Question

In drawing histogram for unequal class intervals the height of the rectangles will be:

• A. Proportional to the frequencies of the class intervals
• B. Proportional to the ratios of the frequencies to the widths of class intervals
• C. Proportional to the frequencies of the succeeding class intervals
• D. Proportional to the ratios of the frequencies to the widths of succeeding class intervals

Hint:

For unequal class intervals in a histogram:
\[ \text{Height} \propto \text{Frequency Density} = \frac{\text{Class Frequency}}{\text{Class Width}} \]
This adjustment ensures the area of the bar correctly represents the frequency.

Answer 1

The Correct Option is B

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

Step 2: Detailed Explanation:
Let us analyze what happens when the class intervals are unequal:
- For equal class intervals, the width of all rectangles is constant, so the height of the rectangles can be plotted directly proportional to the raw frequencies.
- For unequal class intervals, plotting raw frequencies as heights would distort the visual representation because wider bars would have disproportionately larger areas, misleading the viewer.
- To maintain the principle that Area $\propto$ Frequency, we must calculate the frequency density for each class:
\[ \text{Area} = \text{Width} \times \text{Height} \propto \text{Frequency} \]
\[ \text{Height} \propto \frac{\text{Frequency}}{\text{Width}} \]
where:
- Width = class width (class interval size).
- Height = height of the rectangle.
- Thus, the height of each rectangle must be proportional to the ratio of its frequency to its class width (frequency density).

Step 3: Final Answer:
The height of the rectangles will be proportional to the ratios of the frequencies to the widths of class intervals.