📚 Histograms: Area, Frequency and Class Width Problems | 直方图：面积、频率与组距问题
A histogram is one of the most important graphs in statistics because it displays continuous data that has been grouped into class intervals. In a histogram, the area of each bar represents the frequency of that class, and the height of each bar is called the frequency density. This is a common source of confusion: many students treat a histogram like a bar chart, but the rules are different.
直方图是统计学中最重要的图形之一，因为它用来展示已经按组距分组的连续数据。在直方图中，每个长条的面积代表该组的频数，而长条的高度称为频率密度。这里常常产生混淆：许多学生把直方图当作条形图来读，但两者的规则完全不同。
1. What Is a Histogram? | 什么是直方图
A histogram is drawn for continuous data. The horizontal axis is a continuous scale, so the bars must touch each other. The vertical axis is labelled "frequency density" rather than "frequency" when class intervals are unequal.
直方图是为连续数据绘制的。横轴是连续刻度，因此各长条之间必须紧密相连。当各组距不相等时，纵轴应标注为“频率密度”而不是“频数”。
The critical definition is this: a histogram's bar height is the frequency density, not the frequency itself. The frequency is represented by the area of the bar.
关键定义是：直方图中长条的高度是频率密度，而不是频数本身。频数是用长条的面积来表示的。
2. The Central Formula | 核心公式
Every histogram problem is built on one simple relationship between frequency, class width and frequency density.
所有直方图问题都建立在一个简单关系上：频数、组距与频率密度之间的关系。
Frequency density = Frequency ÷ Class width
频率密度 = 频数 ÷ 组距
By rearranging this formula, we can also write:
通过变形，我们还可以写成：
Frequency = Frequency density × Class width
频数 = 频率密度 × 组距
ncy density while the width is class width, the area of each bar is exactly the frequency.
因为矩形面积等于高 × 宽，而高是频率密度、宽是组距，所以每个长条的面积正好等于该组的频数。
3. Why Area Matters | 为什么面积如此重要
Imagine two class intervals: one has width 10 and frequency 20; another has width 20 and frequency 20. If you plotted frequency on the vertical axis, both bars would have the same height even though the second interval covers twice the range of data. This would visually suggest the twenty data points are spread over a wider range, so the "density" of data is lower.
设想两个组距：一组宽度为10，频数为20；另一组宽度为20，频数也为20。如果把频数画在纵轴上，两个长条高度相同，但第二个区间覆盖的数据范围是第一个的两倍。这会让人误以为同样20个数据点在更宽的范围内分布，因此数据“密度”更低。
To compare groups fairly, we need a measure of how crowded each interval is. That measure is the frequency density.
为了公平地比较各组，我们需要一个反映区间内数据密集程度的量，这个量就是频率密度。
If all class intervals have equal width, then height and area are proportional to frequency, and using frequency on the vertical axis will still give a correct-looking histogram. But when widths are unequal, you must use frequency density.
如果所有组距都相等，那么高度和面积都与频数成正比，纵轴直接使用频数也不会影响直方图的形状。但当组距不相等时，就必须使用频率密度。
4. Finding Class Boundaries and Width | 确定组界与组距
The class width is not always obvious from the class limits written in a table. For example, the interval "10 – 20" might have boundaries 10 and 20, but the interval "10 – <20" has the same width while "0 – 10" followed by "10 – 20" also has width 10 at each step.
组距并不总是一眼能从表格中的组限看出来。例如区间“10 – 20”的边界可能是10和20，而“10 – <20”的宽度相同；“0 – 10”后面接“10 – 20”时每一步宽度也都是10。
The class width is calculated from the true boundaries:
组距由真实的组界来计算：
Class width = Upper boundary − Lower boundary
组距 = 上组界 − 下组界
If a table gives intervals as "0 ≤ x < 10", then the boundaries are clearly 0 and 10, and the width is 10. If an interval is "10 < x ≤ 20", the width is again 10. Always check whether the endpoints are included on both sides.
如果表格给出区间“0 ≤ x < 10”，那么边界显然是0和10，宽度为10。如果区间是“10 < x ≤ 20”，宽度同样是10。始终要检查端点是否真正被包含。
egins at 10 and ends at 25. The width of the second interval is then 25 − 10 = 15, not 10.
考试中常见陷阱是：第一个区间从0开始，第二个区间从10到25。此时第二个区间的组距是25 − 10 = 15，而不是10。
5. Reading Frequencies from a Histogram | 从直方图中读取频数
When you are given a histogram, the vertical axis shows frequency density. To find the frequency for any class, multiply the frequency density by the class width.
当题目给出直方图时，纵轴显示的是频率密度。要求某一组的频数，只需将频率密度乘以组距。
Follow these steps:
按以下步骤操作：
Read the frequency density value from the vertical axis for the required bar.
从纵轴读取所需长条的频率密度值。
Identify the class boundaries and calculate the width of that bar.
确定组界并计算该长条的宽度。
Multiply the two numbers to obtain the frequency.
将两个数相乘得到频数。
mple, if a bar has frequency density 3 and a width of 4, then the frequency is 3 × 4 = 12.
例如，如果一个长条的频率密度为3，宽度为4，那么频数就是3 × 4 = 12。
Remember that the total frequency is the sum of all bar areas. If the histogram is drawn on graph paper, you can also count small squares to find areas, but in written exams multiplying density by width is usually faster.
请记住，总频数是所有长条面积之和。如果直方图画在坐标纸上，你也可以通过数格子来求面积，但在笔试中，用密度乘以宽度通常更快。
6. Missing Values and Incomplete Histograms | 缺失数据与不完整直方图
requency density, or a missing bar. You can solve these problems by using the relationship "area = frequency" and the fact that all frequencies add to the total.
考官经常给出一个带有缺失信息的直方图，例如未知频数、未知频率密度或缺少某个长条。你可以利用“面积 = 频数”的关系以及所有频数之和等于总数这一事实来求解。
Suppose the total frequency is known to be 100. From the histogram you can calculate the areas of the visible bars. The missing frequency is simply the total minus the sum of the known areas.
假设总频数为100。你可以从直方图算出可见长条的面积。缺失的频数就是总数减去已知面积之和。
f the missing bar has width 5 and the remaining total frequency is 70 out of 100, that bar's frequency must be 30. Its frequency density is 30 ÷ 5 = 6.
如果缺失的是频率密度，可以建立方程。例如，缺失的长条宽度为5，其余频数合计为70，总数是100，那么这个长条的频数必须是30。其频率密度为30 ÷ 5 = 6。
Always check that your answer is consistent with the vertical scale of the graph before moving on.
在继续做题之前，务必检查你的答案是否与图形纵轴刻度一致。
7. Estimating the Mean from a Histogram | 用直方图估计平均数
Because a histogram groups data into intervals, we cannot know the exact original values. We can only estimate the mean by using the midpoint of each class.
因为直方图把数据分成区间，我们无法知道原始数据的具体值。我们只能使用每组的中点值来估计平均数。
Estimated mean = Σ(midpoint × frequency) ÷ Σ(frequency)
估计平均数 = Σ(中点值 × 频数) ÷ Σ(频数)
In a histogram, the "frequency" for each class is its area. So you first calculate each frequency by multiplying frequency density by class width. Then multiply each midpoint by its frequency, add these products, and divide by the total frequency.
在直方图中，每组“频数”就是它的面积。所以你首先要通过频率密度乘以组距算出每组频数，然后把每组中点值乘以频数，把这些乘积相加，再除以总频数。
20) ÷ 2 = 15. If its frequency is 8, then this class contributes 15 × 8 = 120 to the sum.
例如，如果某组区间为10到20，中点值为(10 + 20) ÷ 2 = 15。如果频数为8，那么该组对总和贡献为15 × 8 = 120。
The word "estimate" is important: the true mean would require all raw data, not grouped data.
“估计”这个词很重要：真正的平均数需要全部原始数据，而不是分组数据。
8. Estimating the Median and Quartiles | 估计中位数与四分位数
The median of grouped data can be estimated from a histogram using areas. Because each area represents frequency, the overall area of the histogram equals the total frequency, and the median is the value that splits this total area into two equal halves.
分组数据的中位数可以利用直方图的面积来估计。因为每个面积都代表频数，直方图的总面积等于总频数，而中位数就是把总面积分成相等两半的那个值。
The standard steps are:
标准步骤如下：
Find the total frequency by summing all bar areas.
把所有长条面积相加得到总频数。
Determine the median position. For grouped data, this is often taken as total frequency ÷ 2.
确定中位数位置。对分组数据，通常取总频数 ÷ 2。
Count areas from the left until you reach the interval containing the median, then use linear interpolation within that interval.
从左边开始累计面积，直到找到包含中位数的区间，然后在该区间内进行线性插值。
In many exam problems, you are only asked to state which class contains the median. In that case, find the class whose cumulative area first reaches or exceeds n/2.
许多考试题目只要求指出中位数落在哪个组。此时只需找到累计面积首次达到或超过 n/2 的组。
For quartiles, use n/4 for Q1 and 3n/4 for Q3. The same cumulative-area method applies.
对于四分位数，Q1 使用 n/4，Q3 使用 3n/4。同样使用累计面积法。
9. Histograms with Unequal Class Widths | 组距不等的直方图
Let us examine a classic situation where class widths are unequal:
我们来看一个组距不等的典型情形：
| Class interval | Frequency | Class width | Frequency density |
| 0 – 10 | 20 | 10 | 2 |
| 10 – 20 | 30 | 10 | 3 |
| 20 – 30 | 15 | 10 | 1.5 |
| 30 – 50 | 20 | 20 | 1 |
| 50 – 80 | 12 | 30 | 0.4 |
Notice that the last two intervals have larger widths. If we used frequency on the vertical axis, the bar for 30 – 50 would be taller than the bar for 20 – 30, which would incorrectly suggest that 20 data points in a wide interval are just as concentrated as 15 data points in a narrow interval. Using frequency density makes the comparison fair.
注意最后两个区间的宽度更大。如果纵轴直接使用频数，那么30 – 50的长条会比20 – 30的长条更高，这会错误地暗示宽区间中的20个数据点与窄区间中的15个数据点有同样的集中程度。使用频率密度才能使比较公平。
10. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Students lose marks on histogram questions for a small number of repeated reasons. Learn these pitfalls now.
学生在直方图题目上失分，往往是因为少数几个重复出现的原因。现在就学会避开这些陷阱。
Treating height as frequency. The height is frequency density, not frequency. Always multiply by class width.
把高度当作频数。高度是频率密度，不是频数。必须乘以组距。
Using wrong class boundaries. For example, using 30 – 50 as width 20 when the interval is actually from 30 to less than 50 is correct, but using 30 – 50 as width 20 is wrong if the boundaries are 25 to 55.
用错组界。例如，区间30 – 50的宽度确实是20，但如果真实边界是25到55，那么把宽度写为20就错了。
Forgetting the total. When estimating a mean, divide by the total frequency, not by the number of classes.
忘记总数。估计平均数时，应该除以总频数，而不是组的个数。
Drawing bars with equal widths. If your histogram has unequal class intervals, the widths on the axis must reflect those intervals.
把所有长条画成等宽。如果直方图的组距不等，横轴上的宽度必须反映这些组距。
Confusing histogram with bar chart. Bars in a histogram touch because the data is continuous. Gaps are only appropriate for categorical bar charts.
把直方图和条形图混淆。直方图的长条之间没有空隙，因为数据是连续的；空隙只适用于分类条形图。
11. Worked Example | 完整例题
Use the table in Section 9 to answer the following questions.
使用第9节中的表格回答下列问题。
Question 1: State the height of the bar for the class 20 – 30.
问题1：写出20 – 30这一组长条的高度。
The height is the frequency density. For 20 – 30, frequency = 15 and width = 10, so density = 15 ÷ 10 = 1.5.
高度就是频率密度。20 – 30组频数为15，宽度为10，所以密度为15 ÷ 10 = 1.5。
Question 2: Find the total frequency represented by the histogram.
问题2：求该直方图表示的总频数。
Add all frequencies: 20 + 30 + 15 + 20 + 12 = 97.
将所有频数相加：20 + 30 + 15 + 20 + 12 = 97。
Question 3: Estimate the mean of the data.
问题3：估计这组数据的平均数。
Use the midpoints: 5, 15, 25, 40, 65. Then:
使用各组中点值：5、15、25、40、65。于是：
(20×5 + 30×15 + 15×25 + 20×40 + 12×65) ÷ 97 = 2505 ÷ 97 ≈ 25.82
Thus the estimated mean is approximately 25.82.
因此估计平均数约为25.82。
Question 4: Which class contains the median?
问题4：中位数落在哪个组？
e cumulative frequency after the first class is 20, after the second class is 50, so the median lies in the class 10 – 20.
中位数位置为97 ÷ 2 = 48.5。第一组累计频数为20，第二组累计频数为50，所以中位数落在10 – 20这一组。
12. Summary and Final Strategy | 总结与最终策略
Histogram problems always reduce to the same core idea: frequency equals area. If you remember this one sentence, you can solve most exam questions.
直方图问题总是归结为同一个核心思想：频数等于面积。只要记住这一句话，你就能解决大多数考试题。
Before starting a histogram question, ask yourself three questions:
开始做直方图问题前，先问自己三个问题：
Is the vertical axis frequency density or frequency?
纵轴是频率密度还是频数？
What are the true class boundaries for each bar?
每个长条的真实组界是什么？
Do I need to find a frequency, a density, a mean, or a median?
我要求的是频数、密度、平均数还是中位数？
frequency relationship and check that all answers are sensible. Histograms are not a place to lose unnecessary marks.
一旦这些问题明确了，就运用“面积 = 频数”的关系，并检查所有答案是否合理。直方图不是白白失分的地方。
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