Re: Mean and Stdev
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Wed, 13 May 2009 20:33:06 GMT
In article <63ddf34a-c4d0-4ea2-8fce-514e7dfcc1fa@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
no_suchluck@xxxxxxxxxxx wrote:
On May 13, 1:27=A0pm, A N Niel <ann...@xxxxxxxxxxxxxxxxxxxxx> wrote:
In article
<53dbe429-cb63-4da1-92ee-f28b9730a...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<no_suchl...@xxxxxxxxxxx> wrote:
Just had a dilemma arise as I was trying to normalize some data.
Here is an extremely simplified example: Given an arbitrary length
list of numbers whose possible value range is from 0 to 10, we would
expect:
max possible mean: 10.0 (i.e, 10, 10, 10, 10)
min possible mean: 0.0 =A0 (i.e, 0, 0, 0, 0)
max possible stdev: 5.0 =A0(i.e, 10, 10, 0, 0)
min possible stdev: 0.0 (i.e, 10, 10, 10, 10)
This issue is, what are the min and max values for combinations of
stdev and mean? For example, for [mean + stdev], one might at first
expect the range to be [0.0, 15.0]. However, what I found out is that
this is NOT the case. The reason is that there are certain mutually
exclusive conditions that exist between mean and stdev. One of which
being that it is impossible for the mean and stdev to simultaneously
be at max. In fact, when the mean is maximum, the stdev is guaranteed
to be minimum. So, the upper bound for [mean + stdev] is somewhere
between 10.0 and 15.0, but I don't know where.
Is there anyway to find the exact upper bound for this example?
Thanks for any help,
This max is at (0,10,10,10) approximately 11.83
Interesting... this seems to be correct. How did you arrive at this
answer? My goal was not to trial/error or brute force the correct
answer, but come up with a generalized solution that would work with
number lists of arbitrary size.
Unfortunately, it seems that the possible max is a function of list
size. A size of 4 seems to yield the highest possible max, as you
indicated, while the possible maxs for list sizes 1-3 and 5+ gradually
decrease from there.
So maybe what I am asking is impossible.
If you have a range of values for each sample, m <= x <= M, the way
to maximize mean + std dev, is to put as close to (2 - sqrt(2))/4 of
the sample at m and as close to (2 + sqrt(2))/4 at M. Then the
mean + std dev is close to M + (M-m) (sqrt(2) - 1)/2. In your case,
with 14.64466% at 0 and 85.35534% at 10, we get the mean + std dev to
be 12.0710678. Of course, we can only approach this high a sum when
we have a very large sample.
For example, with a 1000 samples, 146 at 0 and 854 at 10, we get a
mean of 8.54 and a std dev of 3.5310622 to yield a sum of 12.0710622.
Rob Johnson <rob@xxxxxxxxxxxxxx>
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