Histograms with Unequal Class Widths

1. Why Histograms are Different from Bar Charts

You're used to bar charts, where the height of each bar represents the frequency. This works when all the categories (or bars) have the same width. But what happens when you have grouped data with unequal class widths, like we saw in the last lesson?

Consider this table:

If we drew a bar chart, the second group (frequency 40) would have a bar twice as high as the first (frequency 20). However, the second group's data is spread over an interval that is twice as wide. This is misleading. A histogram solves this problem with one fundamental rule:

In a histogram, the AREA of the bar represents the frequency, not the height.

This means we need a new value for the y-axis that ensures the area relationship holds true. This value is called Frequency Density.

2. Constructing a Histogram: The Role of Frequency Density

Since Area = Height × Width, and we want Frequency = Area, it follows that:
Frequency = Height × Class Width

Rearranging this gives us the formula for the height of each bar:

$\text{Height (Frequency Density)}=\frac{\text{Frequency}}{\text{Class Width}}$

Let's watch a video that explains this core concept and then walks through the process of drawing a histogram from a table.

Here is a static image that summarises the process you just watched. It's a useful reference.

Common Pitfalls

Before you practice, let's highlight some common errors students make. Being aware of these will help you avoid them.

Test your understanding! (Past Paper Question)

This question tests the core principle of frequency density in a typical exam format. You don't need to draw anything, just calculate.

Hint: The height is proportional to the frequency density, but not always equal to it. There might be a scaling factor.

Show answer

1. Analyse the first bar (the one we know everything about):

   • Frequency ($F_1$) = 86
   • Class Width ($C{W}_1$) = 90 - 40 = 50
   • Frequency Density ($F{D}_1$) = $F_1/C{W}_1=86/50=1.72$

2. Find the relationship between Frequency Density and Height:

   • We calculated $F{D}_1=1.72$, but the question states the height is 17.2 cm.
   • This means Height = k × Frequency Density, where k is a scaling factor.
   • $17.2=k\times 1.72$. Solving for k, we find that k = 10.
   • So, for this histogram, the rule is: Height = 10 × Frequency Density.

3. Apply the rule to the second bar:

   • Frequency ($F_2$) = 114
   • Class Width ($C{W}_2$) = 150 - 90 = 60
   • Frequency Density ($F{D}_2$) = $F_2/C{W}_2=114/60=1.9$

4. Calculate the final height:

   • Height of the second bar = $10\times F{D}_2=10\times 1.9=19$ cm.

Final Answer: 19 cm.

3. Interpreting Histograms

Now let's reverse the process. Interpreting histograms involves reading information from the graph to calculate frequencies or complete a table. This often means using the core formula rearranged:

$\text{Frequency}=\text{Frequency Density}\times \text{Class Width}$

Method 1: Using the Frequency Density Scale

The most common task is finding the frequency of one or more bars. You read the height (Frequency Density) from the y-axis and the width from the x-axis, then multiply them. Watch this example where a partially drawn histogram is used to complete a table.

Another common interpretation task is to estimate the number of items within part of a bar. For example, estimating how many animals weigh more than 80 kg when the bar covers the 75-85 kg range.

Method 2: The "Counting Blocks" Strategy (An alternative approach)

Sometimes, a question gives you the total frequency for the entire histogram but provides no scale on the y-axis. In these cases, you can't calculate frequency density directly. This is where a different method is needed, which aligns with your goal of learning more efficient problem-solving strategies.

This method is an excellent example of choosing the right tool for the job, rather than trying to force a single method onto every problem.

Conclusion

You've done an excellent job covering the essential skills for histograms. These questions are a great way to pick up marks on Paper 4 if you are methodical and careful with your calculations.

Key Takeaways:

• The Golden Rule: In a histogram, Area = Frequency.
• Constructing: To draw a histogram, you must first calculate the Frequency Density for each class and use this for the bar's height. The formula is $FD=\frac{\text{Frequency}}{\text{Class Width}}$.
• Interpreting: To find the frequency from a bar, you use the reverse formula: $\text{Frequency}=FD\times \text{Class Width}$.
• Estimation: To estimate the frequency for a part of a bar, you assume the data is evenly spread. For example, the frequency in one-quarter of a class interval is one-quarter of that bar's total frequency.
• Problem-Solving: Always check if the y-axis is scaled. If it isn't, and you are given the total frequency, the "counting blocks" method is an effective strategy.

Preview of the Next Lesson:
We have now concluded the Probability and Statistics module. In the next lesson, we will begin a new topic: Geometric Transformations and Vectors. We will start by learning how to describe and perform single transformations like rotations, reflections, translations, and enlargements.