📚 Averages and Range from Frequency Tables | 从频数表计算平均数与极差
When data is organised in a frequency table, you need to find averages and spread in a slightly different way. This article explains how to calculate the mode, median, mean and range from a discrete frequency table for Cambridge KS3 Mathematics.
当数据以频数表形式整理时，求平均数与数据分布的方法略有不同。本文针对剑桥 KS3 数学，讲解如何从离散频数表计算众数、中位数、平均数和极差。
1. Understanding Frequency Tables | 认识频数表
A frequency table shows data values and how often each value occurs. The frequency tells you the count for each value. For example, a class recorded the number of books read in a month.
频数表列出数据值以及每个值出现的次数。频数表示每个值出现的数量。例如，一个班级记录了一个月内阅读书籍的数量。
| Data value x | Frequency f |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 7 |
| 4 | 3 |
| 5 | 1 |
8, so 18 students were asked. Always identify the total frequency before finding averages.
总频数为 2 + 5 + 7 + 3 + 1 = 18，表示共有 18 名学生参与调查。在求平均数之前，先确认总频数。
2. Finding the Mode from a Frequency Table | 从频数表求众数
The mode is the data value with the highest frequency. Look for the largest number in the frequency column and read off the corresponding data value.
众数是出现频率最高的数据值。只需在频数列中找到最大的数字，并读出对应的数据值。
In the table above, the highest frequency is 7, which belongs to the data value 3. Therefore, the mode is 3 books.
在上表中，最大频数是 7，对应的数据值是 3。因此众数是 3 本书。
3. Finding the Median from a Frequency Table | 从频数表求中位数
The median is the middle value when all data values are placed in order. First find the median position using total frequency n.
中位数是将所有数据值按顺序排列后位于中间的值。首先使用总频数 n 求出中位数的位置。
Median position = (n + 1) ÷ 2
is (18 + 1) ÷ 2 = 9.5. This means the median lies halfway between the 9th and 10th values.
在上表中，n = 18，因此中位数位置为 (18 + 1) ÷ 2 = 9.5。这意味着中位数位于第 9 个和第 10 个值之间。
To locate these positions, add a cumulative frequency column. The cumulative frequencies are 2, 7, 14, 17, 18. The 9th and 10th values both lie in the data value 3 because the cumulative frequency reaches 14 at that point.
为了找到这些位置，需要添加累计频数列。累计频数为 2、7、14、17、18。第 9 个和第 10 个值都在数据值 3 中，因为累计频数在此处达到 14。
Therefore, the median is 3 books.
因此，中位数是 3 本书。
4. Finding the Mean from a Frequency Table | 从频数表求平均数
The mean is found by multiplying each data value by its frequency, adding these products, and then dividing by the total frequency.
求平均数的方法是：将每个数据值乘以其频数，把这些乘积相加，再除以总频数。
Mean = Σfx ÷ Σf
For the table above, calculate Σfx first.
对于上表，先计算 Σfx。
Σfx = (1 × 2) + (2 × 5) + (3 × 7) + (4 × 3) + (5 × 1)
Σfx = 2 + 10 + 21 + 12 + 5 = 50
Now divide by Σf = 18.
现在除以 Σf = 18。
Mean = 50 ÷ 18 ≈ 2.78
So the mean number of books read is approximately 2.78 books per student, correct to two decimal places.
因此，阅读书籍的平均数约为每位学生 2.78 本，保留两位小数。
5. Finding the Range from a Frequency Table | 从频数表求极差
The range measures how spread out the data is. It is the difference between the highest and lowest data values, not the frequencies.
极差衡量数据的分散程度。它是最高数据值与最低数据值之差，而不是频数之差。
Range = highest value − lowest value
our example, the highest value is 5 and the lowest value is 1. So the range is 5 − 1 = 4.
在我们的例子中，最高数据值是 5，最低数据值是 1。因此极差为 5 − 1 = 4。
6. Worked Example: Comparing Two Data Sets | 例题：比较两组数据
Two classes take a short quiz marked out of 5. Class A has a mean of 3.4 and a range of 4. Class B has a mean of 3.1 and a range of 2. Compare the results.
两个班级参加满分 5 分的小测验。A 班的平均分是 3.4，极差是 4。B 班的平均分是 3.1，极差是 2。请比较两组结果。
Class A has a higher mean, so on average the students in Class A performed slightly better. However, Class A has a larger range, meaning their scores are more spread out.
A 班的平均分更高，因此 A 班学生的整体表现略好。但 A 班的极差更大，说明分数更分散。
Class B has a lower mean but a smaller range. This suggests Class B is more consistent, with most scores close together.
B 班平均分较低但极差更小。这表示 B 班成绩更稳定，大多数分数比较接近。
When comparing data, always comment on both an average and the range. Use the average to compare typical performance, and the range to compare consistency.
比较数据时，要同时评价平均数和极差。用平均数比较典型表现，用极差比较稳定性。
7. Common Mistakes | 常见错误
Forgetting to multiply each data value by its frequency before adding.
计算时忘记先将每个数据值乘以其频数再相加。
Dividing by the number of rows instead of the total frequency.
除以行数而不是总频数。
Treating frequencies as data values when finding the mode or range.
求众数或极差时误把频数当作数据值。
Using the cumulative frequency as the median instead of locating the middle position.
直接把累计频数当作中位数，而不是先确定中间位置。
8. Exam Tips | 考试技巧
Always write down the total frequency n first. Show your Σfx column or working clearly if calculating the mean.
始终先写出总频数 n。计算平均数时，清楚地展示 Σfx 列或计算过程。
For the median, create a cumulative frequency column. Underline or circle the middle position so you can locate it quickly.
求中位数时，建立累计频数列。标出或圈出中间位置，以便快速找到对应值。
Check your mean is between the smallest and largest data values. If it is outside that range, you have made an error.
检查平均数是否介于最小和最大数据值之间。如果超出该范围，说明计算出错了。
9. Practice Questions | 练习题
Question 1: Use the frequency table from Section 1 to write down the mode, median, mean and range.
第 1 题：使用第 1 节中的频数表，写出众数、中位数、平均数和极差。
Question 2: A dice is rolled 20 times. The scores are shown below. Find the mean score and the range.
第 2 题：一颗骰子掷了 20 次。得分如下表所示。求平均得分和极差。
| Score x | Frequency f |
|---|---|
| 1 | 3 |
| 2 | 6 |
| 3 | 4 |
| 4 | 2 |
| 5 | 4 |
| 6 | 1 |
ves mode = 3, median = 3, mean ≈ 2.78, range = 4. Question 2 gives mean = 2.95, range = 5.
检查答案：第 1 题的众数 = 3，中位数 = 3，平均数 ≈ 2.78，极差 = 4。第 2 题的平均数 = 2.95，极差 = 5。
10. Summary | 内容总结
Mode is the value with the highest frequency.
众数是出现频率最高的值。
Median position is (n + 1) ÷ 2, located using cumulative frequency.
中位数位置为 (n + 1) ÷ 2，通过累计频数确定。
Mean is Σfx ÷ Σf.
平均数为 Σfx ÷ Σf。
Range is the largest data value minus the smallest data value.
极差等于最大数据值减最小数据值。
Use an average and the range together to compare data sets.
比较数据时同时使用平均数和极差。
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
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