1. Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y .

ANS: \(Y\) is the minimum value of \(Xi\)

Each \(Xi\) has \(k\) possibilities. The total possible number of assignments for the entrie series of random variables \(X1, X2 .. Xn\). If \(Y = 1\) then \(k^n - (k -1)^n\)

Since a “uniform distribution” is a distribution that has constant probability, then one value has an equal probability of being selected. Given that there are a total of \(Xn’s\), the probability of each X would be \(1/n\).

So, the distribution for \(Y\) is:

\(P(k) = 1/(k-1)\) if $1 x k $

  1. Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years.
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric.
x <- 10 #years
p <- 1/x #the machine will fail
q <- 1- p #the machine will not fail

geo <- pgeom(8, p, lower.tail = FALSE)
geo
## [1] 0.3874205
stdev <- round(sqrt((1-p)/(p^2)),2)
stdev
## [1] 9.49
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
x <- 8 #years
p <- 1/10
q <- 1 - p

ev <- (pexp(x, p, lower.tail = F))
ev
## [1] 0.449329
ans2 <- 1 - ev
ans2
## [1] 0.550671
ans3 <- 1/(p)
ans3
## [1] 10
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)
x <- 8 #trails
p <- 1/10
q <- 1 - p
pbio <- pbinom(0, x, p)
pbio
## [1] 0.4304672
ans2 <- 1 - pbio
ans2
## [1] 0.5695328
ans3 <- round(x * p)
ans3
## [1] 1
stdev <- sqrt(ans3 * q)
stdev
## [1] 0.9486833
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
x <- 8 #years
p <- 1/10
q <- 1 - p
lambda <- x * p

pois <- ppois(0, lambda)
pois
## [1] 0.449329
ans2 <- 1 - pois
ans2
## [1] 0.550671
stdev <- sqrt(10)
stdev
## [1] 3.162278