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March 5th, 2019, 12:44 PM  #1 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0  Finding Values of the Mean using Static Values with only Means  What is the formula?
Hey, So I have a question about finding values that make up an average. In the example below each number is an average of three values. in the example below I will explain how the averages are made up. As you can see each A  M is a given mean of three values. The last two values in the mean are carried over and used in the next mean with the first value being dropped and adding one more. The values once they are used are static and cannot change. What I am looking for is a formula to solve for the three values in a mean with only being given a handful of means where the values do not change? The unfortunate part is that we do not know any of the values but we have the means and thay are verified. Now it can be done with guess and check but with so many values and possibilities it would take a while but I'm hoping there is a formula that I don't know of to solve this. C = (X+Y+Z)/3 D = (Y+Z+N)/3 E = (Z+N+P)/3 F = (N+P+Q)/3 and then so on. A=82 B=78 C=77 D=79 E82 F=77 G=81 H=84 I=88 J=82 K=82 L=80 M=75 Last edited by bwilliams632; March 5th, 2019 at 12:56 PM. Reason: Values didn't alight with average unknown 
March 6th, 2019, 03:47 PM  #2 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0 
Bump. Anyone have any ideas? Thanks, 
March 7th, 2019, 07:45 PM  #3 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0 
Bump. Help please? Thanks!, 
March 7th, 2019, 10:51 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,428 Thanks: 1314 
are you saying that $A = \dfrac 1 3 (x_1 + x_2 + x_3)$ $B = \dfrac 1 3 (x_2+x_3+x_4)$ $C= \dfrac 1 3 (x_3 + x_4 + x_5)$ $\dots$ $M = \dfrac 1 3 (x_{13}+x_{14}+x_{15})$ ? and you want to solve for the $x_k$s ? 
March 8th, 2019, 05:52 AM  #5 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0 
That is correct I am trying to solve for the Xk value.

March 8th, 2019, 06:03 AM  #6 
Senior Member Joined: Oct 2009 Posts: 782 Thanks: 280 
But there are 13 equations and 15 unknown values. The $x_k$ will not be unique, there will be multiple solutions. Do you just want one solution, any of them?

March 8th, 2019, 07:52 AM  #7 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0 
Yes just one solution would be perfect and I’d be able to figure the rest out using that example.

March 8th, 2019, 09:32 PM  #8 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0 
So after digging into it for a bit this is what I am getting if I was to solve for C:77=1/3(x3+x4+x5) so trying to solve for x3 I get x3=231x4x5. Which makes sense and eventually you’d solve for each individual value using the main given formula if I was to plug x3 into the main formula. If I’m thinking right the equation would most likely never be solvable? But is there a way to find a definate answer for each value if we were to solve for C? Let me know if I’m completely off and have no idea what I’m taking about!

March 9th, 2019, 10:55 AM  #9 
Newbie Joined: Mar 2019 From: YQR Posts: 7 Thanks: 0 
So I’m assuming this would have to be solved by a matrix?

March 9th, 2019, 11:26 AM  #10 
Senior Member Joined: Sep 2015 From: USA Posts: 2,428 Thanks: 1314  You could but there's no need. You need to select the first two values. They can be anything. Then $3A = x_1+x_2 +x_3$ $x_3 = 3A  x_1  x_2$ Similarly $3B = x_2+x_3+x_4$ $x_4 = 3B  x_2x_3$ and you use the value of $x_3$ you just computed. you can come up with a recursive iteration to do this in a few lines of code. 

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finding, formula, means, static, values 
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