Maximum error and percentage

• May 6th 2009, 06:21 AM
Unmath Sam
Maximum error and percentage
Hello,

I'm having some problems getting the same answer as a text book.

It defines (ignoring units) x as 2.32 and y as 0.45 to the nearest 0.1:
$x = 2.32\pm0.005$
$y = 0.45\pm0.005$
and
$T=(x-y)/y$

For my minimum values subtract 0.005 from both giving me
3.315 and 0.445.

For my maximum values I add 0.005 giving me 2.325 and 0.455.

The part that confuses me is calculating the actual error percentage, what am I dividing by what?

My textbook gives me
$error=((4.22472-4.1556) \div 4.22472)$
...but I don't get where the values are coming from.
• May 6th 2009, 10:45 AM
HallsofIvy
The largest x can be is 2.325 while the smallest it can be is 2.315.
The largest y can be is 0.455 while the smallest it can be is 0.445.

"Largest x- largest y" is 2.325- .455= 1.87.
"Largest x- smallest y" is 2.325- .445= 1.88.
"Smallest x- largest y" is 2.314- .455= 1.859.
"Smallest x- smallest y" is 2.314- .445= 1.869.

The largest of those is 1.88, the smallest is 1.859.

Dividing by a smaller number makes the result larger while dividing by a larger number makes the result smaller, so

Dividing the largest, 1.88, by the smallest y, 1.88/.445= 4.225.
Dividing the smallest, 1.859, by the largest y, 1.859/.455= 4.086.

Those are the largest and smallest possible values for (x- y)/y.

Since the "central" value is (2.32- 45)/.45= 4.156, the possible error are 4.225- 4.156= 0.069 on the "positive side" and 4.156- 4.086= 0.07. To be certain that we are inside the possible range of errors, we have to say $(x-y)/x \le 4.156\pm 0.07$. The maximum error is 0.07.

The percentage error is 0.07/4.156= 0.017 or 1.7%.
• May 6th 2009, 01:00 PM
Unmath Sam
Thank you for explaining this step by step to me.

Math textbooks did not explain that I had to actually try it for both combinations of big x, big y, small x, small y etc...e, it only explained it for area ( $x*y$)