Data on the heights (in cm) of 390 cows was grouped into a frequency distribution, yielding the following results:
Frequencies (fi): 6, 25, 48, 72, 116, 60, 38, 22, 3
Class Marks (CM1 = 112, CM2 = 117) (where CM? represents the class mark of the i-th class).
Determine:
a) The class interval size (class width).
b) The class limits of each class.
c) The class boundaries for each class.
The Correct Answer and Explanation is:
To solve the problem based on the frequency distribution of cow heights, let’s analyze and compute the necessary components:
Given Data:
- Frequencies $f_i$: 6, 25, 48, 72, 116, 60, 38, 22, 3
- Class Marks (CM):
- $\text{CM}_1 = 112$
- $\text{CM}_2 = 117$
a) Class Interval Size (Class Width)
The class width can be found using the difference between two successive class marks:
$$
\text{Class width} = \text{CM}_2 – \text{CM}_1 = 117 – 112 = \boxed{5 \text{ cm}}
$$
b) Class Limits of Each Class
Class marks are the midpoints of class intervals. To find class limits, we use:
$$
\text{Lower limit} = \text{Class mark} – \frac{\text{Class width}}{2}, \quad
\text{Upper limit} = \text{Class mark} + \frac{\text{Class width}}{2}
$$
Since class width is 5:
$$
\frac{5}{2} = 2.5
$$
Thus, for each class mark $\text{CM}_i$, the class limits are:
| CM | Lower Limit | Upper Limit |
|---|---|---|
| 112 | 109.5 | 114.5 |
| 117 | 114.5 | 119.5 |
| 122 | 119.5 | 124.5 |
| 127 | 124.5 | 129.5 |
| 132 | 129.5 | 134.5 |
| 137 | 134.5 | 139.5 |
| 142 | 139.5 | 144.5 |
| 147 | 144.5 | 149.5 |
| 152 | 149.5 | 154.5 |
So the class limits for the 9 classes are:
- 109.5–114.5
- 114.5–119.5
- 119.5–124.5
- 124.5–129.5
- 129.5–134.5
- 134.5–139.5
- 139.5–144.5
- 144.5–149.5
- 149.5–154.5
c) Class Boundaries for Each Class
Class boundaries remove gaps between classes. However, here the classes already touch, so class boundaries = class limits:
- The class boundaries are the same as the limits:
- 109.5–114.5
- 114.5–119.5
- 119.5–124.5
- etc.
A frequency distribution summarizes large data sets by grouping values into intervals (classes). Each class has a midpoint (class mark), and the frequency tells us how many data values fall in that class.
Here, we’re analyzing cow heights. Given class marks of 112 and 117 for the first two groups, we determine the class width as the difference between these: 117 – 112 = 5 cm. This tells us that each class spans 5 cm.
To find class limits, we use the class mark and subtract/add half the class width (2.5). For example, the first class mark is 112. Subtracting 2.5 gives the lower limit (109.5) and adding 2.5 gives the upper limit (114.5). Repeating this for all class marks gives us the full set of class intervals, like 109.5–114.5, 114.5–119.5, and so on.
The class boundaries are typically used to ensure classes do not overlap or leave gaps. But in this case, the computed class limits already have touching boundaries, so no adjustment is needed. Thus, the boundaries are the same as the class limits.
Understanding these aspects is crucial in statistical analysis. Accurate class intervals help in drawing histograms, calculating means and variances, and making inferences. This structured grouping provides a clearer picture of the height distribution among the 390 cows.
