Question

for this frequency table, you should have used the following elements:
min: 8.3, max: 14.6, range: 6.3, class width: 1.3.
if done correctly, your frequency table should have appeared as follows:

lower	upper	freq.	rel. freq.
9.6	10.8	8	0.27
10.9	12.1	8	0.27
12.2	13.4	7	0.23
13.5	14.7	4	0.13

use this information as you continue.
enter the class boundaries below.

#1:	#2:	#3:
#4:	#5:	#6:

Explanation:

Step1: Understand Class Boundaries

Class boundaries are the values that separate classes. For consecutive classes, the upper boundary of one class is the lower boundary of the next, adjusted to avoid gaps. The class width is 1.3, and we start from the min (8.3) and max (14.6).

Step2: Calculate First Lower Boundary

The first lower boundary is the minimum value, 8.3.

Step3: Calculate First Upper Boundary

First upper boundary: \( 8.3 + 1.3 - \epsilon \) (but since the class is 8.3 - 9.5, the boundary between first and second class is \( \frac{9.5 + 9.6}{2} = 9.55 \)? Wait, no—wait, the classes are 8.3 - 9.5, 9.6 - 10.8, etc. So the boundaries between classes are the midpoint of the upper limit of one class and lower limit of the next.

Wait, the classes are:
1. 8.3 – 9.5
2. 9.6 – 10.8
3. 10.9 – 12.1
4. 12.2 – 13.4
5. 13.5 – 14.7

So the class boundaries (the dividing lines between classes) are:

- Below first class: \( 8.3 - \frac{1.3}{2} = 8.3 - 0.65 = 7.65 \)? No, wait, class boundaries for a class with lower limit \( L \) and upper limit \( U \) (where classes are non-overlapping and adjacent) are calculated as:

For class \( [L, U] \), the lower boundary is \( L - \frac{d}{2} \), upper boundary is \( U + \frac{d}{2} \), where \( d \) is the difference between the lower limit of the next class and upper limit of current class. Here, between 9.5 (upper of first) and 9.6 (lower of second), \( d = 0.1 \), so \( \frac{d}{2} = 0.05 \).

So:

1. First class (8.3 – 9.5):
- Lower boundary: \( 8.3 - 0.05 = 8.25 \)
- Upper boundary: \( 9.5 + 0.05 = 9.55 \)

2. Second class (9.6 – 10.8):
- Lower boundary: \( 9.6 - 0.05 = 9.55 \) (same as upper of first)
- Upper boundary: \( 10.8 + 0.05 = 10.85 \)

3. Third class (10.9 – 12.1):
- Lower boundary: \( 10.9 - 0.05 = 10.85 \)
- Upper boundary: \( 12.1 + 0.05 = 12.15 \)

4. Fourth class (12.2 – 13.4):
- Lower boundary: \( 12.2 - 0.05 = 12.15 \)
- Upper boundary: \( 13.4 + 0.05 = 13.45 \)

5. Fifth class (13.5 – 14.7):
- Lower boundary: \( 13.5 - 0.05 = 13.45 \)
- Upper boundary: \( 14.7 + 0.05 = 14.75 \)

But the question asks for "class boundaries"—maybe the boundaries between the classes (the six boundaries? Wait, the table has 5 classes, so there are 6 boundaries: from below first class to above last class.

Wait, the min is 8.3, max is 14.6. The class width is 1.3. Let's list all boundaries:

- Start: \( 8.3 - \frac{1.3}{2} = 8.3 - 0.65 = 7.65 \)? No, earlier approach with the class limits (8.3, 9.5; 9.6, 10.8; etc.)—the difference between upper of first (9.5) and lower of second (9.6) is 0.1, so the boundary between them is 9.55 (midpoint). So the boundaries are:

1. Below first class: \( 8.3 - 0.05 = 8.25 \) (lower boundary of first class)
2. Between first and second: \( 9.55 \)
3. Between second and third: \( 10.85 \)
4. Between third and fourth: \( 12.15 \)
5. Between fourth and fifth: \( 13.45 \)
6. Above fifth class: \( 14.75 \)

Wait, but the problem says "Enter the class boundaries below" with 6 boxes (#1 to #6). So these are the six boundaries:

#1: 8.25 (lower boundary of first class)
#2: 9.55 (between first and second)
#3: 10.85 (between second and third)
#4: 12.15 (between third and fourth)
#5: 13.45 (between fourth and fifth)
#6: 14.75 (upper boundary of fifth class)

Let's verify:

- First class: 8.25 – 9.55 (contains 8.3–9.5)
- Second: 9.55 – 10.85 (contains 9.6–10.8)
- Third: 10.85 – 12.15 (contains 10.9–12.1)
- Fourth: 12.15 – 13.45 (contains 12.2–13.4)
- Fifth: 13.45 – 14.75 (contains 13.5–14.7, and max is 14.6 which is within)

Yes, that makes sense. The class boundaries are calculated by taking the midpoint between the upper limit of one class and lower limit of the next. Since the upper limit of a class and lower limit of the next differ by 0.1 (e.g., 9.5 and 9.6), the midpoint is \( \frac{9.5 + 9.6}{2} = 9.55 \), and so on.

Answer:

#1: 8.25
#2: 9.55
#3: 10.85
#4: 12.15
#5: 13.45
#6: 14.75