Probabilities, averages? I don't know.

  • Thread starter Thread starter RickyDee
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RickyDee

Ok, here's the scenario. I hope I word this correctly,
forgive me if I don't.

I have a list (one column) of random numbers that could
change, ranging from 87.3 (cell B17) and 223 (cell B1).
For simplicity (I Hope) I'll use these two cells. I can
come up with the total of these two cells, and the
difference of these two cells. PROBLEM IS: All of these
numbers in this whole column are all AVERAGES.

My way of thinking is this: If (for instance) 87.3 is an
average, the actual 'high and low' could be 79-93. So, if
this is an average, and the '223' is an average, then the
sum of the difference between these two numbers is NOT an
average. In other words, the difference between 223 and
87.3 is NOT 185.7. It is a little bit higher.

If you don't understand what I'm trying to do, please just
laugh and bypass this. Something is bugging me about this
and I just can't put my finger on it. It isn't right!

Thank you for your time, and I hope I didn't bother anyone
much with this.
 
If you have two averages, say 16 and 72, and the former is the average of 12
and 20 while the latter is the average of 60 and 84, then anyway you slice
it the difference of the averages is 56.

72 - 16 = 56

[(84 - 20) + (60 - 12)] / 2 = [64 + 48] / 2 = 112 / 2 = 56

[(84 - 12) + (60 - 20)] / 2 = [72 + 40] / 2 = 112 / 2 = 56

As long as your averages are all based on the same number of observed
values, no matter how you pair up the observed values that produce two
different averages, the difference of the averages is the average of the
differences.
 
Ricky
As Harlan stated, if your averages are based on the same number of inputs your average of the averages is the same as the total average

However, If your averages are weighted you can use something like this to get the averall average

=((B1*COUNTA(A1:A10))+(B17*COUNTA(A17:A27)))/(COUNTA(A1:A10)+COUNTA(A17:A27)

Assuming B1 is the average of numbers in A1 through A10, and B17 is the avergae of numbers in A17 to A27. If some cells are left blank this will still weight properly

If the averages are input manually, then you could also input the number of items in used to create each average, say in C1 and C17, and then use

=((B1*C1)+(B17*C17))/(C1+C17

Good Luck
Mark Graesse
(e-mail address removed)

----- RickyDee wrote: ----

Ok, here's the scenario. I hope I word this correctly,
forgive me if I don't

I have a list (one column) of random numbers that could
change, ranging from 87.3 (cell B17) and 223 (cell B1).
For simplicity (I Hope) I'll use these two cells. I can
come up with the total of these two cells, and the
difference of these two cells. PROBLEM IS: All of these
numbers in this whole column are all AVERAGES.

My way of thinking is this: If (for instance) 87.3 is an
average, the actual 'high and low' could be 79-93. So, if
this is an average, and the '223' is an average, then the
sum of the difference between these two numbers is NOT an
average. In other words, the difference between 223 and
87.3 is NOT 185.7. It is a little bit higher

If you don't understand what I'm trying to do, please just
laugh and bypass this. Something is bugging me about this
and I just can't put my finger on it. It isn't right

Thank you for your time, and I hope I didn't bother anyone
much with this.
 
I think what you need to deal with are weighted averages. For
example, if you have 3 values that average to 50, and 6 values that
average to 100, you don't get 75 as an overall average, you get 83.33.
Inversely, if you have 6 values that average to 50 and 3 that average
to 100, you get 66.67. You need to multiply each average by the
number of values used to figure it, and then divide by the total
number of values ((3*50+6*100)/9) = 83.33; ((6*50+3*100)/9) = 66.67.
As someone else said, if you have the same number of values for each
average, then the simple subtraction works.
 
Hey guys, thank you very, very much. I knew I ran across
this in my studies at the local university here, but night
school really doesn't stay in my mind much.

I really thought I'd get laughed at! But you guys helped
immensely, to put it (VERY) lightly.

Thanks again. (THIS IS COOL!)
Have a safe day,
RickyDee
 
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