Histograms | 直方图

📚 Histograms | 直方图

A histogram is a powerful graphical tool used to display the distribution of continuous or grouped data. Unlike a bar chart, it uses the area of bars to represent frequency, which allows meaningful comparisons even when class widths vary. In Edexcel A-Level Mathematics, mastery of histograms is essential for data representation, estimation of averages, and interpretation of spread.

直方图是一种强大的图形工具，用于展示连续或分组数据的分布。与条形图不同，它利用条形的面积来表示频率，即使在组距不等的情况下也能进行有意义的比较。在 Edexcel A-Level 数学中，掌握直方图对于数据展示、平均值估计以及离散程度的解释至关重要。

1. What is a Histogram? | 什么是直方图？

A histogram is a diagram consisting of adjacent rectangles whose areas are proportional to the frequency of observations falling into each class interval. The horizontal axis represents the variable being measured, while the vertical axis displays frequency density – not the raw frequency. This distinction is fundamental because it ensures the total area reflects the total frequency of the data set.

直方图是由相邻矩形组成的图形，每个矩形的面积与落入相应区间的观测值频率成正比。横轴表示被测量的变量，纵轴则显示频率密度——而不是原始频率。这一区别至关重要，因为它确保了总面积能够反映数据集的总频率。

In Edexcel examinations, you may be given a histogram and asked to find frequencies, or you may need to construct one from a frequency table where class widths are not equal. Understanding that frequency = frequency density × class width is the key to unlocking histogram problems.

在 Edexcel 考试中，你可能会拿到一个直方图并被要求求频率，或者需要根据组距不等的频数表画出直方图。理解频率 = 频率密度 × 组距这一关系是解决直方图问题的关键。

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2. Histograms vs Bar Charts | 直方图与条形图

It is a common misconception that histograms are simply bar charts for grouped data. However, the fundamental difference lies in how information is encoded. A bar chart uses the height of each bar to represent frequency or count, and the bars are often separated by gaps to indicate distinct categories. A histogram, by contrast, uses area to represent frequency and the bars touch each other to reflect the continuous nature of the data.

一个常见的误解是以为直方图不过是用于分组数据的条形图。然而，根本区别在于信息的编码方式。条形图用每个条形的高度来表示频率或计数，并且条形之间通常留有间隙以表明不同的类别。与此相反，直方图使用面积来表示频率，并且条形彼此接触以反映数据的连续性。

In a bar chart, all categories are usually of equal width, so height alone gives a fair visual comparison. In a histogram, class intervals may have different widths; plotting raw frequency on the vertical axis would exaggerate the importance of wider intervals and mislead the reader. That is why we scale the vertical axis in frequency density.

在条形图中，所有类别通常宽度相等，因此仅凭高度就能进行公正的视觉比较。而在直方图中，组距可能不同；若将原始频率标在纵轴上，会夸大较宽区间的重要性并误导读者。这就是为什么纵轴要用频率密度来刻度。

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3. Frequency Density: The Core Concept | 频率密度：核心概念

Frequency density is the frequency per unit of the class interval. It is defined as frequency divided by class width. Every bar in a histogram has a height equal to its frequency density, and its area equals frequency density × class width, which simplifies to frequency. This relationship ensures that the total area of all bars represents the sum of all frequencies in the data set.

频率密度是指组距中每单位区间所对应的频率。它定义为频率除以组距。直方图中每个条形的高度等于其频率密度，而它的面积等于频率密度 × 组距，这就简化为频率。这个关系确保了所有条形的总面积代表了数据集中所有频率的总和。

Frequency density = Frequency ÷ Class width

频率密度 = 频率 ÷ 组距

Grasping this formula is the first step towards accurate histogram construction and interpretation. Without it, students often plot raw frequencies and then struggle to answer exam questions that involve unequal class widths.

掌握这个公式是准确构建和解读直方图的第一步。没有它，学生们经常会绘制原始频率，然后在涉及不等组距的考题中遇到困难。

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4. Constructing a Histogram Step by Step | 逐步绘制直方图

To draw a histogram from a grouped frequency table, start by identifying the class boundaries (the exact edges of each interval where data can fall). Then, for each class, calculate the frequency density using the formula. Plot the vertical axis labelled ‘Frequency density’ and the horizontal axis with the continuous variable. Draw each bar from the lower class boundary to the upper class boundary, with height equal to the calculated frequency density.

要根据分组频数表画出直方图，首先确定组边界（每个区间数据可能落入的精确边缘）。然后，对每一组用公式计算频率密度。标出纵轴为“频率密度”，水平轴为连续变量。从组下限到组上限绘制每个条形，高度等于计算出的频率密度。

Always check that neighbouring bars share a common vertical edge – there should be no gaps. If a class has zero frequency, its frequency density is zero, so no bar is drawn. In Edexcel exams, you may be given an incomplete histogram and asked to complete it, or to find missing frequencies from plotted bars.

始终检查相邻条形是否共享相同的垂直边——不应有间隙。如果某一组的频率为零，其频率密度为零，因此不绘制条形。在 Edexcel 考试中，你可能会拿到一张不完整的直方图并被要求补全，或者根据已绘制的条形找出缺失的频率。

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5. Working with Unequal Class Widths | 处理不等组距

The real power of a histogram emerges when class widths are unequal. If raw frequencies were used as bar heights, a class of width 20 containing 10 observations would appear to have the same importance as a class of width 10 with 10 observations, even though the first class is twice as wide and the data is more spread out. By using frequency density, the wider class gets a lower bar, restoring visual fairness.

当组距不等时，直方图的真正优势显现出来。如果使用原始频率作为条形高度，一个宽度为20并有10个观测值的组会显得与宽度为10并有10个观测值的组具有同等重要性，尽管第一组的宽度是第二组的两倍且数据更为分散。通过使用频率密度，较宽的组会得到较低的条形，从而恢复视觉上的公正。

For example, suppose a class 10 ≤ x < 20 has frequency 12 (width 10), and a class 20 ≤ x < 40 has frequency 18 (width 20). Their frequency densities are 12/10 = 1.2 and 18/20 = 0.9 respectively. The wider class appears shorter, correctly reflecting that the data is less concentrated. This prevents misinterpretation in the exam and in real data analysis.

举例来说，假设一组 10 ≤ x < 20 的频率为12（组距为10），而另一组 20 ≤ x < 40 的频率为18（组距为20）。它们的频率密度分别为 12/10 = 1.2 和 18/20 = 0.9。较宽的组显得较矮，正确地反映出数据集中度较低。这能防止在考试和实际数据分析中产生误读。

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6. Finding Frequencies from a Histogram | 从直方图求频率

If you are given a histogram, you can work backwards to recover the frequency for any class. The key is to read the height (frequency density) from the vertical axis and multiply it by the class width. Many Edexcel exam questions involve completing a frequency table or comparing frequencies between intervals using areas.

如果给定一幅直方图，你可以逆向求出任何一组的频率。关键是从纵轴上读出高度（频率密度）并将其乘以组距。许多 Edexcel 考题都涉及补全频数表，或者利用面积比较不同区间的频率。

Sometimes the vertical scale is not directly labelled, but you will be given the area or frequency for one bar. Use that reference bar to find the scale factor: if a bar of height h and width w has frequency f, then each unit of frequency density corresponds to f / (h × w) which should simplify to 1 when the correct scaling is applied. Be careful with units and read axis scales accurately.

有时纵轴并未直接标出刻度，但你会得到一个条形的面积或频率。利用该参考条求出比例因子：若某高度为 h、宽度为 w 的条形频率为 f，则在正确标度下每个频率密度单位对应的频率应能简化为1。注意单位并准确读取坐标轴刻度。

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7. Estimating the Median Using Linear Interpolation | 用线性插值法估计中位数

Since a histogram displays grouped data, we cannot determine the exact median; we must estimate it. The method of linear interpolation assumes that data values within the median class are evenly distributed. First, find which class contains the middle data point (total frequency n, so median position is at the n/2th value). Then use the formula:

由于直方图展示的是分组数据，我们无法求出确切的中位数，必须进行估计。线性插值法假设中位数所在组内的数据值是均匀分布的。首先，找到包含中间数据点的那个组（总频率为 n，因此中位数位于第 n/2 个值）。然后使用公式：

Median = L + ( (n/2 − F) / f ) × w

中位数 = L + ( (n/2 − F) / f ) × w

Here, L is the lower class boundary of the median class, F is the cumulative frequency before the median class, f is the frequency of the median class, and w is its class width. This calculation is a staple in Edexcel S1 and the Statistics parts of the new specification.

其中，L 是中位数组的组下限，F 是中位数组之前的累积频率，f 是中位数组的频率，w 是该组的组距。这个计算是 Edexcel S1 和新大纲统计部分的常见内容。

To avoid errors, always draw a small diagram or annotate the histogram with cumulative frequencies. Confirm that the median lies within the interval and that your estimate is sensible given the shape of the distribution.

为避免错误，可以画一个简图或在直方图上标注累积频率。确认中位数落于该区间内，并且你的估计值在分布形状下是合理的。

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8. Estimating Quartiles and the Interquartile Range | 估计四分位数和四分位距

The same interpolation technique is used to estimate the lower quartile (Q1, at the n/4th value) and the upper quartile (Q3, at the 3n/4th value). Once you have Q1 and Q3, the interquartile range (IQR) = Q3 − Q1. The IQR provides a measure of spread that is resistant to outliers.

相同的插值方法用于估计下四分位数（Q1，位于第 n/4 个值）和上四分位数（Q3，位于第 3n/4 个值）。得到 Q1 和 Q3 后，四分位距（IQR）= Q3 − Q1。IQR 提供了一种抗离群值影响的离散程度度量。

For each quartile, identify the class containing the relevant cumulative frequency position, note its lower boundary, the cumulative frequency before that class, the class frequency, and the width. Then apply the same linear interpolation formula. On a histogram, you can visually approximate quartiles by shading areas that correspond to a quarter of the total area, but exact estimation requires algebraic working.

对每个四分位数，找到包含相应累积频率位置的那个组，记下其下限、该组之前的累积频率、组频率以及组距。然后应用相同的线性插值公式。在直方图上，可以通过阴影标出对应总面积四分之一的部分来目测近似值，但精确估计需要代数运算。

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9. Interpreting Distribution Shape and Skewness | 解释分布形状和偏度

A histogram reveals the shape of the data distribution at a glance. If the bars taper off symmetrically on both sides of a central peak, the distribution is roughly symmetric. If the right tail is longer and the peak is shifted left, the data is positively skewed – mean > median > mode typically. If the left tail is longer, we have negative skew – median > mean.

直方图可以一眼揭示数据分布的形状。如果条形在中心峰两侧对称地逐渐减小，则分布大致对称。若右尾较长且峰偏左，数据呈正偏态——通常均值 > 中位数 > 众数。若左尾较长，则为负偏态——中位数 > 均值。

In the exam, you may be asked to comment on skewness from a given histogram and to link it to comparisons of mean, median, and mode. Remember that for grouped data, the mode is often approximated as the midpoint of the class with the highest frequency density, not the highest frequency. With unequal class widths, the modal class is the one with the tallest bar on the histogram.

考试中可能要求根据给定直方图评论偏度，并将其与均值、中位数和众数的比较联系起来。请记住，对于分组数据，众数通常近似为频率密度最高组的中点，而不是最高频率组。在组距不等的情况下，众数组是直方图上条形最高的那个组。

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10. Common Errors and Exam Technique | 常见错误与应试技巧

One frequent mistake is plotting frequency instead of frequency density when class widths are unequal. Always pause and ask: ‘Have I divided by class width?’ Another error is misreading or misinterpreting class boundaries – ensure you understand whether intervals are given as 0–10, 10–20, and whether they are inclusive or exclusive, so that the correct width is used.

一个常见错误是在组距不等时画出频率而不是频率密度。每次停顿并问自己：“我除以组距了吗？”另一个错误是误读或曲解组边界——确保你理解给出的区间是 0–10, 10–20，以及它们是包含上界还是下界，以便使用正确的组距。

When interpolating, many students lose marks by using the wrong cumulative frequencies or by mixing up n/2, n/4, and 3n/4. To stay safe, write down all relevant values clearly: total frequency, cumulative frequency column, position, and the four components of the interpolation formula. Show full working – even if the final answer is slightly off, method marks can be generous in Edexcel papers.

进行插值时，很多学生因使用错误的累积频率，或混淆 n/2、n/4 和 3n/4 而失分。为了安全起见，清楚地写下所有相关值：总频率、累积频率列、位置，以及插值公式的四个组成部分。展示完整的计算过程——即使最终答案略有偏差，Edexcel 的评分也可能会给足过程分。

Finally, when drawing a histogram, label axes clearly, use a sensible scale, and ensure bars are drawn with a ruler. Untidy graphs can lead to inaccurate readings and lost accuracy marks.

最后，在画直方图时，要清晰地标注坐标轴，使用合理的比例，并确保用尺子绘制条形。不整洁的图形会导致读数不准并丢掉精确分。

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11. Worked Example: From Histogram to Summary Statistics | 例题：从直方图到汇总统计量

Consider a histogram representing the times (in minutes) taken by 80 students to complete a puzzle. The classes and frequency densities are: 0–10 min (fd = 0.8), 10–15 min (fd = 3.2), 15–25 min (fd = 2.1), 25–40 min (fd = 0.6). First, compute frequencies: 0–10 class width = 10, so freq = 0.8×10 = 8. For 10–15, width=5, freq=3.2×5=16. For 15–25, width=10, freq=2.1×10=21. For 25–40, width=15, freq=0.6×15=9. Total = 54? Wait, 8+16+21+9=54, not 80. This signals that the example is hypothetical and may not add to 80. Adjust so that total is 80: suppose 0–10 freq=12, 10–15 freq=20, 15–25 freq=30, 25–40 freq=18, then fd = 1.2, 4.0, 3.0, 1.2 respectively. Now median position = 40th value. Cumulative frequencies: 12, 32, 62, 80. Median class is 15–25. Lower boundary L=15, F=32, f=30, w=10. Median = 15 + ((40−32)/30)×10 = 15 + (8/30)×10 = 15 + 2.667 = 17.667 min.

考虑一幅直方图，表示80名学生完成拼图所用的时间（分钟）。各组及频率密度为：0–10分钟（fd = 1.2），10–15分钟（fd = 4.0），15–25分钟（fd = 3.0），25–40分钟（fd = 1.2）。首先计算频率：0–10组距=10，频率=1.2×10=12。10–15组距=5，频率=4.0×5=20。15–25组距=10，频率=3.0×10=30。25–40组距=15，频率=1.2×15=18。总频率=80。中位数位置=第40个值。累积频率：12, 32, 62, 80。中位数组为15–25。下限 L=15, F=32, f=30, w=10。中位数 = 15 + ((40−32)/30)×10 = 15 + (8/30)×10 ≈ 17.667分钟。

This step-by-step approach illustrates how every part of the histogram ties back to frequency and position. Practise similar problems from past Edexcel papers to build speed and confidence.

这种逐步解题的方法说明了直方图的每一部分如何与频率和位置相联系。通过练习以往的 Edexcel 真题来建立速度和信心。

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