📚 Histograms | 直方图
Histograms are a specialised type of diagram used to display the distribution of grouped continuous data. Unlike bar charts, histograms can handle class intervals of different widths by using area – not just height – to represent frequency. In IGCSE Edexcel Mathematics, mastering frequency density is essential for both constructing and interpreting these graphs accurately.
直方图是一种用于展示分组连续数据分布的专业图表。与条形图不同，直方图通过面积而不仅仅是高度来表示频数，因此能够处理不同宽度的组距。在IGCSE爱德思数学中，掌握频数密度是准确绘制和解读此类图形的关键。
1. What Is a Histogram? | 什么是直方图？
A histogram is a graphical representation of grouped continuous data. The data is divided into intervals (classes), and a rectangle is drawn for each class. The area of each rectangle is proportional to the frequency of that class. When class widths are equal, the height also reflects frequency, but when widths vary, height alone is misleading.
直方图是分组连续数据的图形表示。数据被划分为多个区间（组），每个组对应一个矩形。矩形的面积与该组的频数成正比。当组距宽度相等时，高度也能反映频数；但当宽度不等时，仅凭高度会产生误导。
2. Frequency Density Formula | 频数密度公式
The key to histograms is frequency density. It tells you how many data values are packed into each unit of the horizontal axis.
直方图的核心在于频数密度，它反映了横轴每个单位上聚集了多少数据值。
Frequency Density = Frequency ÷ Class Width
频数密度 = 频数 ÷ 组距宽度
Class width is the difference between the upper and lower boundaries of the interval. Always use the true boundaries for continuous data, not the rounded endpoints.
组距宽度是区间上限与下限的差值。对于连续数据，务必使用真实边界，而非舍入后的端点值。
3. Calculating Frequency Density | 计算频数密度
Consider a data set with the following grouped frequencies. We will calculate the frequency density for each interval.
考虑以下分组频数数据集，我们将为每个区间计算频数密度。
| Class Interval (组距) | Frequency (频数) | Class Width (组宽) | Frequency Density (频数密度) |
|---|---|---|---|
| 0 < w ≤ 5 | 10 | 5 | 10 ÷ 5 = 2 |
| 5 < w ≤ 10 | 16 | 5 | 16 ÷ 5 = 3.2 |
| 10 < w ≤ 20 | 24 | 10 | 24 ÷ 10 = 2.4 |
| 20 < w ≤ 40 | 18 | 20 | 18 ÷ 20 = 0.9 |
n though the frequency is 18. This prevents the bar from being disproportionately tall and incorrectly suggesting a concentration of values.
请注意，像20 < w ≤ 40这样较宽的区间，尽管频数为18，但其频数密度很低。这可以防止矩形过高而错误地暗示数据集中在此区间。
4. Drawing a Histogram: Step by Step | 绘制直方图：分步讲解
Step 1: Find the class boundaries. For example, if intervals are given as 10–19, 20–29, the boundaries for the first class are 9.5 and 19.5, giving a width of 10.
第一步：找出组边界。例如，若区间给定为10–19、20–29，则第一组的边界为9.5和19.5，宽度为10。
Step 2: Calculate frequency density for each class using the formula.
第二步：用公式计算每组的频数密度。
Step 3: Draw axes. The horizontal axis shows the continuous variable, marked with the class boundaries. The vertical axis is labelled 'Frequency Density'.
第三步：画出坐标轴。横轴表示连续变量，标记组边界；纵轴标注为“频数密度”。
Step 4: For each class, draw a rectangle whose width equals the class width and whose height equals the frequency density. The bars should touch one another because the data is continuous.
第四步：为每一组画一个矩形，宽度等于组距宽度，高度等于频数密度。由于数据是连续的，矩形之间应互相接触。
Step 5: Check your graph. The total area of all bars should be proportional to the total frequency. If you sum the products of width x height for each bar, you will get the total frequency (or a scaled version if you used a scaling factor).
第五步：检查图形。所有矩形的总面积应与总频数成正比。如果将所有矩形的宽度×高度乘积相加，会得到总频数（若使用比例尺，则为按比例缩放的值）。
5. Interpreting Histograms: Finding Frequency | 解读直方图：求频数
Given a histogram, you can recover the frequency of a class by multiplying the bar's width by its height (frequency density).
已知直方图，你可以通过将矩形宽度乘以高度（频数密度）来恢复该组的频数。
Frequency = Frequency Density × Class Width
频数 = 频数密度 × 组距宽度
If the histogram shows a bar for the interval 30–50 with a height of 1.5 on the frequency density axis, then the frequency is 1.5 × (50 – 30) = 1.5 × 20 = 30.
若直方图上30–50的矩形在频数密度轴上高度为1.5，则频数为1.5 × (50 – 30) = 1.5 × 20 = 30。
Always read the class boundaries accurately from the horizontal axis. If the boundaries are not explicitly labelled, use the grid to determine the exact start and end points.
务必从横轴上准确读取组边界。若边界未明确标注，可借助网格确定精确的起始点和结束点。
6. Unequal Class Widths: Why Frequency Density Matters | 不等宽组距：为何频数密度重要
Suppose you have two classes: 0–4 with frequency 8, and 4– , 12÷16=0.75), the first bar is taller, correctly showing that data is more densely packed in the narrower interval.
假设有两个组：0–4 频数为8，4–20 频数为12。若用频数作纵轴，第二个矩形会比第一个高1.5倍，但它的宽度是第一个的4倍。这会让数据看起来更集中于第二个区间，尽管频数仅稍高一些。通过绘制频数密度（8÷4=2、12÷16=0.75），第一个矩形更高，正确表明在较窄区间内数据更为密集。
Frequency density compensates for unequal widths, ensuring that a larger area always means a larger frequency.
频数密度可补偿宽度不等的效应，确保面积越大始终意味着频数越大。
7. Histograms vs Bar Charts | 直方图与条形图的区别
Many students confuse histograms with bar charts, but they serve different purposes. A bar chart is used for discrete or categorical data, with gaps between the bars and no meaningful horizontal scale. A histogram represents continuous data, with no gaps (unless a class has zero frequency) and area proportional to frequency.
许多学生容易混淆直方图和条形图，但它们的用途不同。条形图用于离散或分类数据，柱形之间有间隙，且横轴刻度无实际含义。直方图表示连续数据，无间隙（除非某组频数为零），且面积与频数成正比。
| Feature (特征) | Bar Chart (条形图) | Histogram (直方图) |
|---|---|---|
| Data type (数据类型) | Discrete/categorical (离散/分类) | Continuous (连续) |
| Gaps between bars (间隙) | Yes (是) | No (否) |
| Vertical axis (纵轴) | Frequency (频数) | Frequency density (频数密度) |
| Meaning of area (面积意义) | Not used (无) | Proportional to frequency (与频数成正比) |
In an exam, spotting the difference is crucial for choosing the correct method to analyse a given diagram.
在考试中，识别两者的区别对于选择正确的分析方法来解读给定图表至关重要。
8. Finding the Mode from a Histogram | 从直方图求众数
The modal class is the class interval with the highest frequency, not necessarily the highest bar on the histogram. To find the modal class, calculate the frequency (area) of each bar, then pick the class with the largest frequency. If a histogram is drawn with equal-width bars, the tallest bar indicates the modal class directly, but for unequal widths you must check the area.
众数所在的组是频数最高的组，而不一定是直方图上最高的矩形。要找出众数组，先计算每个矩形的频数（面积），然后选出频数最大的组。如果直方图是等宽矩形，最高的矩形直接指示出众数组；但对于不等宽矩形，必须检查面积。
Once you identify the modal class, you can give this interval as the answer – you do not need to estimate a single value unless specifically asked.
一旦确定众数组，将其区间作为答案即可——除非题目有特定要求，否则无需进一步估计单一数值。
9. Estimating the Median from a Histogram | 从直方图估算中位数
The median is the value that splits the total frequency into two equal halves. In a histogram, you can estimate the median by finding the point where the cumulative frequency reaches half of the total frequency.
中位数是将总频数平分为两半的数值。在直方图中，可以通过寻找累积频数达到总频数一半的位置来估算中位数。
First, calculate the total frequency. Then, starting from the left, add up frequencies (areas) until you reach or exceed half the total. The median lies within that class. Use linear interpolation within the interval to find a more precise estimate: find how far into the class you need to go to reach the halfway point, then scale this distance to the class width.
首先计算总频数。然后从左侧开始累加频数（面积），直至达到或超过总频数的一半。中位数即位于该组内。通过组内线性插值可获得更精确的估计值：确定还需进入该组多大比例才能达到半数点，并将该比例乘以组宽。
Median = L + ( (N/2 – F) / f ) × w
中位数 = L + ( (N/2 – F) / f ) × w
where L is the lower boundary of the median class, N is total frequency, F is the cumulative frequency before the median class, f is the frequency of the median class, and w is the class width.
其中 L 为中位数所在组的下边界，N 为总频数，F 为该组之前的累积频数，f 为该组的频数，w 为组宽。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Mistake 1: Using frequency instead of frequency density on the vertical axis when class widths are unequal. This can distort the appearance of the distribution and lose marks.
错误一：当组距不等时，纵轴使用频数而非频数密度。这会扭曲分布特征并被扣分。
Mistake 2: Forgetting to use true class boundaries. Given intervals like 1–10, 11–20, the width is actually 10 if the data is continuous (0.5–10.5, 10.5–20.5), not 9.
错误二：忘记使用真实组边界。对于如1–10、11–20的区间，若数据连续，实际宽度为10（0.5–10.5、10.5–20.5），而不是9。
Mistake 3: Leaving gaps between bars. Histograms for continuous data must have bars that touch, except where a class has zero frequency.
错误三：矩形之间留有空隙。连续数据的直方图矩形必须相互接触，除非某组频数为零。
Exam tip: Always label the vertical axis as 'Frequency density' and include the units if applicable. When calculating class width, double-check whether the boundary values are inclusive or exclusive.
考试技巧：纵轴务必标注“频数密度”，并在必要时带上单位。计算组宽时，反复确认边界值是包含还是不包含。
Exam tip: If asked to complete a histogram from a table, first calculate all frequency densities and draw a table of values before plotting. Use a ruler and ensure the height of each bar corresponds accurately to the scale on the frequency density axis.
考试技巧：若要求根据表格补全直方图，应先计算所有频数密度并列出数值表再绘图。使用直尺，确保每个矩形的高度精确对应频数密度轴的刻度。
11. Practice with a Worked Example | 实例精练
The table below shows the distances travelled to school by a group of students. Draw a histogram and estimate the median distance.
下表为一组学生上学的路程数据。请绘制直方图并估算中位数路程。
| Distance d (km) | Frequency | Class width | Frequency density |
|---|---|---|---|
| 0 < d ≤ 2 | 12 | 2 | 12 ÷ 2 = 6 |
| 2 < d ≤ 5 | 30 | 3 | 30 ÷ 3 = 10 |
| 5 < d ≤ 8 | 24 | 3 | 24 ÷ 3 = 8 |
| 8 < d ≤ 15 | 14 | 7 | 14 ÷ 7 = 2 |
quency up to the second class is 12 + 30 = 42, so the median falls in the 2 < d ≤ 5 class.
总频数 = 12 + 30 + 24 + 14 = 80，半数即40。第二组及之前的累积频数为12 + 30 = 42，因此中位数位于 2 < d ≤ 5 组内。
Median = 2 + ( (40 – 12) / 30 ) × 3 = 2 + (28/30)×3 = 2 + 2.8 = 4.8 km
中位数 = 2 + ( (40 – 12) / 30 ) × 3 = 2 + (28/30)×3 = 2 + 2.8 = 4.8 公里
12. Final Check: Area and Total Frequency | 最终检查：面积与总频数
ple above: (6 × 2) + (10 × 3) + (8 × 3) + (2 × 7) = 12 + 30 + 24 + 14 = 80, which matches.
为验证直方图，将各组的频数密度乘以组宽并求和，结果应等于总频数。以上例而言：(6 × 2) + (10 × 3) + (8 × 3) + (2 × 7) = 12 + 30 + 24 + 14 = 80，完全吻合。
This simple check can prevent careless errors and ensure your histogram is accurately scaled.
这个简单核对可避免粗心错误，并确保直方图比例精确。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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