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Work out the standard deviation

A problem on brilliant was answered by 236 users in \( \frac{45}{24}\) days, giving an average number of answers per day, \( \mu \) \( (\approx 125.9) \). Looking at this problem now, I see that three users answered it at times, \( \frac{15}{24*60},\frac{31}{24*60},\frac{69}{24*60}\) 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 \( 7 \) 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 \(\frac{t}{24*60}\) days, and the 4th last posted \(\frac{T}{24*60}\) ago, the differences are \(\frac{t+15}{24*60},\frac{16}{24*60},\frac{38}{24*60},\frac{T-69}{24*60} \).

Note by A L
4 years, 1 month ago

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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. Calvin Lin Staff · 4 years, 1 month ago

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@Calvin Lin 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/ Namrata Haribal · 3 years, 10 months ago

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Which Brilliant problem was this? I would very much like to try it. Tim Ye · 4 years, 1 month ago

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@Tim Ye Either geometry/combinatorics or number theory (although only level 3- I'm new here!). A L · 4 years, 1 month ago

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@Tim Ye I think It's just a setting for the problem. Sridhar Thiagarajan · 4 years, 1 month ago

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