Question

14.6 8.3 13.3 18.0 12.8 9.4 7.8 8.1 16.8 12.1

12.1 10.6 11.8 11.2 8.7 8.3 9.4 13.1 17.0 11.3

7.3 9.6 10.3 6.5 17.6 9.6 13.7 9.5 8.3 5.6

7.8 17.8 14.0 13.1 11.9 10.1 14.7 11.2 9.2 10.3

13.5 10.2 14.2 15.3 9.3 9.9 8.7 10.8 16.0 8.6

9.1

I'm supposed to make a frequency distribution table using 7 classes and 5 being the lower limit in the first class

I understand that the class width should be 2 because (18-5.7)/7=2 rounded up, but some of the values in the raw data don't fit in the ranges

Answer 1

Lower limit = 5
class width = 2
7 classes, total width = 14
Upper limit of last class = 5+14=19

Data:
min: 5.6
max: 17.8
So everything fits into the 7 classes.

Answer 2

The seven classes are:
5-6.9 (min. 5.6)
7-8.9
9-10.9
11-12.9
13-14.9
15-16.9
17-18.9 (max. 17.9)
So everything fits tightly, right?

Answer 3

Yea everything fits, but how did you get the upper limit for each class, and to find the class boundaries would you just subract 0.05 from the lower limit and add 0.05 to the upper limit?

Answer 4

Yes, you're right, the upper limit is kind of arbitrary.
Since data here is given to 0.1 unit, so 6.9 is the highest you can do for the first class. The next greater value would be 7.0, which goes in the next class.
I cold have put the upper limit as 6.99, or 6.95, but with the given data, I was just plain lazy! :)

Answer 5

How would you find the class boundaries?

Answer 6

As you have posted at the beginning.
From the max. and min. values, (5.6 and 17.8), we have lower limit of first class = 5.
There are 41 pieces of data, the choice of 7 classes is kind of arbitrary, but it gives 41/7=6 entries per class, which is reasonable.
Also, we can have a class size of 1 (13 classes) , 1.5 (9 classes, not preferred) or 2 (7 classes), so choose 2. This ends up with the same as what you gave earlier.

Answer 7

Since class size=2, the lower limits are 5,7,9,11,13,15 and 17.
The upper boundary is a small number less than the lower boundary of the next higher class, so upper for 5 is 7-0.05=6.95, etc.
The small number must be smaller than (or equal to) the precision of the measurements (0.1 in this case)

Answer 8

@austin23
Sorry I have to go, perhaps other helper could continue to answer any questions you still have.
@Hero @jabez177

Answer 9

I'm about to go too ... to dinner that is.

Answer 10

@austin23
To help you check your results, I have the sorted dataset below:
5.6,6.5,7.3,7.8,8.3,8.3,8.6,8.7,8.7,9.1,
9.2,9.3,9.4,9.5,9.6,9.6,9.9,10.1,10.2,10.3,
10.3,10.6,10.8,11.2,11.2,11.3,11.8,11.9,12.1,13.1,
13.1,13.5,13.7,14.0,14.2,14.7,15.3,16.0,17.0,17.6,
17.8