Question

In drawing Histogram for unequal class intervals, the height of the rectangle will be:

• A. Proportional to the frequencies
• B. Proportional to the class width
• C. Proportional to the ratio of the frequencies to the width of the classes
• D. Proportional to the ratio of the width of the classes to the frequencies

Hint:

Remember the gold standard rule of histograms:
- Area \(\propto\) Frequency
- If you change the width of a bar, you must adjust its height using the formula \(\mathbf{\text{Height} = \text{Frequency} / \text{Width}}\) to keep the area mathematically correct.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A histogram is a graphical representation of a grouped frequency distribution.
The fundamental rule of a histogram is that the area of each rectangle must be proportional to its class frequency.

Step 2: Key Formula or Approach:
For any rectangle in a histogram:
\[ \text{Area} = \text{Width (Class Interval)} \times \text{Height (Frequency Density)} \] To maintain the relationship where \(\text{Area} \propto \text{Frequency}\):
\[ \text{Height} \propto \frac{\text{Frequency}}{\text{Class Width}} \]

Step 3: Detailed Explanation:
Let us analyze how histograms are drawn for different class intervals:
1. When class intervals are equal, the width of each rectangle is constant.
Consequently, the height can be plotted directly proportional to the class frequencies.
2. When class intervals are unequal, plotting height directly from frequencies would distort the graph, making wider classes appear to contain more data than they actually do.
3. To prevent this distortion, we must adjust the height using the frequency density of each class.
4. Frequency density is calculated as:
\[ \text{Frequency Density} = \frac{\text{Class Frequency}}{\text{Class Width}} \] 5. The height of each rectangle is then plotted proportional to this frequency density (the ratio of the frequency to the class width).
This adjustment ensures that the area of the rectangle accurately represents the frequency of that class.
Therefore, the height must be proportional to the ratio of the frequencies to the width of the classes.

Step 4: Final Answer:
For unequal class intervals, the rectangle height is proportional to the ratio of frequencies to class widths, matching Option (C).