- #1
carnivalcougar
- 40
- 0
1. Homework Statement
An average value of N measurements is defined as:
ravg = 1N ∑ ri where i = 1 and it sums up to N measurements
By using this expression in the master formula for a general function z = (x1, x2,...) error propagation, we find the δravg error propagation:
(δravg)2 = ( (∂ravg (ra, r2,...) / ∂r1 ) (δr1))2 + ( (∂ravg (ra, r2,...) / ∂r1) (δr1))2 + ...
Finally, we arrive at the following conclusion:
δravg = δrN√
since all measurements are independent (δr1)2 = (δr2)2 = ... = (δr)2. In your present investigation the counts of the geiger counter are integer numbers. They can become zero, navg = 0, if the shiled is thick or the time interval becomes small. Does this mean that standard deviation vanishes, [itex]\sqrt{navg}[/itex] --> 0, and statistical error becomes infinite, δnavg ∝ navg-1/2 → ∞ ? Or, does the presented statistical error analysis become inapplicable in this case? Is it better to have one longer time interval or many short intervals of the same cumulative duration to minimize the standard deviation (or does it make no difference)?
2. Homework Equations
Provided in question
3. The Attempt at a Solution
I think that the error analysis becomes inapplicable. I don't think the standard deviation vanishes and the statistical error becomes infinite. This may be completely wrong though. I'm not sure if the interval matters though.
An average value of N measurements is defined as:
ravg = 1N ∑ ri where i = 1 and it sums up to N measurements
By using this expression in the master formula for a general function z = (x1, x2,...) error propagation, we find the δravg error propagation:
(δravg)2 = ( (∂ravg (ra, r2,...) / ∂r1 ) (δr1))2 + ( (∂ravg (ra, r2,...) / ∂r1) (δr1))2 + ...
Finally, we arrive at the following conclusion:
δravg = δrN√
since all measurements are independent (δr1)2 = (δr2)2 = ... = (δr)2. In your present investigation the counts of the geiger counter are integer numbers. They can become zero, navg = 0, if the shiled is thick or the time interval becomes small. Does this mean that standard deviation vanishes, [itex]\sqrt{navg}[/itex] --> 0, and statistical error becomes infinite, δnavg ∝ navg-1/2 → ∞ ? Or, does the presented statistical error analysis become inapplicable in this case? Is it better to have one longer time interval or many short intervals of the same cumulative duration to minimize the standard deviation (or does it make no difference)?
2. Homework Equations
Provided in question
3. The Attempt at a Solution
I think that the error analysis becomes inapplicable. I don't think the standard deviation vanishes and the statistical error becomes infinite. This may be completely wrong though. I'm not sure if the interval matters though.