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Lana
Head Banana


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Any math gurus in the house?
#5894135 - 07/24/06 11:52 AM (17 years, 6 months ago) |
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I need a little help finding a formula. The formula basically finds a variable within an average (for lack of a better way of saying it)
For example. If I have a range of numbers between 0 and 10
Then I'm given 5, 2, 7, and 9 the average of all of those numbers is 5.75 (no brain teaser there).
But what I want to know is the formula for finding the total number of times a number can be used until only the tenth place is effected.
In the example above there are 11 possibilites. 0 through 10.
If I use all 11 variables X amount of times, it will get to a point where the average itself will only be affected in the tenths place. Now if a huge increase in one specific number is used then the average will move from the tenths to the next highest place.
I wish I could explain this better but if I knew the name of the formula I wouldn't be posting
And help would be great! 
Thanks, Lana
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Ythan
ᕕ( ᐛ )ᕗ


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Re: Any math gurus in the house? [Re: Lana]
#5895892 - 07/24/06 09:26 PM (17 years, 6 months ago) |
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Who needs math when you have programming? I don't have an answer does this describe your problem though?
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Seuss
Error: divide byzero


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Re: Any math gurus in the house? [Re: Ythan]
#5897250 - 07/25/06 06:32 AM (17 years, 6 months ago) |
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Grrr... I came up with a solution, but it is incorrect. It works for a single value, but not for a series of values.
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RandalFlagg
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Re: Any math gurus in the house? [Re: Lana]
#5898200 - 07/25/06 01:16 PM (17 years, 6 months ago) |
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I'll tell you the answer if you go out on a date with me. 
Ok....I admit that I don't know the answer. I might be able to figure it out if I tried though.
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Lana
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Re: Any math gurus in the house? [Re: RandalFlagg]
#5898221 - 07/25/06 01:24 PM (17 years, 6 months ago) |
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Hey guys, thanks for trying
And Randall... if I wasn"t already seeing some (and if you had a case of Penn Dark) I'd think about it 
Lana
-------------------- Myco Supply - Distributors of Mycological Products http://www.MycoSupply.com The Premiere Source for Mushroom Growing Supplies. Visit us online or call us toll free
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daba
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Re: Any math gurus in the house? [Re: Lana]
#5901094 - 07/26/06 05:09 AM (17 years, 6 months ago) |
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I'd love to help, but I don't quite understand what you're saying...
If you use a set of integers from an interval, and you use each number x times, then your average will always be the same as if x = 1?
What variable are you looking for exactly?
To the best of my understanding, it seems that you're looking for some variable y such that given a set of numbers {a1,a2,a3,...,an} that the average, for some integer x, of [x*(a1+a2+...+an)] / n (call this A) will change in the tenths place when adding an additional y of some number in your set. That is, the average of [x*(a1+a2+...+an) + y*ai]/ (n+y) will have its digit in the tenths place changed.
E.g.: if you take Ythan's example the average stays at 5 until you add a certain number of times of some number. For the record, Ythan, 4 is not the right answer to your own question; your floating point arithmetic installed on your computer rounded 5.05 up to 5.1. So are you looking for how many times you really need to add 6 to effect the tenths place?
Edit: OK after staring at your wording a little while I think this is what you mean:
Say you have a set of numbers {5,2,7,9} which has an average of 5.75 according to you. Then you want to know by continuing to add a certain number in that set, say 5, how many times can you add it to obtain an average that only changes in the hundredths place?
In the example above, let's choose the number 5. Then how many times can we add 5 to our average calculation such that the new average will change only in the hundredths place?
Anyways, the answer to that question can be studied as follows. You have your average, A, and you know theoretically that the number of 5's you add, say x of them, will force the average to approach 5 as x tends to infinity.
Meaning that the average of A + 5 + 5 + 5 + ... (up until some huge number) will have its limit at 5. This should make sense since when x grows huge, then the addition of a bunch of other numbers A will be negligible.
So then basically what you're looking for is how many times you can add 5 until you get 5.0xxxx.
Here's some MATLAB code I've written. The answer is 26.
function lana
i = 0 A = [5 2 7 9]
while abs(mean(A) - 5) > .1 A = [A 5] i = i+1 end
i
Edit: Oh I should explain. When you run that code, i is the minimum number of additions of 5 you can add to your set of numbers such that the tenths digit will keep changing. After i = 26, that is after you add your 25th 5 into your average, the tenth digit will no longer change.
Edited by daba (07/26/06 05:57 AM)
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Seuss
Error: divide byzero


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Re: Any math gurus in the house? [Re: daba]
#5901113 - 07/26/06 05:24 AM (17 years, 6 months ago) |
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Quote:
Say you have a set of numbers {5,2,7,9} which has an average of 5.75 according to you. Then you want to know by continuing to add a certain number in that set, say 5, how many times can you add it to obtain an average that only changes in the hundredths place?
In the example above, let's choose the number 5. Then how many times can we add 5 to our average calculation such that the new average will change only in the hundredths place?
That was my final understanding. I came up with a solution to tell me the maximum value I can add into the average keeping the change in average under ε, but I couldn't come up with a general case of adding a number n times. If I remember correctly, it was something close to: v=ε*(n+1)-An*n+An*(n+1) (where An is your current average, n is the number of items in the average)
Edited by Seuss (07/26/06 05:29 AM)
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daba
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Re: Any math gurus in the house? [Re: Seuss]
#5901155 - 07/26/06 05:48 AM (17 years, 6 months ago) |
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OK so what it seems you are looking for is some number p such that given a set of n numbers which you have chosen to multiply each by some number m, will this expression hold (if you choose to keep adding xi):
|(m*x1 + m*x2 + ... + m*xi + ... + m*xn + p*xi) / (m*n+p) - xi| > 0.1
Now you need to solve this expression for p. Just do it like:
(m*x1 + m*x2 + ... + m*xi + ... + m*xn + p*xi) / (m*n+p) - xi > 0.1
or
(m*x1 + m*x2 + ... + m*xi + ... + m*xn + p*xi) / (m*n+p) - xi < -0.1
Er, forgot the n. Obviously, you want p>=0.
So if we use your example up there, wiht 5 2 7 9 and adding 5, we see that the correct solution (do the algebra) will give you p = 26, which is consistent with the MATLAB output. You can set up the problem like this:
|(5+2+7+9+5*p)/(4*1+p) - 5| > 0.1
and solve.
Seuss: The subscripts do work. Then I'm guessing this must be superscript.
Edited by daba (07/26/06 02:41 PM)
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daba
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Re: Any math gurus in the house? [Re: daba]
#5901166 - 07/26/06 05:55 AM (17 years, 6 months ago) |
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Um, you're also looking for the minimum of such p that satisfies the two inequalities. I'm not sure if there is an easier way but the MATLAB code runs fine... you should be able to program that in most any language. You also might want to tidy it up a bit since I wrote it pretty ad hoc for your example.
Seuss, how do you write in subscripts? I ran into a forum that allowed LaTeX code but I didn't know this feature was available here!
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Seuss
Error: divide byzero


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Re: Any math gurus in the house? [Re: daba]
#5901196 - 07/26/06 06:14 AM (17 years, 6 months ago) |
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I don't know if it is an admin only thing or not. I use sub and /sub in square brackets around the item to subscript. Hehe... why does it not surprise me that you know LaTeX.
(I have a picture somewhere of a little store in Thaiwan with a sign that reads "Dum LaTex".)
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daba
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Re: Any math gurus in the house? [Re: Seuss]
#5902061 - 07/26/06 02:36 PM (17 years, 6 months ago) |
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Yea, LATEX really is great for typing up anything that has math in it. Also great for just typesetting in general; much more professional and clean than Microsoft Word or any other WYSIWYG type text editors.
Anyways, Lana... did that answer your question? It doesn't even matter how many times you multiply your average before you start adding lots of the same number. You can just treat that as some residue. What matters is what number you start with and how precise you want your "average" to be; that is, do you want only the hundredths digit to be changing, thousandths, etc.
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Lana
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Re: Any math gurus in the house? [Re: daba]
#5902146 - 07/26/06 03:09 PM (17 years, 6 months ago) |
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Thanks Seuss and daba...
Did it answer my question... kind of
Let me make it more applicable to what I'm doing. I work in real estate. The numbers that I crunch can range between 0 and 250 (sometimes up to 450)
In the course of any given time frame, my numbers may be 35, 95, 127, 71, 13, 8, 353 so on and so forth.
What I'm trying to do is, when I average out all of my numbers, how much or how many more numbers will I need in order to affect the overall average by only a tenth?
A cheap example may be this.. If I have 10,000 numbers, each one ranging between 0 and 450, chances are that I could add the number 55 to the list and the overall average would change very little.
Year ago I once read up on how averages after a certain point barely change value due to the large amount of numbers in the average.
Then again, that was years ago, maybe numbers have changed since then
Lana
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daba
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Re: Any math gurus in the house? [Re: Lana]
#5902255 - 07/26/06 03:50 PM (17 years, 6 months ago) |
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This sounds like a statistics sensitivity problem... anyways... so let me get this straight...
You have a bunch of numbers and you take their average. Then you keep adding random numbers within your specified range, and you want to know how many numbers, randomly chosen, will result in a change in the tenths place?
For example if I have the numbers 0,1,...,10 and I just randomly pick numbers from that set and keep taking their average... I want to know after what number p of randomly chosen numbers from the set does the average hardly move anymore?
This seems a bit strange though; because this problem might necessarily not have a limit... that is the average can bounce around forever.
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Seuss
Error: divide byzero


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Re: Any math gurus in the house? [Re: Lana]
#5902361 - 07/26/06 04:41 PM (17 years, 6 months ago) |
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> What I'm trying to do is, when I average out all of my numbers, how much or how many more numbers will I need in order to affect the overall average by only a tenth?
Alrighty... see if this is what you mean. I am taking this slow... the path to the answer is a bit unorganized:
Lets say you have the original four numbers: 2, 5, 7, 9
Let the number of items be n (in our example, n=4) Let the average at n be denoted as An (in our exampe, An=5.75 for n=4)
What happens when we add a new value into the average? Lets call the new value v and the new average An+1.
Given the current average, and n, and the new value v, we can calculate the new average:
An+1=(An*n+v)/(n+1)
If we pretend our new value is 8, then the new average would be (5.75*4+8)/5 = (23+7)/5 = (2+5+7+9+8)/5
Ok, now back to the problem presented. We want to find a value such that the difference between An+1 and An is less than a constant.
Let the maximum change allowable be ε (in our example, 1/10)
Writing the problem (ignoring absolute value for now):
An+1-An<ε
We already figured out how to calculate An+1, so lets drop it into the above equation:
[(An*n+v)/(n+1)]-An<ε
If you run through the algebra, solving for v:
v<ε*(n+1)-(An*n)+An*(n+1)
Back to our example:
v<[(1/10)*(4+1)]-(5.75*4)+(5.75*5) v<0.5-23+28.75 v<6.25
So as long as we add less than 6.25, our average should not change by more than 1/10. Lets see:
(2+5+7+9+6.25)/5=29.25/5=5.85 (1/10 more than 5.75)
So as long as we are under 6.25, we will be adding less than 1/10 to the average.
I ignored the negative case (An-An+1<ε)
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ChuangTzu
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Re: Any math gurus in the house? [Re: Seuss]
#5902761 - 07/26/06 07:42 PM (17 years, 6 months ago) |
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I haven't been thinking about this as much as you guys have. But I interpreted her question as something like the following:
"Assuming a collection of n numbers in the range [a,b], how large does n need to be in order that adding any one number in [a,b] can only change the average of the new collection with respect to the old by a maximum of 0.1?"
Obviously adding the number in [a,b] which is the most different from the current average will have the biggest effect on the new average. And the amount this number can affect the average depends on the size of n, and on the "size" of the range (b-a). Now re-read the statement of the problem I wrote above and see if that makes sense. : )
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