Exact probability of a 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.

[Rik]

This sounds like homework. Is it?

[Burhan Elaldi]

yes

[Burhan Elaldi]

 I asked for a point where I was stuck in my homework.

Answer 1

[James Tursa]

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

[Burhan Elaldi]

Actually, I found the probability that triangle random variable is greater than its mean (54) , from 1000 random variables that I transferred from Matlab to Excel as you see on the above. This value which I found from excel is 0.511. Now I have to compare 0.511 with the exact one.

What is the exact one? Is 0.5191 which you find?

Why did you divide by 1000000? I have to perform simulation 1000 times and I think we should divide by 1000 not 1000000.

What is the mean of 0.5194? I did not get it.

Thank you for your helping :)

[James Tursa]

Why did you divide by 1000000?

This:

sum(random(pd,1000000,1)>mean(pd))/1000000

is a one-line simulation of a random sampling of 1000000 trials. I was just picking a big number for example.

In your case, yes you would use 1000 not 1000000.

What is the exact one?

Just as I have stated, this is the exact one:

1 - cdf(pd,mean(pd))

If you want to see more digits, then

>> format longg
>> 1 - cdf(pd,mean(pd))
ans =
          0.51909090909091

[Burhan Elaldi]

Thank you so much :)