📚 IGCSE Mathematics: Reading and Interpreting Histograms | IGCSE数学：直方图的读取与信息解读
Histograms are a fundamental topic in the Edexcel IGCSE Mathematics syllabus. Unlike bar charts, histograms represent continuous data and require a different method of reading and interpretation. This article explains everything you need to know, from the basic definitions to advanced problem-solving techniques.
直方图是 Edexcel IGCSE 数学课程中的基础考点。与条形图不同，直方图表示连续数据，其读取和解读方法也截然不同。本文将带你掌握从基本定义到高级解题技巧的全部内容。
1. What Is a Histogram? | 什么是直方图？
A histogram is a graphical representation of a frequency distribution for continuous data. It uses adjacent rectangular bars to show how data are distributed across class intervals. The horizontal axis represents the variable measured, such as height, time, or weight, and the vertical axis represents the frequency density.
直方图是对连续数据的频率分布进行图形化表示的工具。它使用相邻的矩形条来展示数据在各个组距（class interval）内的分布情况。横轴表示所测量的变量，例如身高、时间或体重；纵轴表示频率密度（frequency density）。
In a histogram, the bars touch each other because there are no gaps between class intervals for continuous data. This is the key visual difference between a histogram and a bar chart.
在直方图中，各矩形条是彼此相邻的，因为连续数据的各组距之间没有空隙。这是直方图与条形图最直观的区别。
For grouped continuous data, the frequency of each class interval is proportional to the area of the corresponding bar, not its height alone. This is a crucial idea that underpins many exam questions.
对于分组连续数据，每个组距的频率与该组距对应矩形的面积成正比，而不仅仅是由高度决定。这是支撑许多考试题目的关键概念。
2. Continuous Data and Class Intervals | 连续数据与组距
Continuous data can take any value within a range. Examples include time taken to run a race, the height of a plant, or the temperature at noon. Because these data can be measured to any degree of precision, we must group them into class intervals.
连续数据可以在某个范围内取任意值。例如完成赛跑所用的时间、植物的高度或正午的气温。由于这些数据可以测量到任意精度，我们必须将它们归入不同的组距。
Each class interval has a lower class boundary and an upper class boundary. For example, if the intervals are 0–10, 10–20, 20–30 (in seconds), the boundaries are exactly 0, 10, 20, and 30. The class width is the difference between the upper and lower boundaries.
每个组距都有下边界和上边界。例如，若间隔为 0–10 秒、10–20 秒、20–30 秒，那么边界就是 0、10、20 和 30。组距宽度就是上边界与下边界之差。
Notice that the boundary values belong to the next interval in the conventional notation. For instance, a value of exactly 10 goes into the 10–20 interval, not the 0–10 interval. This convention ensures every data point is counted exactly once.
注意，在常规书写中，边界值归入下一个间隔。例如，恰好为 10 的值应计入 10–20 组，而不是 0–10 组。这一约定确保每个数据点只被计数一次。
3. The Axes: Frequency Density vs Frequency | 坐标轴：频率密度与频率
In a histogram, the y-axis is labelled ‘frequency density’ or sometimes ‘frequency per unit width’. The x-axis is the variable in question.
在直方图中，纵轴标为“频率密度”（或“单位宽度频率”），横轴表示所研究的变量。
Frequency density is defined by the formula:
Frequency Density = Frequency ÷ Class Width
频率密度的定义公式为：
频率密度 = 频率 ÷ 组距宽度
The reason we use frequency density instead of frequency on the vertical axis is that class intervals may have different widths. The height of each bar must be adjusted so that the area of the bar represents the frequency.
之所以在纵轴上使用频率密度而不是频率本身，是因为各组距的宽度可能不同。必须调整每个矩形条的高度，使得矩形的面积代表频率。
If all class intervals are of equal width, the histogram bars will have heights directly proportional to frequencies, but you should still use frequency density to construct the histogram correctly.
如果所有组距宽度相同，矩形条的高度将直接与频率成正比，但为了正确构建直方图，你仍然应当使用频率密度。
4. Area = Frequency | 面积 = 频率
The most important relationship in a histogram is that the area of each rectangle equals the frequency of that class interval.
直方图中最重要的关系是：每个矩形的面积等于该组距的频率。
For a rectangle, area = height × width, and since height is frequency density and width is class width, we get:
Frequency = Frequency Density × Class Width
对矩形而言，面积 = 高 × 宽。由于高为频率密度，宽为组距宽度，因此得到：
频率 = 频率密度 × 组距宽度
For example, suppose a bar has a frequency density of 4 and a class width of 5. Then the frequency is 4 × 5 = 20. This simple calculation is used repeatedly in IGCSE exam questions.
例如，假设某矩形条的频率密度为 4，组距宽度为 5，则频率为 4 × 5 = 20。这个简单计算在 IGCSE 考试题目中反复出现。
If you need to find the total frequency, sum the areas of all the rectangles. This is often needed to find the total number of observations, especially when the histogram is all you are given.
如需计算总频率，只需将所有矩形的面积相加。这在仅给定直方图的情况下常常用于求观察值总数，是考试中的常见题型。
5. Reading Histograms: Extracting Data from Bars | 读取直方图：从矩形条中提取数据
To read a histogram, you must be able to find the frequency of any class interval by looking at the bar’s dimensions.
要读取直方图，你必须能够通过观察矩形条的尺寸找出任意组距的频率。
There are three quantities involved: frequency density (height), class width (horizontal length), and frequency (area). Given any two, you can find the third.
这里涉及三个量：频率密度（高度）、组距宽度（水平长度）和频率（面积）。已知其中任意两个量，就能求出第三个量。
For example, if in a histogram a bar representing 40–50 kg has a frequency density of 3, the class width is 50 − 40 = 10 kg, so the frequency is 3 × 10 = 30 people.
例如，在直方图中，表示 40–50 公斤的矩形条频率密度为 3，则组距宽度为 50 − 40 = 10 公斤，因此频率为 3 × 10 = 30 人。
Sometimes the axis scales are not obvious. You should always check the labels and the scale values carefully before attempting any calculation. A common error is to read the frequency directly from the y-axis scale, forgetting that the y-axis is frequency density.
有时坐标轴刻度并不明显。在开始任何计算之前，务必仔细检查标签和刻度值。一个常见错误是直接读取纵轴刻度作为频率，而忘记了纵轴是频率密度。
Always write down the class boundaries clearly, especially when intervals are written in a form like ‘0–10’ where the exact limits may need to be inferred from the context.
始终清晰写出组距的边界，特别是当间隔以“0–10”这类形式书写时，可能需要根据上下文推断确切的边界值。
6. Finding Class Width and Frequency Density | 求组距宽度与频率密度
Class width is the difference between the upper class boundary and the lower class boundary. If you are given intervals like ’10 ≤ x < 15', the boundaries are 10 and 15, so the width is 5.
组距宽度是上边界减去下边界。若给你形如“10 ≤ x < 15”的区间，边界为 10 和 15，则宽度为 5。
If the intervals are written as ’10–15′ in a histogram, there may be an ambiguity, but in most IGCSE questions the boundaries are continuous. For intervals like ’10–15′, ’15–20′, the class width is 5 because the boundary between them is exactly 15.
如果直方图中的区间写成“10–15”，可能会有歧义，但在大多数 IGCSE 题目中边界是连续的。对于“10–15”“15–20”这类区间，宽度为 5，因为它们之间的边界恰好是 15。
Frequency density is calculated by dividing the frequency by the class width. Conversely, if you know the frequency density and the class width, you can find the frequency by multiplication.
频率密度等于频率除以组距宽度。反之，若已知频率密度和组距宽度，则可通过乘法求出频率。
In exam questions, you may be asked to complete a table given a histogram. For each class interval, read the frequency density from the graph, multiply by the class width, and enter the frequency in the table.
在考试题中，给定直方图后可能要求你补全表格。对每个组距，从图中读取频率密度，乘以组距宽度，然后在表格中填入频率。
7. Estimating the Number of Items in an Interval | 估算区间内的数据个数
Sometimes the class intervals in a histogram are not the same as the intervals you are asked about. For example, the histogram may be drawn with intervals 0–10, 10–20, and 20–30, but you need to find the frequency between 5 and 15.
有时直方图所用的组距与题目要问的区间并不一致。例如，直方图以 0–10、10–20、20–30 绘制，但你需要求 5 到 15 之间的频率。
In such cases, you must split the bars proportionally. For a bar covering 0–10, the portion from 5 to 10 is half of that bar. If the frequency for 0–10 is 12, then the frequency from 5 to 10 is 12 × 5 ÷ 10 = 6.
在这种情况下，你必须按比例拆分矩形条。对于覆盖 0–10 的矩形条，从 5 到 10 的部分是该条的一半。若 0–10 的频率为 12，则 5 到 10 的频率为 12 × 5 ÷ 10 = 6。
Similarly, for the 10–20 interval, if the frequency is 16, then the portion from 10 to 15 is half of that bar, giving a frequency of 8. Adding these together, the frequency between 5 and 15 is 6 + 8 = 14.
同理，对于 10–20 区间，若频率为 16，则 10 到 15 的部分占一半，得到频率 8。将两者相加，5 到 15 之间的频率为 6 + 8 = 14。
This technique relies on the assumption that data within each class interval are evenly distributed. In IGCSE questions this assumption is generally expected.
这种技术依赖于一个假设：每个组距内的数据呈均匀分布。在 IGCSE 题目中，通常默认使用这一假设。
8. Finding the Median from a Histogram | 从直方图中求中位数
The median is the middle value of a data set. In a histogram, the total frequency is the sum of the areas of all bars. The median is the value at which the cumulative area reaches half of the total area.
中位数是一组数据的中间值。在直方图中，总频率等于所有矩形条面积之和。中位数是指累计面积达到总面积一半时的变量值。
To find the median, first find the total frequency, say N. Then locate the value where the 50th percentile lies, i.e. where the cumulative area equals N ÷ 2.
求中位数时，首先要找到总频率 N。然后确定第 50 百分位所在的位置，即累计面积等于 N ÷ 2 的位置。
If the median falls inside a particular class interval, use linear interpolation. For example, if the median is in the interval 20–30 with a lower class boundary of 20, and you need to cover an additional area of 5 within that bar, then the median is approximately 20 + (5 ÷ frequency density of that bar).
如果中位数落在某个组距内，则使用线性插值。例如，若中位数在 20–30 区间内，下边界为 20，且需要在该矩形条内额外覆盖面积为 5，则中位数约为 20 + (5 ÷ 该条频率密度)。
Always remember that the median divides the area of the histogram into two equal parts. This idea is simple but powerful in exam problems.
始终记住：中位数将直方图的面积分成相等的两部分。这一概念简单但非常有用。
9. Mode and Modal Class | 众数与模态组
For continuous grouped data, we cannot find an exact mode from a histogram alone. Instead, we identify the modal class, which is the class interval with the highest frequency density.
对于连续分组数据，我们无法仅凭直方图找到精确的众数，而是确定模态组（modal class），即频率密度最高的组距。
Why do we use frequency density rather than frequency? Because the modal class is the class with the greatest ‘density’ of data. A wide class with a larger total frequency might not have the highest frequency per unit width.
为什么使用频率密度而不是频率？因为模态组是数据“密度”最大的组。一个组距较宽且总频率较大的组，其单位宽度频率不一定最高。
In an exam, the modal class is simply the bar with the greatest height. Be careful if two bars have the same height – then both are modal classes.
在考试中，模态组就是高度最高的那个矩形条。注意若有两个矩形条高度相同，则两者都是模态组。
Some questions may ask for an estimate of the mode using a formula such as: mode ≈ lower boundary + (Δ₁ / (Δ₁ + Δ₂)) × width, where Δ₁ is the difference in frequency density between the modal bar and the previous bar, and Δ₂ is the difference between the modal bar and the next bar. This is less common in IGCSE but may appear in some papers.
有些题目可能要求使用公式估算众数，例如：众数 ≈ 下边界 + (Δ₁ / (Δ₁ + Δ₂)) × 宽度，其中 Δ₁ 是模态组与前一组的频率密度之差，Δ₂ 是模态组与后一组的频率密度之差。这类公式在 IGCSE 中出现较少，但在部分试卷中可能出现。
10. Comparing Data from Histograms | 通过直方图比较数据
Histograms allow us to compare distributions. You can compare the central tendency by looking at where the median lies, and the spread by looking at the range of the data, from the smallest lower boundary to the largest upper boundary.
直方图使我们可以比较分布。你可以通过观察中位数的位置来比较集中趋势，并通过从最小下边界到最大上边界的范围来比较离散程度。
Skewness is also visible. If the histogram has a long tail to the right, the distribution is positively skewed; if the tail is to the left, it is negatively skewed.
偏态也是可见的。如果直方图右侧有长尾，则分布呈正偏；如果左侧有长尾，则呈负偏。
When comparing two histograms, be aware of differing axis scales. If the total frequencies are different, compare frequency densities rather than raw frequencies if you are looking at the shape of distributions.
在比较两个直方图时，要注意坐标轴刻度可能不同。如果总频率不同，若要观察分布形状，应比较频率密度而非原始频率。
In an IGCSE question, you may be asked to say whether one distribution has a higher median or a greater spread. Use the graph to justify your answer with reference to areas and boundaries.
在 IGCSE 题目中，可能要求你说明某个分布是否具有更高的中位数或更大的离散程度。应参考面积和边界，用图表来证明你的答案。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Here are some typical mistakes students make with histograms, and how to avoid them:
以下是学生在直方图题目中常犯的错误以及如何避免它们：
- Reading frequency from the y-axis: Always remember that the y-axis is frequency density, not frequency.
- 忽略纵轴是频率密度：永远记住纵轴是频率密度，不是频率。
- Confusing histograms with bar charts: Bar charts have gaps, histograms do not. Use area, not just height.
- 混淆直方图与条形图：条形图有间隙，直方图没有。要使用面积而不是仅看高度。
- Forgetting to use class boundaries: For example, an interval of ’10–15′ has a width of 5, not 15 − 10 = 5, which is correct, but if written as ’10 < x ≤ 15' you must use exact boundaries.
- 忘记使用组距边界：例如区间“10–15”的宽度为 5，这是正确的；但若写成“10 < x ≤ 15”，你必须使用精确边界。
- Misinterpreting the median: The median is found by dividing the area, not by taking the midpoint of the class with the largest height.
- 误读中位数：中位数是通过面积分割找到的，而不是取最高条的中点。
Check whether your answer makes sense. If the total frequency is large, a frequency of 0.5 suggests you may have misread a scale. Many histogram scales are written with decimal intervals, so read them carefully.
检查答案是否合理。如果总频率很大，而某个频率算出来只有 0.5，那很可能读错了刻度。许多直方图的刻度带有小数间隔，要仔细阅读。
12. Worked Example | 例题精讲
Now let’s attempt a complete exam-style question based on the Edexcel IGCSE specification.
现在我们来做一道基于 Edexcel IGCSE 考纲的完整考试风格题目。
Question: The histogram below shows the time, in minutes, that 200 students spend on homework each day. The first three bars are: 0–10 (frequency density 2), 10–20 (frequency density 3), 20–30 (frequency density 4). The bar 30–60 has a frequency density of 1. Find: (a) the frequencies of the first three intervals; (b) the frequency of the 30–60 interval; (c) the percentage of students who spend at least 20 minutes on homework.
题目：下面的直方图显示了 200 名学生每天花在家庭作业上的时间（单位：分钟）。前三根矩形条为：0–10（频率密度 2）、10–20（频率密度 3）、20–30（频率密度 4）。30–60 的矩形条频率密度为 1。求：(a) 前三组距的频率； (b) 30–60 组距的频率； (c) 花至少 20 分钟做作业的学生百分比。
Solution (a): For 0–10, width = 10, frequency density = 2, so frequency = 2 × 10 = 20. For 10–20, width = 10, density = 3, frequency = 30. For 20–30, width = 10, density = 4, frequency = 40.
解答 (a)：0–10 的宽度为 10，频率密度为 2，因此频率 = 2 × 10 = 20。10–20 的宽度为 10，密度为 3，频率 = 30。20–30 的宽度为 10，密度为 4，频率 = 40。
Solution (b): The 30–60 interval has a width of 30 and a frequency density of 1. Frequency = 30 × 1 = 30. To check the total: 20 + 30 + 40 + 30 = 120, but the question says 200 students. There must be another bar (60–120) not described; its frequency is 200 − 120 = 80. Thus the 30–60 interval has frequency 30.
解答 (b)：30–60 的宽度为 30，频率密度为 1，频率 = 30 × 1 = 30。验证总数：20 + 30 + 40 + 30 = 120，但题目说有 200 名学生。因此必然还有另一根矩形条（60–120）未被描述；其频率为 200 − 120 = 80。因此 30–60 的频率为 30。
Solution (c): Students who spend at least 20 minutes are those in the 20–30 and 30–60 and 60–120 intervals. Frequency = 40 + 30 + 80 = 150. Percentage = 150 ÷ 200 × 100 = 75%.
解答 (c)：花至少 20 分钟的学生位于 20–30、30–60 和 60–120 组距内。频率 = 40 + 30 + 80 = 150。百分比 = 150 ÷ 200 × 100 = 75%。
Notice how we used the area rule throughout. In part (c), we added frequencies rather than areas directly, but since each area equals a frequency, the result is the same.
注意我们全程使用了面积规则。在 (c) 部分，我们直接将频率相加，其实面积相加也可以，因为每个面积就等于一个频率。
In a real exam, the histogram would have a complete set of bars. Always verify that the total area matches the total frequency given in the question.
在真实考试中，直方图会包含所有的矩形条。始终验证所有面积之和与题目给出的总频率是否一致。
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