📚 EnVision G5 17-06: Mean, Median, Mode, and Range | enVision G5 17-06：均值、中位数、众数与极差
In IGCSE Mathematics and enVision Grade 5 Lesson 17-6, you meet three measures of central tendency: mean, median, and mode, together with the range as a measure of spread. These tools summarise a data set and help you compare everyday data such as test scores, temperatures, sports results, and prices.
在 IGCSE 数学和 enVision 五年级第 17-6 课中，你将学习三种集中趋势量数：平均数（均值）、中位数和众数，以及表示离散程度的极差。这些工具能够概括一组数据，帮助你比较考试成绩、气温、运动成绩和价格等日常数据。
1. Understanding the Mean | 理解均值
The mean is the arithmetic average. To find it, add all the values together, then divide the total by the number of values. It is useful when data are fairly evenly spread and have no extreme outliers.
均值就是算术平均数。求均值时，先把所有数值相加，再用总和除以数据的个数。当数据分布比较均匀且没有极端异常值时，均值很实用。
Mean = (Sum of all values) ÷ (Number of values)
or example, the data set 4, 5, 6, 7, 8 has sum 30 and 5 values, so the mean is 30 ÷ 5 = 6.
例如，数据集 4、5、6、7、8 的总和是 30，数据个数是 5，因此均值是 30 ÷ 5 = 6。
2. Understanding the Median | 理解中位数
The median is the middle value when the data are arranged in order from smallest to largest. If there is an odd number of values, the median is the exact middle value. If there is an even number of values, the median is the mean of the two middle values.
中位数是把数据从小到大排列后位于中间位置的数值。如果数据个数是奇数，中位数就是正中间的那个数；如果数据个数是偶数，中位数就是中间两个数的平均数。
- Odd example: 3, 5, 7, 9, 11. The median is 7.
- : 2, 4, 6, 8, 10, 12. The two middle values are 6 and 8, so the median is (6 + 8) ÷ 2 = 7.
奇数个例子：3、5、7、9、11，中位数是 7。偶数个例子：2、4、6、8、10、12，中间两个数是 6 和 8，因此中位数是 (6 + 8) ÷ 2 = 7。
3. Understanding the Mode | 理解众数
The mode is the value that occurs most often in a data set. A set may have one mode, more than one mode, or no mode at all. The mode is especially useful for non-numerical data, such as favourite colours or shoe sizes.
众数是一组数据中出现次数最多的数值。一组数据可能有一个众数、多个众数，也可能没有众数。众数特别适用于非数值数据，例如最喜欢的颜色或鞋码。
- Data: 2, 3, 3, 4, 5, 3 → Mode = 3.
- Data: 1, 2, 2, 3, 3, 4 → Modes = 2 and 3, so it is bimodal.
- Data: 5, 6, 7, 8 → No mode.
例：2、3、3、4、5、3 → 众数是 3。例：1、2、2、3、3、4 → 众数是 2 和 3，称为双众数。例：5、6、7、8 → 没有众数。
4. Understanding the Range | 理解极差
The range measures how spread out the data are. It is found by subtracting the smallest value from the largest value. A small range means the data are close together; a large range means they are more spread out.
极差衡量数据的离散程度。它的求法是用最大值减去最小值。极差小说明数据比较集中；极差大说明数据比较分散。
Range = Largest value − Smallest value
For the data set 3, 8, 5, 11, 7, the range is 11 − 3 = 8.
对于数据集 3、8、5、11、7，极差是 11 − 3 = 8。
5. Choosing the Right Measure | 选择合适的量数
Use the mean when the data are symmetric and have no outliers. Use the median when the data are skewed or contain outliers, because the median is not affected by extreme values. Use the mode when you need the most common category or value.
当数据对称且没有异常值时使用均值。当数据偏斜或含有异常值时使用中位数，因为中位数不受极端值影响。当需要知道最常见的类别或数值时使用众数。
| Situation | 情况 | Best measure | 最佳量数 |
|---|---|
| Test scores with one very low mark | 有一个很低分的考试成绩 | Median | 中位数 |
| Heights of students in a class | 班级学生身高 | Mean | 均值 |
| Favourite fruit in a survey | 调查中最喜欢的水果 | Mode | 众数 |
6. Worked Example: Test Scores | 例题：考试成绩
A student has these test scores: 8, 9, 7, 10, 8, 6, 9, 8. Find the mean, median, mode, and range.
一名学生的考试成绩为：8、9、7、10、8、6、9、8。求均值、中位数、众数和极差。
- scores = 8 + 9 + 7 + 10 + 8 + 6 + 9 + 8 = 65. Number of scores = 8. Mean = 65 ÷ 8 = 8.125.
- : 6, 7, 8, 8, 8, 9, 9, 10. The two middle values are 8 and 8, so median = (8 + 8) ÷ 2 = 8.
- Mode = 8, because 8 appears three times.
- Range = 10 − 6 = 4.
成绩总和 = 8 + 9 + 7 + 10 + 8 + 6 + 9 + 8 = 65，成绩个数 = 8，均值 = 65 ÷ 8 = 8.125。排序后：6、7、8、8、8、9、9、10，中间两个数是 8 和 8，中位数 = (8 + 8) ÷ 2 = 8。众数 = 8，因为 8 出现了三次。极差 = 10 − 6 = 4。
7. Worked Example: Line Plot Data | 例题：线图数据
A line plot shows the number of books read by ten students. The values and frequencies are shown in the table below.
一幅线图显示了十名学生阅读书籍的数量。数值和频数如下表所示。
| Number of books | 书籍数量 | Frequency | 频数 |
|---|---|
| 2 | 1 |
| 3 | 2 |
| 4 | 4 |
| 5 | 2 |
| 6 | 1 |
an is (4 + 4) ÷ 2 = 4. The mode is 4 because it occurs four times. The range is 6 − 2 = 4.
学生总数为 1 + 2 + 4 + 2 + 1 = 10。均值 = (2×1 + 3×2 + 4×4 + 5×2 + 6×1) ÷ 10 = 40 ÷ 10 = 4。按顺序排列为：2、3、3、4、4、4、4、5、5、6，中位数是 (4 + 4) ÷ 2 = 4。众数是 4，因为 4 出现了四次。极差是 6 − 2 = 4。
8. Common Mistakes | 常见错误
Students often forget to put the data in order before finding the median. They also confuse the mode with its frequency, or forget that the mean must be divided by the total number of values, not by another number. Another error is choosing the mean when a very large or very small outlier makes the median more representative.
学生常在求中位数前忘记对数据排序。他们还容易把众数和它的频数混淆，或者忘记均值必须除以数据的总个数，而不是除以其他数字。另一个错误是：当存在非常大或非常小的异常值时，仍然选择均值，而实际上中位数更能代表整体情况。
- Always order the data for the median.
- Write the mode as the data value, not its frequency.
- Check the divisor in the mean calculation.
- Look for outliers before deciding which average to use.
求中位数时一定要先排序。写众数时要写数据值本身，而不是它的频数。检查均值计算中的除数。在决定使用哪个平均数之前，先检查是否有异常值。
9. Exam-Style Practice | 考试风格练习
Try these two practice questions, then check your answers below.
请尝试以下两道练习题，然后对照下面的答案。
- Question 1: Find the mean, median, mode, and range of the data set 12, 15, 12, 18, 13.
- Question 2: A line plot has values 1, 2, 3, 4 with frequencies 3, 1, 2, 4. Find the median, mode, and range.
问题 1：求数据集 12、15、12、18、13 的均值、中位数、众数和极差。问题 2：某线图的数值为 1、2、3、4，频
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