Let's break down the problem and find the class boundaries, midpoint, and width for the given class interval:

Given:

Class interval: $8.24 - 12.36$

1. Class Boundaries:

Class boundaries are the lower and upper limits of the class, adjusted to ensure that there are no gaps between successive classes.

• The lower boundary is found by subtracting 0.5 units from the lower limit of the class.
• The upper boundary is found by adding 0.5 units to the upper limit of the class.

For the class $8.24 - 12.36$:

• Lower boundary = $8.24 - 0.5 = 7.74$
• Upper boundary = $12.36 + 0.5 = 12.86$

So, the class boundaries are 7.74 and 12.86.

2. Midpoint:

The midpoint is the average of the upper and lower limits of the class.

$\text{Midpoint} = \frac{8.24 + 12.36}{2} = \frac{20.6}{2} = 10.3$

So, the midpoint is 10.3.

3. Class Width:

The class width is the difference between the upper and lower limits of the class.

$\text{Class Width} = 12.36 - 8.24 = 4.12$

So, the class width is 4.12.

Final Answer:

• Class Boundaries: 7.74 and 12.86
• Midpoint: 10.3
• Class Width: 4.12

Would you like any further details or clarifications on these concepts?

Here are some related questions you might consider:

1. How would you calculate class boundaries if the class interval were a whole number?
2. What is the significance of class boundaries in a frequency distribution?
3. How would overlapping class intervals affect the calculation of class boundaries?
4. Can class width vary within a data set, and how would that impact the analysis?
5. What is the relationship between the midpoint and the median in a grouped frequency distribution?

Tip: Always check for any gaps or overlaps between class intervals, especially when preparing a frequency distribution table. This ensures accurate representation of the data.