of 50 people, the mean wealth of the room suddenly becomes millions, but this does not accurately represent the typical person in the room.

平均数的一个重要局限是它对异常值敏感——比其他值大得多或小得多的极端值。例如，如果一个亿万富翁走进一个有50人的房间，房间的平均财富突然变成数百万，但这并不能准确代表房间里的典型人群。在这种情况下，中位数可能是更好的度量。

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2. Median | 中位数

What is the Median? | 什么是中位数？

The median is the middle value when the data is arranged in order from smallest to largest. If there are two middle values, the median is the mean of those two values. The median splits the dataset exactly in half — 50% of values lie below it and 50% above it.

中位数是将数据按从小到大的顺序排列后的中间值。如果有两个中间值，中位数是这两个值的平均数。中位数将数据集精确地分成两半——50%的值低于它，50%高于它。

How to Find the Median | 如何求中位数

For an odd number of values: The median is the value at position (n+1)/2

For an even number of values: The median is the mean of the values at positions n/2 and (n/2)+1

奇数个值：中位数位于第(n+1)/2个位置的值
偶数个值：中位数是第n/2个和第(n/2)+1个位置值的平均数

Worked Examples | 示例

Example 1 (Odd count): The test scores of 7 students are: 45, 72, 68, 91, 55, 83, 60. Find the median score. Step 1: Arrange in order → 45, 55, 60, 68, 72, 83, 91. Step 2: Position = (7+1)/2 = 4th value. Step 3: The 4th value is 68. Answer: The median score is 68.

示例1（奇数个）：7名学生的考试成绩为：45, 72, 68, 91, 55, 83, 60。求中位数分数。第一步：按顺序排列 → 45, 55, 60, 68, 72, 83, 91；第二步：位置 = (7+1)/2 = 第4个值；第三步：第4个值是68。答案：中位数分数为68。

Example 2 (Even count): The heights (in cm) of 8 basketball players: 185, 192, 178, 201, 188, 195, 182, 190. Find the median height. Step 1: Arrange in order → 178, 182, 185, 188, 190, 192, 195, 201. Step 2: Two middle positions → n/2 = 4th and (n/2)+1 = 5th. Step 3: 4th value = 188, 5th value = 190. Step 4: Mean of 188 and 190 = (188+190)/2 = 189. Answer: The median height is 189 cm.

示例2（偶数个）：8名篮球运动员的身高（厘米）：185, 192, 178, 201, 188, 195, 182, 190。求中位数身高。第一步：按顺序排列 → 178, 182, 185, 188, 190, 192, 195, 201；第二步：两个中间位置 → 第4个和第5个；第三步：第4个值=188，第5个值=190；第四步：188和190的平均数 = (188+190)/2 = 189。答案：中位数身高为189厘米。

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3. Mode | 众数

What is the Mode? | 什么是众数？

The mode is the value that appears most frequently in a dataset. A dataset may have one mode (unimodal), two modes (bimodal), or many modes (multimodal). If no value repeats, the dataset has no mode. The mode is particularly useful for categorical or non-numerical data where the mean and median cannot be calculated.

众数是数据集中出现频率最高的值。数据集可以有一个众数（单峰）、两个众数（双峰）或多个众数（多峰）。如果没有值重复，数据集就没有众数。众数对于无法计算平均数和中位数的分类或非数值数据特别有用。

Worked Example | 示例

Question: A class of 30 students was surveyed about their favourite fruit. The results were: Apple (8 students), Banana (5 students), Orange (8 students), Grape (6 students), Mango (3 students). What is the mode?

问题：对30名学生的班级进行了最喜欢的水果调查。结果为：苹果（8名学生）、香蕉（5名学生）、橙子（8名学生）、葡萄（6名学生）、芒果（3名学生）。众数是什么？

Solution: Both Apple and Orange appear 8 times each (the highest frequency). This dataset is bimodal with modes: Apple and Orange.

解答：苹果和橙子各出现8次（最高频次）。此数据集是双峰的，众数为：苹果和橙子。

Mode from Frequency Tables | 从频率表中找众数

When data is presented in a frequency table, the mode is simply the value with the highest frequency. For grouped data, we identify the modal class — the class interval with the highest frequency.

当数据以频率表呈现时，众数就是频率最高的值。对于分组数据，我们识别众数类——频率最高的组区间。

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4. Range | 极差

What is the Range? | 什么是极差？

The range is a measure of how spread out the data is. It is calculated as the difference between the largest and smallest values in a dataset.

极差是衡量数据分散程度的指标。它计算为数据集中最大值与最小值之间的差。

Formula | 公式

Range = Largest value − Smallest value

极差 = 最大值 − 最小值

Worked Example | 示例

Question: The daily maximum temperatures (in °C) in London for one week in July were: 22, 28, 19, 31, 25, 24, 27. Find the range.

问题：伦敦七月一周的每日最高气温（摄氏度）为：22, 28, 19, 31, 25, 24, 27。求极差。

Solution: Largest value = 31°C | Smallest value = 19°C. Range = 31 − 19 = 12°C.

解答：最大值 = 31°C | 最小值 = 19°C；极差 = 31 − 19 = 12°C。

Limitations of the Range | 极差的局限性

While the range is simple to calculate, it only considers the two extreme values and ignores how the data is distributed between them. Two datasets with identical ranges can have very different distributions. For a more robust measure of spread, students will later learn about the interquartile range (IQR) and standard deviation.

虽然极差计算简单，但它只考虑两个极端值，忽略了数据在它们之间的分布情况。两个极差相同的数据集可能有非常不同的分布。为了获得更稳健的离散度度量，学生将在后续学习中接触四分位距(IQR)和标准差。

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Comparing Datasets Using All Four Measures | 使用四种度量比较数据集

Cambridge KS3 exam questions often ask students to compare two datasets using the mean, median, mode, and range. Here is a structured approach:

剑桥KS3考试题目经常要求学生使用平均数、中位数、众数和极差来比较两个数据集。以下是结构化方法：

Example: Compare the test scores of Class A and Class B.

示例：比较A班和B班的考试成绩。

Measure | 度量	Class A | A班	Class B | B班
Mean | 平均数	72	72
Median | 中位数	74	68
Mode | 众数	78	65
Range | 极差	20	45

Analysis: Both classes have the same mean (72), suggesting similar overall performance. However, Class A has a higher median (74 vs 68) and mode (78 vs 65), indicating that more students in Class A scored at the higher end. Class B has a much larger range (45 vs 20), suggesting greater variability in performance — some students scored very high and others very low. Class A shows more consistent performance overall.

分析：两个班的平均数相同（72），表明整体表现相似。然而，A班的中位数（74 vs 68）和众数（78 vs 65）更高，表明A班更多学生得分较高。B班的极差大得多（45 vs 20），表明成绩差异更大——有些学生得分很高，有些很低。A班整体表现更一致。

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Practice Questions | 练习题

Test your understanding with these Cambridge Checkpoint-style questions:

用这些剑桥Checkpoint风格的题目测试你的理解：

Q1: The ages of 9 children at a birthday party are: 7, 6, 8, 7, 9, 6, 8, 7, 10. Find the mean, median, mode, and range.

第1题：生日派对上9个孩子的年龄为：7, 6, 8, 7, 9, 6, 8, 7, 10。求平均数、中位数、众数和极差。

Q2: A dice is rolled 12 times. The results are: 3, 5, 2, 1, 4, 6, 3, 2, 5, 3, 6, 3. Which number is the mode? What does this tell us?

第2题：一个骰子掷了12次。结果为：3, 5, 2, 1, 4, 6, 3, 2, 5, 3, 6, 3。哪个数字是众数？这告诉我们什么？

Q3: Alex says: “The mean of five consecutive integers is always equal to the median.” Is Alex correct? Justify your answer with an example.

第3题：Alex说：”五个连续整数的平均数总是等于中位数。”Alex说得对吗？用一个例子证明你的答案。

Q4: Two football teams recorded their goals per match over a season. Team X: mean = 2.1, range = 4. Team Y: mean = 2.1, range = 2. Which team is more consistent in scoring? Explain.

第4题：两支足球队记录了他们一个赛季每场比赛的进球数。X队：平均数=2.1，极差=4。Y队：平均数=2.1，极差=2。哪支球队得分更稳定？解释你的答案。

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Answers | 答案

A1: Ordered: 6, 6, 7, 7, 7, 8, 8, 9, 10. Mean = 68/9 ≈ 7.56. Median = 5th value = 7. Mode = 7 (appears 3 times). Range = 10 − 6 = 4.

答1：排序：6, 6, 7, 7, 7, 8, 8, 9, 10。平均数=68/9≈7.56。中位数=第5个值=7。众数=7（出现3次）。极差=10−6=4。

A2: Counting: 1 appears 1 time, 2 appears 2 times, 3 appears 4 times, 4 appears 1 time, 5 appears 2 times, 6 appears 2 times. Mode = 3 (appears 4 times). This suggests the dice may be biased towards 3.

答2：计数：1出现1次，2出现2次，3出现4次，4出现1次，5出现2次，6出现2次。众数=3（出现4次）。这暗示骰子可能偏向数字3。

A3: Yes, Alex is correct. Example: 10, 11, 12, 13, 14. Mean = (10+11+12+13+14)/5 = 60/5 = 12. Median = 3rd value = 12. The mean and median are equal because five consecutive integers form a symmetric arithmetic sequence.

答3：是的，Alex说得对。示例：10, 11, 12, 13, 14。平均数=(10+11+12+13+14)/5=60/5=12。中位数=第3个值=12。平均数和中位数相等，因为五个连续整数构成对称的等差数列。

A4: Team Y is more consistent. Both teams have the same mean (2.1 goals per match), but Team Y has a smaller range (2 vs 4), meaning their goal counts per match vary less from game to game.

答4：Y队更稳定。两队有相同的平均数（每场2.1个进球），但Y队的极差更小（2 vs 4），意味着他们每场比赛的进球数波动更小。

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Key Takeaways | 关键要点

• Mean — the arithmetic average; sensitive to outliers; best for symmetric distributions without extreme values. | 平均数 — 算术平均值；对异常值敏感；最适合没有极端值的对称分布。
• Median — the middle value; robust against outliers; best for skewed distributions. | 中位数 — 中间值；对异常值有抵抗力；最适合偏态分布。
• Mode — the most frequent value; useful for categorical data; a dataset can have multiple modes or none. | 众数 — 最频繁出现的值；适用于分类数据；数据集可以有多个众数或没有众数。
• Range — the difference between max and min; a simple measure of spread; does not reflect internal distribution. | 极差 — 最大值与最小值的差；简单的离散度度量；不反映内部分布。

Mastering these four statistical measures is essential for KS3 Cambridge Mathematics and provides a strong foundation for more advanced statistical analysis at IGCSE and beyond. Remember: always arrange data in order before finding the median, and always check for outliers when interpreting the mean. With regular practice, these concepts will become second nature.

掌握这四种统计度量对KS3剑桥数学至关重要，并为IGCSE及更高级的统计分析打下坚实基础。记住：在求中位数之前始终将数据排序，在解释平均数时始终检查异常值。通过定期练习，这些概念将成为你的第二天性。

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