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How to Calculate Class Width for a Frequency Distribution

How to calculate class width when constructing a grouped frequency distribution or histogram from quantitative data?

Subject: Statistics Chapter: Organizing and Graphing Data Topic: Organizing and Graphing Quantitative Data Answer included
how to calculate class width class width grouped frequency distribution histogram bins class intervals class limits class boundaries range
Accepted answer Answer included

The task how to calculate class width arises when quantitative data are grouped into equal-size classes to form a frequency table or histogram. Class width is the numerical “size” of each class interval.

Definition (most reliable form)

The class width is the difference between the upper and lower class boundaries:

\[ w = (\text{upper class boundary}) - (\text{lower class boundary}) \]

When classes are written using lower class limits (for example, 50–59, 60–69, 70–79), the class width is also the difference between consecutive lower class limits:

\[ w = L_{2} - L_{1} \]

Step-by-step: how to calculate class width from raw data

  1. Find the minimum and maximum of the data, then compute the range:
    \[ R = x_{\max} - x_{\min} \]
  2. Choose the number of classes \(k\) (common choices are 5–10 for many introductory tables; larger samples may use more).
  3. Compute a preliminary class width:
    \[ w_0 = \frac{R}{k} \]
  4. Round up to a convenient value for readability (often to the next integer or to 1, 2, 5, 10, 20, … depending on scale):
    \[ w = \lceil w_0 \rceil \quad \text{(if integer-width classes are desired)} \]
  5. Build classes of width \(w\), selecting a starting lower limit at or below \(x_{\min}\) so all observations are included without gaps.

Worked example (grouped frequency distribution)

Consider 20 exam scores: 52, 55, 56, 58, 61, 62, 64, 66, 67, 70, 72, 73, 75, 78, 81, 83, 85, 88, 90, 93.

1) Range

\[ x_{\min} = 52,\quad x_{\max} = 93,\quad R = 93 - 52 = 41 \]

2) Choose number of classes

Take \(k = 5\) classes (a compact summary for a small dataset).

3) Compute and round class width

\[ w_0 = \frac{R}{k} = \frac{41}{5} = 8.2 \]

Round up to a convenient width \(w = 10\). This yields easy-to-read class intervals such as 50–59, 60–69, and so on.

4) Form the classes and count frequencies

Class limits Class boundaries Class width Frequency
50–59 \(49.5\) to \(59.5\) \(59.5 - 49.5 = 10\) 4
60–69 \(59.5\) to \(69.5\) \(69.5 - 59.5 = 10\) 5
70–79 \(69.5\) to \(79.5\) \(79.5 - 69.5 = 10\) 5
80–89 \(79.5\) to \(89.5\) \(89.5 - 79.5 = 10\) 4
90–99 \(89.5\) to \(99.5\) \(99.5 - 89.5 = 10\) 2

Visualization: class width as a repeated interval

50 60 70 80 90 100 class width \(w = 10\) 50–59 60–69 70–79 80–89 90–99 Equal-width classes create uniform histogram bins and comparable frequencies across intervals.
Each class interval spans the same distance on the number line; the repeated distance is the class width.

Two quick checks that prevent mistakes

  • No gaps, no overlaps: class boundaries should touch (for example, 59.5 is both the upper boundary of 50–59 and the lower boundary of 60–69).
  • Width consistency: verify \(L_{2}-L_{1} = w\) across all classes (for example, \(60-50=10\), \(70-60=10\), etc.).

Summary

To calculate class width, compute the range \(R=x_{\max}-x_{\min}\), divide by the chosen number of classes \(k\) to get \(w_0=R/k\), then round up to a convenient \(w\) and build equal-size class intervals whose boundaries differ by \(w\).

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