A problem on brilliant was answered by 236 users in days, giving an average number of answers per day, . Looking at this problem now, I see that three users answered it at times, days ago from now.
Taking into account the next person to post ( who could post at any time from right now to the end of days (very unlikely)) and the 4th last person to solve the problem (assuming, say, a constant probability of posting (although feel free to use a more realistic model)), estimate the standard deviation for the time intervals between people solving the problem.
For clarity: if the next person posts in days, and the 4th last posted ago, the differences are .
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
How familiar are you with probability models? For example, if you assume that the events are independent and you can estimate the rate, then you can model it with the Poisson distribution.
A unique feature of the Poisson distribution, is that the mean is equal to the variance, which allows you to calculate the standard deviation.
Log in to reply
Hey Calvin, can you help me with another standard deviation problem please? I've even started the discussion but I have had no responses..
here is the discussion: https://brilliant.org/discussions/thread/calculation-of-standard-deviation-of-coordinates/
Which Brilliant problem was this? I would very much like to try it.
Log in to reply
I think It's just a setting for the problem.
Either geometry/combinatorics or number theory (although only level 3- I'm new here!).