Group of 20: Class Width 20 in Grouped Frequency | 组距20的分组频率分布

📚 Group of 20: Class Width 20 in Grouped Frequency | 组距20的分组频率分布

In Edexcel A-Level Mathematics, grouped frequency distributions often use a fixed class width to summarise continuous data. The phrase “Group of 20” can be interpreted as choosing a class width of 20 units, such as 0-20, 20-40, 40-60, and so on. This article explains how to handle grouped data when the class width is 20, covering frequency density, histograms, cumulative frequency, interpolation, and estimation of the mean and standard deviation.

在 Edexcel A-Level 数学中，分组频率分布常使用固定组距来概括连续数据。”Group of 20″ 可以理解为选择 20 个单位为组距，例如 0-20、20-40、40-60 等。本文讲解当组距为 20 时如何处理分组数据，涵盖频率密度、直方图、累积频率、插值法，以及均值和标准差的估计。

1. Grouped Data and Class Width 20 | 分组数据与组距20

Continuous data are often grouped into intervals called classes. If each class has a width of 20, we write intervals such as 0 ≤ x < 20, 20 ≤ x < 40, and so on. The class width is the difference between the upper and lower class boundaries, here 20 - 0 = 20.

连续数据通常被分成若干区间，称为组。如果每个组的组距为 20，我们会写成 0 ≤ x < 20、20 ≤ x < 40 等。组距是上组界与下组界之差，此处为 20 - 0 = 20。

Using a class width of 20 is common when the data range is large, such as test scores from 0 to 100, or ages from 0 to 80. It gives a manageable number of classes: for example, five classes for 0-100.

当数据范围较大时，使用组距 20 很常见，例如 0 到 100 的考试成绩，或 0 到 80 的年龄。这样可以得到数量适中的组数：例如 0-100 可分为五个组。

2. Frequency Density and Histograms | 频率密度与直方图

In Edexcel A-Level Statistics, histogram bars must have area proportional to frequency. Since the class width may not be 1, we calculate frequency density using:

在 Edexcel A-Level 统计中，直方图的条形面积必须与频率成正比。由于组距不一定为 1，我们需要计算频率密度：

Frequency density = Frequency ÷ Class width

For class width 20, frequency density = frequency ÷ 20. This means a frequency of 30 in a 0-20 class has a frequency density of 30 ÷ 20 = 1.5.

当组距为 20 时，频率密度 = 频率 ÷ 20。这意味着 0-20 组内频率为 30 时，频率密度为 30 ÷ 20 = 1.5。

If all class widths are equal, you may draw a bar chart, but exam questions often include unequal widths to test the frequency density concept. Always check the width before plotting.

如果所有组距都相等，可以画条形图；但考试题常包含不等组距，以考查频率密度概念。绘图前务必检查组距。

3. Calculating Frequency from a Histogram | 由直方图计算频率

Given a histogram, the frequency of a class is the area of its bar: frequency = frequency density × class width. For a class width of 20, multiply the frequency density by 20 to recover the frequency.

已知直方图时，某组的频率等于其条形面积：频率 = 频率密度 × 组距。当组距为 20 时，将频率密度乘以 20 即可还原频率。

Example: if a histogram bar for 20 ≤ x < 40 has height 2.5, then frequency = 2.5 × 20 = 50. This reverses the frequency density calculation.

例如：若 20 ≤ x < 40 的直方图条形高度为 2.5，则频率 = 2.5 × 20 = 50。这是频率密度计算的逆运算。

4. Cumulative Frequency and the Ogive | 累积频率与累积频率曲线

Cumulative frequency is the running total of frequencies. For classes with width 20, add frequencies in order: 0-20, 20-40, 40-60, etc. Plot cumulative frequency against the upper class boundary.

累积频率是频率的累计总和。对于组距为 20 的数据，按 0-20、20-40、40-60 等顺序累加频率。以累积频率对上组界作图。

The upper class boundaries for width 20 classes are 20, 40, 60, 80, 100. Plot points at (20, cumulative frequency up to 20), (40, cumulative frequency up to 40), and join with a smooth curve.

组距 20 的各组上组界为 20、40、60、80、100。在 (20, 截至 20 的累积频率)、(40, 截至 40 的累积频率) 等处描点，并用平滑曲线连接。

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

For grouped data, the median is estimated by linear interpolation because individual data values are unknown. The median is the value at the n/2-th position.

对于分组数据，由于单个数据值未知，中位数需要通过线性插值来估计。中位数位于第 n/2 个位置。

Median ≈ 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 the class width (20 in our case).

其中 L 是中位数所在组的下组界，F 是中位数所在组之前的累积频率，f 是中位数所在组的频率，w 是组距（此处为 20）。

Example: if n = 120, n/2 = 60, the median class is 40-60 with L = 40, F = 45, f = 30, then median ≈ 40 + ((60 – 45)/30) × 20 = 40 + 10 = 50.

示例：若 n = 120，n

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