If the likelihood for a flower opening is the same every minute, and such that the average time interval between two successive openings is equal to one day, then we get the following statistical distribution for the time interval between two successive openings:
Probability that time interval is < 0.9 days/0.5 days/0.25 days/1 hour = 59.6/39.7/22.4/4.1 %
Probability that time interval is > 1.0 days/1.5 days/2.0 days/3.0 days = 36.6/22.3/ 13.5/5.0 %
This makes more sense than my previous post with the Rayleigh distribution.
Matlab/Octave code:
N=100000;% number of days to simulate
increment = 1/(24*60); % 1 min granularity
T=zeros(1,N); % init memory
for k=1:N,
T_inc = 0;
while rand(1,1)>increment,% An event occurs if random number in [0;1] is <= increment
T_inc = T_inc + increment;
end
T(k) = T_inc;
end
mu = mean(T),
value_below_090__050__025__004 = [100*length(T(T<0.9))/length(T), 100*length(T(T<0.5))/length(T), 100*length(T(T<0.25))/length(T), 100*length(T(T<1/24))/length(T)]
value_above_100__150__200__300 = [100*length(T(T>1.0))/length(T), 100*length(T(T>1.5))/length(T), 100*length(T(T>2.0))/length(T), 100*length(T(T>3.0))/length(T)]
figure
hist(T,50); title(['Histogram: Same probability for an event in each time instant']);
xlabel(['Time T between two successive events, mean=',num2str(mu)])
