Question

for this frequency table, you should have used the following elements:
min: 6.7, max: 15.1, range: 8.4, class width: 1.7.
if done correctly, your frequency table should have appeared as follows:

lower	upper	freq.	rel. freq.
8.4	10.0	11	0.367
10.1	11.7	9	0.300
11.8	13.4	3	0.100
13.5	15.1	2	0.067

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

#1:	#2:	#3:

Explanation:

Step1: Determine the first lower boundary

The first lower class boundary is the minimum value, which is \( 6.7 \).

Step2: Find the first upper boundary (and second lower boundary)

The class width is \( 1.7 \). To find the upper boundary of the first class, add the class width to the lower boundary minus a small epsilon (but since the classes are continuous here, we can use the next lower boundary as the upper boundary of the previous class plus a tiny amount, but in this case, the upper of the first class is \( 8.3 \), so the lower of the second class is \( 8.3 + 0.1 = 8.4 \) (wait, actually, for continuous data, the class boundaries are such that the upper boundary of one class is the lower boundary of the next class. Wait, the given classes have lower and upper as 6.7 - 8.3, 8.4 - 10.0, etc. So the class boundaries (the true boundaries) would be: for the first class, lower boundary is \( 6.7 \), upper boundary is \( 8.3 + 0.1 = 8.4 \)? No, wait, the difference between 8.3 and 8.4 is 0.1, so the class boundaries are:

First class: \( 6.7 - 8.3 \) (but the true boundary between first and second class is \( 8.35 \)? Wait, no, the given classes are 6.7 - 8.3, 8.4 - 10.0, so the gap between 8.3 and 8.4 is 0.1. So to make them continuous, we take the midpoint between 8.3 and 8.4, which is \( \frac{8.3 + 8.4}{2} = 8.35 \). Wait, maybe the problem is using the given lower and upper as the class limits, and we need to find the class boundaries.

Class boundaries are calculated as: lower boundary of a class = lower limit - 0.05 (if decimal is to one decimal place), upper boundary = upper limit + 0.05? Wait, no, the class width is 1.7. Let's check: 8.3 - 6.7 = 1.6, but the class width is given as 1.7. Wait, maybe the classes are:

First class: 6.7 to 8.3 (width 1.6? No, 8.3 - 6.7 = 1.6, but class width is 1.7. Wait, maybe the lower limit of the first class is 6.7, upper limit is 6.7 + 1.7 = 8.4? But the table says upper is 8.3. Maybe there's a typo, but we'll use the given table.

The class boundaries (the boundaries between classes) are:

Between first and second class: \( \frac{8.3 + 8.4}{2} = 8.35 \)

Between second and third class: \( \frac{10.0 + 10.1}{2} = 10.05 \)

Between third and fourth class: \( \frac{11.7 + 11.8}{2} = 11.75 \)

Between fourth and fifth class: \( \frac{13.4 + 13.5}{2} = 13.45 \)

And the lower boundary of the first class is \( 6.7 - 0.05 = 6.65 \) (if we consider the lower limit as 6.7, the boundary is 6.7 - 0.05 = 6.65), upper boundary of the last class is \( 15.1 + 0.05 = 15.15 \)

Wait, maybe the problem is just asking for the lower and upper limits as boundaries? No, class boundaries are different from class limits.

Wait, the problem says "Enter the class boundaries below." The table has 6 boxes, so probably the 6 boundaries: lower of first, upper of first (lower of second), upper of second (lower of third), upper of third (lower of fourth), upper of fourth (lower of fifth), upper of fifth (upper of last).

From the frequency table:

First class: Lower 6.7, Upper 8.3

Second class: Lower 8.4, Upper 10.0

Third class: Lower 10.1, Upper 11.7

Fourth class: Lower 11.8, Upper 13.4

Fifth class: Lower 13.5, Upper 15.1

So the class boundaries (the points between classes) are:

1. Lower boundary of first class: 6.7 (or 6.65 if we adjust, but since the first lower limit is 6.7, maybe the first boundary is 6.7)

Wait, no, class boundaries for a class with lower limit \( L \) and upper limit \( U \) (where \( U = L + \text{class width} - \text{small} \)) are \( L - \frac{d}{2} \) and \( U + \frac{d}{2} \), where \( d \) is the decimal precision. Here, the data is to one decimal place, so \( d = 0.1 \), so \( \frac{d}{2} = 0.05 \).

So for first class:

Lower boundary: \( 6.7 - 0.05 = 6.65 \)

Upper boundary: \( 8.3 + 0.05 = 8.35 \)

Second class:

Lower boundary: \( 8.4 - 0.05 = 8.35 \) (which matches the upper boundary of first class)

Upper boundary: \( 10.0 + 0.05 = 10.05 \)

Third class:

Lower boundary: \( 10.1 - 0.05 = 10.05 \)

Upper boundary: \( 11.7 + 0.05 = 11.75 \)

Fourth class:

Lower boundary: \( 11.8 - 0.05 = 11.75 \)

Upper boundary: \( 13.4 + 0.05 = 13.45 \)

Fifth class:

Lower boundary: \( 13.5 - 0.05 = 13.45 \)

Upper boundary: \( 15.1 + 0.05 = 15.15 \)

So the six class boundaries are: 6.65, 8.35, 10.05, 11.75, 13.45, 15.15

Let's verify the class width: 8.35 - 6.65 = 1.7, 10.05 - 8.35 = 1.7, etc. Perfect, matches the given class width of 1.7.

So the six boundaries are:

#1: 6.65

#2: 8.35

#3: 10.05

#4: 11.75

#5: 13.45

#6: 15.15

Answer:

#1: 6.65, #2: 8.35, #3: 10.05, #4: 11.75, #5: 13.45, #6: 15.15