Mfcttrf

I am trying to generate a sample of 2000 rows. I have the following values
min = 80
max = 12000
mean = 500

I want to generate only positive numbers. I tried using triangular distribution and range rule(sd = max-min/4). The values were negative.

Is there anyway I can generate only positive numbers?

Thanks,
Anusha

While the problem is very much ill-posed, since there is an infinite range of distributions satisfying these constraints, a possible solution is to find the maximum entropy distribution under the constraint of a support of $(80,12000)$ [thus using the uniform measure on that interval as the reference measure] and a mean of $mathbb E[X]=500$ is of the form
$$p(x)=expalpha+beta x,mathbb I_(80,1200)(x)$$
with
$$int_80^12000 expalpha+beta x,text dx=1qquadtextandqquad
int_80^12000 xexpalpha+beta x,text dx=500$$
which leads to
$$exp-alpha=beta^-1[exp12000beta-exp80beta]$$
and$$beta^-1expalpha[12000exp12000beta-80exp80beta]-beta^-1=500$$which can be solved numerically in $beta$. Leading to
$$beta^*=-.00238quadtextandquadalpha^*=-5.850$$which can be easily simulated as a truncated exponential distribution, by inversion of the cdf, e.g., using qexp() in R. For instance,

function(n=1)
 return(qexp(pexp(80,.00238)+runif(n)*
 (pexp(12000,.00238)-pexp(80,.00238)),.00238))

If the question is instead about simulating a sample $X_1:2000$ such that $$min(X_1:2000)=80,quadmax(X_1:2000)=1200,quadbar X_1:2000=500$$
there is again an infinite range of solutions, the simplest being a multinomial distribution constrained by its minimum being 80 and its maximum being 12000 since
$$underbraceX_(1)_80+cdots+underbraceX_(2000)_12000 = 80 + 1998times 987920 + 12000= 2000times 500$$

If you don't care about the distribution aside from min, max, and mean, then there is a simple answer.

Take 96.476510067114100 percent of draws as 80 and 3.523489932885910 percent of draws as 12000. On average, you get 500, and you have your min and max. I calculated the percentages by solving a system of equations

$$a + b =1$$ $$80a + 12000b = 500$$

The first equation establishes the the values must sum to one, making sure that we are dealing with probabilities. The second equation get us our average of 500.

D <- rep(NA,2000) # define a vector of NAs to hold your sampled values
for (i in 1:2000)
 X <- rbinom(1,1,0.96476510067114100) # determine which value you'll take, 80 or 12000
 if (X==0)D[i] <- 12000 # declare observation i as 12000
 if (X==1)D[i] <- 80 # declare observation i as 80