MATLAB: Exact probability of a triangular distribution

triangular distribution

How can I find exact probability of triangular distribution? Actually my exact question is: use simulation to estimate the probability that a triangular random variable is greater than its mean. You can take the parameters of the triangular random variable as follows: the minimum value as the sum of last digits of IDs of your group members, the mode value as the sum of the last two digits of of IDs of your group members and the maximum value as the sum of the last three digits of of IDs of your group members. Please, perform the simulation 1000 times and compare the estimated probability with the exact one. But I could not find the exact probabilty?
parameters = [31 56 75];
lower = min(parameters);
peak = median(parameters);
upper = max(parameters);
pd = makedist('Triangular','a',lower,'b',peak,'c',upper)
x = 0:.1:230;
y = pdf(pd,x);
plot(x,y)
title('Triangular Distribution')
xlim([0 100])
rng('default');  % For reproducibility

r = random(pd,1000,1)
sort_time = sort(time,'descend');
secondLargest = sort_time(2);
figure
pd2 = makedist('Triangular','a',lower,'b',peak,'c',secondLargest);
y2 = pdf(pd2,x);
plot(x,y2,'LineWidth',2)
title('Triangular Distribution') 
xlim([0 100])
rng('default'); % For reproducibility
r2 = random(pd2,1000,1)
I transferred the random numbers I produced in Matlab to Excel.

Best Answer

The exact probability of getting a number greater than the mean is simply the sum of the probabily to the right of the mean. Since the cumulative distribution function gives the sum of the probablity to the left, you simply need to take 1-cdf evaluated at your point of interest to get your exact probability. E.g.,
>> exact_probability = 1 - cdf(pd,mean(pd))
exact_probability =
    0.5191
And this seems to match a large simulation result pretty well
>> sum(random(pd,1000000,1)>mean(pd))/1000000
ans =
    0.5194
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