Understanding Statements on Histograms

A histogram is a graphical representation of the distribution of numerical data. It is an estimate of the probability distribution of a continuous variable (quantitative variable). To construct a histogram, the first step is to "bin" the range of values—that is, divide the entire range of values into a series of intervals—and then count how many values fall into each interval.

Analyzing Statement 1: Area of Histogram vs. Frequency Polygon

Statement 1 says that the total area of the rectangles in a histogram is equal to the total area bounded by the corresponding frequency polygon and the x-axis.

A frequency polygon is constructed by joining the midpoints of the top edges of the adjacent rectangles of a histogram with straight lines. If we also join the midpoint of the top edge of the first rectangle to the midpoint of the class interval just before the first class (with zero frequency) on the x-axis, and similarly, the midpoint of the top edge of the last rectangle to the midpoint of the class interval just after the last class (with zero frequency) on the x-axis, we get a closed figure.

When comparing the area of the histogram and the frequency polygon constructed this way, it can be shown that the area added by the frequency polygon outside the histogram bars in some parts is equal to the area removed from inside the histogram bars in other parts. Therefore, the total area under the frequency polygon (bounded by the polygon and the x-axis) is indeed equal to the total area of the rectangles in the histogram.

So, Statement 1 is correct.

Analyzing Statement 2: Area Proportionality in Histograms with Unequal Class Intervals

Statement 2 says that when class intervals are unequal in a frequency distribution, the area of the rectangle is proportional to the frequency.

In a histogram, when the class intervals are of equal width, the height of each rectangle is made proportional to the frequency of the corresponding class. In this case, the area of the rectangle (width \(\times\) height) is also proportional to the frequency (since width is constant).

However, when the class intervals are of unequal width, simply making the height proportional to the frequency would make the areas of wider classes disproportionately large compared to their frequency. To ensure that the histogram accurately represents the distribution, it is the area of the rectangle that must be made proportional to the frequency. This means the height of the rectangle for a class with unequal width is calculated as:

\text{Height} = \frac{\text{Frequency}}{\text{Width of the class}} \times \text{Constant (often the minimum class width)}

This constant factor ensures that the area of the rectangle is proportional to the frequency. The quantity \(\frac{\text{Frequency}}{\text{Width of the class}}\) is called the frequency density.

Thus, when class intervals are unequal, the principle is that the area of the rectangle represents the frequency, meaning the area is proportional to the frequency.

So, Statement 2 is correct.

Conclusion

Based on the analysis of both statements:

• Statement 1 regarding the equality of areas between a histogram and its corresponding frequency polygon is correct.
• Statement 2 regarding the proportionality of area to frequency when class intervals are unequal is correct.

Therefore, both statements are correct.