sim_count = 1000
samp_size = 30
#Distribution of Means
set.seed(7)
hist(replicate(sim_count, mean(rexp(samp_size))),
main = "Distribution of Means of a Exponential Distibution")
#Distribution of Mins
set.seed(7)
hist(replicate(sim_count, min(rexp(samp_size))),
main = "Distribution of Minimums of a Exponential Distibution")
Within the function rexp, the rate, or lambda, defaults to 1. Lambda is the inverse of the mean as well as the standard deviation (mean = std = 1/lambda = 1). For each simulation of the the exponential distribution, the mean should tend towards 1 for large sample sizes. The distribution of the means, as shown above, does follow this tendency.
The distribution of the minimum is also following the behavior of exponential distribution, with labmda = 1. Since the standard deviation should tend towards 1 with a mean of 1 and the sample will only produce positive numbers, the minimum expected value should be 0, as is evident from the distribution of minimums above.