hi
If we consider major earthquakes,7+,for prediction accuracy for the year 2010,it would be like this
1) There are 45 dates predicted for 6+ quakes.(some of them stated as 7+ also)
2)There were 22 quakes of 7+ in year 2010 (till 26th December 2010)
3)7+ quakes occured in window period of prediction are 12.
4) odds of getting 7+ quake in window period of + or - one day ( ie 3 day window ) is 0.18
Now ,if we consider 8+ quakes,there is only one 8+ quakes on 27th February (chile).The date predicted was 28th february 2010. so ,it is in window period.
what is odd of getting 8+ quake in 3 day window?
regards
Amit
3 comments:
Amit;
Based on one 8+ per year the odds are about .008.
However, you made 43 predictions, any one of which might have caught the quake so your chances of success are 0.344
Roger
Yes I think 1/122=0.8% for 8+
You made 45 predictitions so 45x0.8% = 36.8% chance of randomly predicting 8+
Amit,
To help you out on the numbers. Assuming binomial distribution, or Poisson but the difference is small between them, (and a problem here is that the events are sequential and not simultaneously occurring, so much better to use probabilistic approach, but that is another discussion) then for the 7+ case, where p=0.123 per year, 3 day slot,
then
P(1 success) = 20.3%
P(2 successes) = 27%
P(3 successes) = 22.7%
P(4 successes) = 13.58%
etc etc
So if we get 4 correct 7+ in a year this has prob of 13.58% according to a Binomial Distribution...Hope this helps.
Remi
Post a Comment