Class Width Midpoints B39756

1. **State the problem:** We are given a frequency distribution of temperatures and need to find: (a) the class width, (b) the class midpoints, (c) the class boundaries. 2. **Find the class width:** The class width is the difference between the lower limits of two consecutive classes. For example, between classes $50-53$ and $54-57$, the lower limits are $50$ and $54$. $$\text{Class width} = 54 - 50 = 4$$ 3. **Find the class midpoints:** The class midpoint is the average of the lower and upper class limits. Formula: $$\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$ Calculate midpoints for each class: - $50-53$: $\frac{50 + 53}{2} = \frac{103}{2} = 51.5$ - $54-57$: $\frac{54 + 57}{2} = \frac{111}{2} = 55.5$ - $58-61$: $\frac{58 + 61}{2} = \frac{119}{2} = 59.5$ - $62-65$: $\frac{62 + 65}{2} = \frac{127}{2} = 63.5$ - $66-69$: $\frac{66 + 69}{2} = \frac{135}{2} = 67.5$ - $70-73$: $\frac{70 + 73}{2} = \frac{143}{2} = 71.5$ - $74-77$: $\frac{74 + 77}{2} = \frac{151}{2} = 75.5$ 4. **Find the class boundaries:** Class boundaries are the values that separate classes without gaps. They are found by subtracting $0.5$ from the lower limit and adding $0.4$ to the upper limit of each class. Given the classes and boundaries: - $50-53$: $49.5 - 53.4$ - $54-57$: $53.5 - 57.4$ - $58-61$: $57.5 - 61.4$ - $62-65$: $61.5 - 65.4$ - $66-69$: $65.5 - 69.4$ - $70-73$: $69.5 - 73.4$ - $74-77$: $73.5 - 77.4$ **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 above in the table.