1

Take a rectangle with length $A \pm a$ and width $B \pm b$.

Taking the max value as $(A+a)(B+b)$ and min value as $(A-a)(B-b)$ and uncertainty range as $(max-min)/2$ we get the formula: $$Area=(AB+ab)\ \pm (Ab+aB)$$

I find the presence of the $ab$ interesting. Why is it there? Have I made an error?

Kantura
  • 2,721

1 Answers1

2

Your calculation is correct if the error bars are defined to be something like "maximal deviation from the center value" without specifying how the error-distribution looks like. In particular the result value $AB+ab$ is just the center of all possible values. it is NOT "the most likely value" nor is it "the mean value" (nor "median value" either). Its not very meaningful in a statistical sense.

In order to get any of these properties, you have to specify how the error-bars on $A$ and $B$ should be interpreted. Common choices include "normal distrubtion with mean $a$ and standard deviation $a$" or maybe "uniform distribution over the range $[A-a,A+a]$". Depending on this choice, the result will change.

Simon
  • 5,061