I'm willing to give help, but in proportion to the amount of effort you show.The question I'd like to answer is: calculate the average deviation for the following distribution: 1, 2, 3, 4, 5, 6, 7, 8, 9.
I actually didn't know that. I apologize, and thanks for the informationYou know by now that we have a no work shown, no help given policy.
I don't like to state it quite as baldly as Jomo, but it looks like you haven't yet been told to read this:I actually didn't know that. I apologize, and thanks for the information![]()
I don't like to state it quite as baldly as Jomo, but it looks like you haven't yet been told to read this:
READ BEFORE POSTING
Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...www.freemathhelp.com
Show some of your work or explain where you're stuck. Don't worry that your work might be wrong; learning is a process of making mistakes. We want to check your efforts, to see where you're at. If you show no work at all, some tutors may assume that you need links to lessons or just the first step. Other tutors might think you're looking for somebody to do your homework and might not help at all.
Now please respond to what I said, including showing whatever work you have done!
Sounds good. Incidentally, I prefer the term "mean absolute deviation", which makes the meaning clear. It's conceivable that your book could have meant something different by "average deviation".I may have read that when I first joined, but I probably just forgot. The reminder was needed. Thanks!! For this problem, I said "chronological numbers" because I didn't realize that it was just giving the individual values. The order of the numbers didn't end up being important to the problem. Once I figured out I had to take the absolute value for each one, I found the answer pretty quickly. I had just missed the table in my book where it had shown the process I was looking for. I had to subtract the population mean from each score, then take the absolute value, then divide by N to find the average deviation. I'll still have to work at solidifying the process in my mind, but it wasn't too bad. The problems I have real challenges with are those requiring z and t scores, of which I've posted a couple.
For this problem, the average deviation ended up being 2.22 because the absolute value summed is 20. Divide 20 by 9 (N), and the answer is 2.22. I'm probably explaining everything you already know, but I'm just showing that I did the work.