Question

Two mechanics are changing oil filters for the arriving customers.
The service time has an Exponential distribution with mean 12 minutes for the first mechanic,
and mean 3 minutes for the second mechanic. When you arrive to have your oil filter changed,
your probability of being served by the faster mechanic is 0.8.
Use simulation to generate 10000 service times and estimate the mean service time for you.

Using R code

Answer 1

Let the service time for first mechanic be $X_1$ and the service time for first mechanic be $X_2$.

Then

X1~ Exp( 1/12), X2~ Exp(A = 1/3)

The resulting service time is $X_2$ with probability 0.8 and $X_1$ with probability 0.2.

The R code for simulating 10,000 resultant service times and finding the mean is given below.

set.seed(7657)
N <- 10000
X <- array(dim=N)
for(i in 1:N)
{
u <- runif(1)
if(u<0.8)
{
X[i] <- rexp(1,rate=1/3)
}
else
{
X[i] <- rexp(1,rate=1/12)
}
}
mean(X)

The mean service time is

4816 minutes