📚 A-Level Maths: Histograms and Data Analysis | A-Level 数学：直方图绘制与数据分析
A histogram is one of the most important graphical tools for summarising continuous data in A-Level statistics. It looks similar to a bar chart, but the key difference is that the area of each bar, not its height, represents the frequency.
直方图是 A-Level 统计中概括连续数据最重要的图形工具之一。它看起来与条形图相似，但关键区别在于：每个条形的面积表示频数，而不是条形的“高度”。
1. What Is a Histogram? | 什么是直方图？
A histogram is used for continuous data that has been grouped into class intervals. The horizontal axis shows the variable, such as height, time or speed, and the vertical axis shows frequency density.
直方图用于已按组距分组后的连续数据。横轴表示变量，例如身高、时间或速度；纵轴表示频率密度（frequency density）。
In a correct histogram, bars are drawn touching each other. This is because the data is continuous and class intervals have no gaps. A bar chart, by contrast, is used for categorical or discrete data and normally has gaps between bars.
在正确的直方图中，条形之间是紧密相连的。这是因为数据是连续的，各组距之间没有空隙。而条形图则用于分类数据或无顺序的离散数据，条形之间通常有间隙。
2. Frequency Density and Why Area Matters | 频率密度与面积的意义
When class intervals are all the same width, bar height can be used to compare frequencies directly. However, when class widths are unequal, using the frequency itself as the height would make the visual picture misleading.
当所有组距相等时，可以直接用条形高度比较频数。但当组距不相等时，如果用频数本身作为条形高度，图形会给人错误的印象。
Therefore, the vertical axis must be frequency density. The formula is:
因此，纵轴必须是频率密度，其公式为：
Frequency Density = Frequency ÷ Class Width
Because frequency equals class width multiplied by frequency density, the area of each bar represents the frequency:
由于频数等于组距乘以频率密度，因此每个条形的面积就代表了该组的频数：
Frequency = Class Width × Frequency Density
If you are given a histogram and asked to find an unknown frequency, you must find the area of the corresponding bar, not simply read its height.
如果题目给出一个直方图并要求求未知频数，你必须计算对应条形的面积，而不能只读取高度。
3. Constructing a Histogram from a Frequency Table | 由频数表绘制直方图
To draw a histogram correctly, follow a consistent set of steps. First, identify the class boundaries and decide the scale for the horizontal axis.
要正确绘制直方图，需要遵循一套连贯的步骤。首先，确定各组的下限和上限，并设定横轴的比例尺。
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Calculate the class width for every interval by subtracting the lower boundary from the upper boundary.
用每个区间的上限减去下限，计算出各组组距。
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For each class, divide the frequency by the class width to obtain the frequency density.
对每一组，用频数除以组距，得到频率密度。
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Draw bars of width equal to the class width and height equal to the frequency density.
画出宽度等于组距、高度等于频率密度的条形。
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Ensure bars touch, and label both axes clearly.
确保条形相接，并清楚标注两个坐标轴。
When the class widths are equal, the frequency density is proportional to the frequency, so the shape of the histogram looks the same whether you plot frequency or frequency density.
当组距相等时，频率密度与频数成正比，所以无论纵轴绘制频数还是频率密度，直方图的形状都一样。
4. Worked Example: Histogram with Unequal Class Widths | 实例：不等组距直方图
The table below shows the mass, in kilograms, of 45 students. Notice that the class widths are not all the same.
下表显示了 45 名学生的体重（单位：千克）。注意各组组距并不完全相同。
| Mass (kg) | 0 ≤ x < 10 | 10 ≤ x < 20 | 20 ≤ x < 40 | 40 ≤ x < 60 |
| Frequency | 5 | 10 | 24 | 6 |
| Class width | 10 | 10 | 20 | 20 |
| Frequency density | 0.5 | 1 | 1.2 | 0.3 |
which matches the frequency. The third bar has width 20 and height 1.2, so its area is 24.
第一个条形宽为 10、高为 0.5，面积为 10 × 0.5 = 5，正好等于该组频数。第三个条形宽为 20、高为 1.2，面积为 24。
When drawing the histogram, the horizontal axis must show the boundaries 0, 10, 20, 40 and 60, and the bars must be adjacent. The vertical axis should be labelled “Frequency density”.
绘制直方图时，横轴必须标出界限 0、10、20、40 和 60，条形必须相邻。纵轴应标注为“频率密度”。
5. Reading a Histogram: Modal Class and Skewness | 读取直方图：众数类与偏态
The modal class is the class interval with the highest frequency density. It is not necessarily the class with the largest frequency when widths are unequal, but with equal widths it is simply the tallest bar.
众数类（modal class）是频率密度最大的组。在组距不相等时，它不一定是频数最大的组；而在组距相等时，它通常就是最高的条形。
above, the modal class is 20 ≤ x < 40 because its frequency density of 1.2 is the largest.
在上面例子中，众数类是 20 ≤ x < 40，因为它的频率密度 1.2 最大。
A histogram also shows skewness. If the right tail is longer than the left tail, the distribution is positively skewed. If the left tail is longer, it is negatively skewed.
直方图还可以显示偏态。如果右侧尾部比左侧更长，则分布为右偏（正偏）；如果左侧尾部更长，则为左偏（负偏）。
Many A-Level data sets are positively skewed, especially data such as income, waiting times and reaction times.
许多 A-Level 考试中的数据集都呈正偏态，例如收入、等待时间和反应时间等数据。
6. Estimating the Mean from a Histogram | 由直方图估计平均数
To estimate the mean from grouped data, we use the midpoint of each class interval as an approximation for every value in that class.
要用分组数据估计平均数，我们通常用每个区间的中点（midpoint）代表该组中的每一个值。
For the mass example, the midpoints are 5, 15, 30 and 50. The estimate of the mean is calculated as follows:
对于上述体重例子，各区间中点为 5、15、30 和 50。平均数的估计值计算如下：
Mean ≈ (5×5 + 15×10 + 30×24 + 50×6) ÷ 45
= 1195 ÷ 45 = 26.6 kg (to 1 d.p.)
This is only an estimate, because we assume that values are evenly distributed within each class. The actual mean could be slightly different.
这只是一个估计值，因为我们假设每个区间内的数据大致均匀分布。真实平均数可能会略有不同。
7. Estimating the Median and Quartiles | 估计中位数与四分位数
The median is the value that splits the data into two equal halves. To estimate it from a histogram, we need to use the area of the bars, or equivalently the cumulative frequency.
中位数是将数据一分为二的值。要从直方图估计中位数，我们需要利用条形面积，也就是累积频数。
se we use the position (45 + 1) ÷ 2 = 23. The cumulative frequencies are 5, 15, 39 and 45.
对于体重例子，总频数为 45。中位数是第 23 个数据，因为位置为 (45 + 1) ÷ 2 = 23。累积频数分别为 5、15、39 和 45。
ow 20, we need the 8th value from the 24 values in this class. Using linear interpolation:
第 23 个值落在区间 20 ≤ x < 40 内。因为 20 以下已有 15 个值，我们需要在该类的 24 个值中取第 8 个。使用线性插值：
Median ≈ 20 + (8 ÷ 24) × 20 = 20 + 6.67 = 26.7 kg
The same method can be used for quartiles. The lower quartile Q1 is at position (n + 1) ÷ 4, and the upper quartile Q3 is at position 3(n + 1) ÷ 4.
同样的方法可用于四分位数。下四分位数 Q1 位于第 (n + 1) ÷ 4 个数据，上四分位数 Q3 位于第 3(n + 1) ÷ 4 个数据。
8. Linking Histograms, Cumulative Frequency and Box Plots | 直方图、累积频数与箱线图的联系
A histogram can be converted into a cumulative frequency table by adding frequencies across the class intervals. The cumulative frequency is plotted against the upper class boundary to produce an S-shaped cumulative frequency graph.
我们可以把直方图转换为累积频数表，
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