Subjects statistics

Class Intervals 49200E

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Question: Use the given frequency distribution to find the (a) class width. (b) class midpoints. (c) class boundaries. Temperature (°F) Frequency 50-53 1 54-57 3 58-61 5 62-65 11 66-69 7 70-73 7 74-77 1 (c) What are the class boundaries? Complete the table below. (Type integers or decimals.) Temperature (°F) Frequency Class boundaries 50-53 1 54-57 3 58-61 5 62-65 11 66-69 7 70-73 7 74-77 1
1. **State the problem:** We are given a frequency distribution with temperature classes and frequencies. We need to find: (a) Class width (b) Class midpoints (c) Class boundaries 2. **Find the class width:** The class width is the difference between the lower limits of two consecutive classes. For example, between the first two classes: $$54 - 50 = 4$$ So, the class width is $4$. 3. **Find the class midpoints:** The midpoint of a class is the average of the lower and upper class limits. Formula: $$\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$ Calculate for each class: - $\frac{50 + 53}{2} = \frac{103}{2} = 51.5$ - $\frac{54 + 57}{2} = \frac{111}{2} = 55.5$ - $\frac{58 + 61}{2} = \frac{119}{2} = 59.5$ - $\frac{62 + 65}{2} = \frac{127}{2} = 63.5$ - $\frac{66 + 69}{2} = \frac{135}{2} = 67.5$ - $\frac{70 + 73}{2} = \frac{143}{2} = 71.5$ - $\frac{74 + 77}{2} = \frac{151}{2} = 75.5$ 4. **Find the class boundaries:** Class boundaries are the values that separate classes without gaps. Since classes are continuous and the upper limit of one class is $53$ and the next class starts at $54$, the boundary is the midpoint between $53$ and $54$: $$\frac{53 + 54}{2} = 53.5$$ Similarly, for other boundaries: - Between $57$ and $58$: $\frac{57 + 58}{2} = 57.5$ - Between $61$ and $62$: $\frac{61 + 62}{2} = 61.5$ - Between $65$ and $66$: $\frac{65 + 66}{2} = 65.5$ - Between $69$ and $70$: $\frac{69 + 70}{2} = 69.5$ - Between $73$ and $74$: $\frac{73 + 74}{2} = 73.5$ Now, the class boundaries for each class are: - First class: lower boundary = $49.5$ (one half less than $50$), upper boundary = $53.5$ - Second class: $53.5$ to $57.5$ - Third class: $57.5$ to $61.5$ - Fourth class: $61.5$ to $65.5$ - Fifth class: $65.5$ to $69.5$ - Sixth class: $69.5$ to $73.5$ - Seventh class: $73.5$ to $77.5$ (one half more than $77$) 5. **Complete the table with class boundaries:** | Temperature (°F) | Frequency | Class Boundaries | |------------------|-----------|------------------| | 50-53 | 1 | 49.5 - 53.5 | | 54-57 | 3 | 53.5 - 57.5 | | 58-61 | 5 | 57.5 - 61.5 | | 62-65 | 11 | 61.5 - 65.5 | | 66-69 | 7 | 65.5 - 69.5 | | 70-73 | 7 | 69.5 - 73.5 | | 74-77 | 1 | 73.5 - 77.5 | **Final answers:** - (a) Class width = $4$ - (b) Class midpoints = $51.5, 55.5, 59.5, 63.5, 67.5, 71.5, 75.5$ - (c) Class boundaries as in the table above.