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 analyze the frequency distribution data, we must understand three key components: class interval size (class width), class limits, and class boundaries. Let’s solve each part step-by-step.
Given:
- Frequencies:
fi=6,25,48,72,116,60,38,22,3f_i = 6, 25, 48, 72, 116, 60, 38, 22, 3 - Class Marks (CM):
CM1=112CM_1 = 112, CM2=117CM_2 = 117
a) Determine the class interval size (class width):
Class width is the difference between the class marks (or midpoints) of two consecutive classes. Class Width=CM2−CM1=117−112=5\text{Class Width} = CM_2 – CM_1 = 117 – 112 = \boxed{5}
b) Determine the class limits of each class:
The class mark is the midpoint of the class, calculated by: Class Mark=Lower Limit+Upper Limit2\text{Class Mark} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}
To find the limits, we can use the class mark and width.
If the class width is 5, then each class spans 5 units from lower to upper limit. That means each class covers 5 cm, centered around the class mark.
So for the first class (CM = 112): Lower Limit=112−52=109.5Upper Limit=112+52=114.5\text{Lower Limit} = 112 – \frac{5}{2} = 109.5 \\ \text{Upper Limit} = 112 + \frac{5}{2} = 114.5
But class limits are usually written in whole numbers. Hence: Lower Limit=110,Upper Limit=114\text{Lower Limit} = 110, \quad \text{Upper Limit} = 114
So all class limits will be:
- 110–114 (CM=112)
- 115–119 (CM=117)
- 120–124
- 125–129
- 130–134
- 135–139
- 140–144
- 145–149
- 150–154
c) Determine the class boundaries:
Class boundaries remove gaps between classes. They are found by averaging adjacent class limits.
For example, between 114 (end of class 1) and 115 (start of class 2): Boundary=114+1152=114.5\text{Boundary} = \frac{114 + 115}{2} = 114.5
Thus:
- Class 1: 109.5–114.5
- Class 2: 114.5–119.5
- Class 3: 119.5–124.5
- Class 4: 124.5–129.5
- Class 5: 129.5–134.5
- Class 6: 134.5–139.5
- Class 7: 139.5–144.5
- Class 8: 144.5–149.5
- Class 9: 149.5–154.5
Summary of Answers:
a) Class Width: 5
b) Class Limits:
- 110–114, 115–119, 120–124, 125–129, 130–134, 135–139, 140–144, 145–149, 150–154
c) Class Boundaries:
- 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
Explanation (300+ Words):
When analyzing grouped frequency distributions, understanding class width, limits, and boundaries is crucial for further statistical calculations like the mean or standard deviation.
Class width defines how broad each group or class is in a frequency table. In this case, it was determined from the difference between consecutive class marks (117 − 112), yielding a class width of 5. This width tells us that each class spans a 5 cm height range.
Class limits are the smallest and largest values that belong to a particular class. Using the class mark and width, we calculated each class’s limits. For instance, the class mark of 112 implies the class includes values from 110 to 114 (centered around 112). Repeating this process, we identified all 9 class intervals.
Class boundaries are used to remove any gaps between classes, especially for continuous data. For example, the upper limit of one class is 114, and the lower limit of the next is 115, creating a gap. To ensure smooth continuity, we average the two to get the boundary (114.5). Boundaries help avoid ambiguity when data lies exactly at the edges and are especially important when plotting histograms or calculating cumulative frequencies.
In sum, these three values—width, limits, and boundaries—ensure data is organized correctly for both graphical representation and numerical analysis, ensuring accurate and meaningful results in data interpretation.