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Science Forum Index » Statistics - Math Forum » Basic Question: Average Greater then Maximum?
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| Uliari Moriarty |
 Posted: Thu May 01, 2008 9:44 am |
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Hello All,
I was reading a graph generated by a software application and one of
my users asked a very simple question about the units on the graph and
how it related to the min. 0.0 max. .82 and avg. 7.82. The units were
said to be in percentage but the range of values for the scale was 0
to 5.0. Now I asked the development team this question and a day later
I got the answer was that " labels and units in our graphs are
statically set" but that got me onto the idea of the values shown.
Now, I have taken stats classes and I can honestly say I have no idea
how it is possible for the average to be greater then the maximum. If
I remember correctly average is calculated as:
Avg = (V1+V2+....+Vn)/n where V# is a given value and n is the total
number of values.
Is there any way that this could be correct?
Thanks for your time... |
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| Luis A. Afonso |
| Posted: Thu May 01, 2008 2:28 pm |
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Let the values be U1, U2, …, Un the mean is
_______m = (1/n) * SUM Uj
Put the values X1, X2, …, Xn by decreasing order
Then
_______SUM Xj = SUM Uj
Therefore the mean is UNCHANGED!
One wants to prove that
_______X1 + X2 + …+ Xn <= n*X1 … [a]
Note that, by construction,
_______X2 <= X1
_______X3 <= X1
…
_______Xn <= X1
Then [a] I proved, as it was asked.
In [a] the equality SHOULD HOLD ONLY if
__X2=X1, X3=X1, …
Luis Amaral Afonso |
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| Richard Ulrich |
| Posted: Thu May 01, 2008 4:27 pm |
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On Thu, 1 May 2008 12:44:12 -0700 (PDT), Uliari Moriarty
<uliari.moriarty@gmail.com> wrote:
Quote: Hello All,
I was reading a graph generated by a software application and one of
my users asked a very simple question about the units on the graph and
how it related to the min. 0.0 max. .82 and avg. 7.82. The units were
said to be in percentage but the range of values for the scale was 0
to 5.0. Now I asked the development team this question and a day later
I got the answer was that " labels and units in our graphs are
statically set" but that got me onto the idea of the values shown.
Now, I have taken stats classes and I can honestly say I have no idea
how it is possible for the average to be greater then the maximum. If
You are right. There is no way that the average
can be greater than the max. And it does not matter
what kind of "average" or "mean" that someone is
talking about, be it geometric mean, reciprocal mean,
median, or what-have-you.
A predefined graph might omit extreme points, so that
the "mean" could lie outside of a range that is *displayed*.
Quote: I remember correctly average is calculated as:
Avg = (V1+V2+....+Vn)/n where V# is a given value and n is the total
number of values.
Is there any way that this could be correct?
--
Rich Ulrich
http://www.pitt.edu/~wpilib/index.html |
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| John Smith |
| Posted: Thu May 01, 2008 5:48 pm |
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If you want the average greater than the maximum, simply ask Adumbo to write a program to do it.
In the past, he has written programs so that
(1) the upper 2.5 percentile of the F distribution with 15 and 15 degrees of freedom is 3.71679 (correct is 2.86209
(2) calculate the sample standard deviation using the raw sum of squares instead of squared deviations from mean
(3) writes programs for confidence intervals that do not depend on sample data
(4) confuses signficance level with p-value
oh, just search his posts and find the errors yourself. It's much fun to have Adumbo giving us humor.
John |
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