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Given the following baby weights (in kg), construct a grouped frequency distribution and determine the class width: [list of weights]
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Step-by-step solution
3 stepsClassify the data into a grouped frequency distribution.
First, list the given baby weights:
2.57, 4.21, 1.05, 3.06, 4.50, 5.05, 3.45, 0.92
3.12, 2.87, 0.76, 4.18, 5.93, 4.18, 2.03, 0.57, 1.85
4.16, 3.41, 1.86, 2.53, 1.48, 3.85, 5.12, 3.24, 1.89
2.51, 0.93, 1.24, 3.21, 5.86, 3.57, 2.18, 4.29, 3.30
0.91, 0.82, 1.47, 4.25, 3.81, 2.48, 1.27, 5.35, 3.33The classes are: 0.01 – 1.00, 1.01 – 2.00, 2.01 – 3.00, 3.01 – 4.00, 4.01 – 5.00, 5.01 – 6.00.
Now, count the frequency for each class:
- Class 0.01 – 1.00: 0.92, 0.76, 0.57, 0.93, 0.91, 0.82 (Frequency = 6)
- Class 1.01 – 2.00: 1.05, 1.85, 1.86, 1.48, 1.89, 1.24, 1.47, 1.27 (Frequency = 8)
- Class 2.01 – 3.00: 2.57, 2.87, 2.03, 2.53, 2.51, 2.18, 2.48 (Frequency = 7)
- Class 3.01 – 4.00: 3.06, 3.45, 3.12, 3.41, 3.85, 3.24, 3.21, 3.57, 3.30, 3.81, 3.33 (Frequency = 11)
- Class 4.01 – 5.00: 4.21, 4.50, 4.18, 4.18, 4.16, 4.29, 4.25 (Frequency = 7)
- Class 5.01 – 6.00: 5.05, 5.93, 5.12, 5.86, 5.35 (Frequency = 5)
The grouped frequency distribution is:
Find the class width.
The class width is the difference between consecutive lower class limits (or upper class limits).
Using the first two classes:
Lower limit of the second class = 1.01
Lower limit of the first class = 0.01
Class width =Alternatively, using class boundaries:
For the class 0.01 – 1.00, the lower class boundary is and the upper class boundary is .
Class width = .The class width is .
For the class 4.01 – 5.00, name the value of
- The class center
- The class limit
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The class center (or midpoint) is the average of the lower and upper class limits.
For the class 4.01 – 5.00:
The class center is .
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The class limits are the smallest and largest values that can belong to the class. For the class 4.01 – 5.00, the class limits are 4.01 and 5.00. The class limits are .
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