Question 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.

Question 2

Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturers expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.)

Part A

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)

# 10 year of the copier
prob_failure <- 1/10
no_issue_prob <- 1 - prob_failure
num_year <- 8

# first 8 years
no_issue_prob_8_geom <- 1 - pgeom(num_year-1, prob_failure)

no_issue_prob_8_geom
## [1] 0.4304672

Part B

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

# expected value
expected_failure <- 1/prob_failure
expected_failure
## [1] 10
# standard deviation
stand_dev <- sqrt((1 - prob_failure)/(prob_failure^2))

stand_dev
## [1] 9.486833
# exponential model
no_issue_prob_8_exp <- pexp(num_year, prob_failure, lower.tail = FALSE)

no_issue_prob_8_exp
## [1] 0.449329

Part C

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)

# expected value
expected_failure2 <- num_year * prob_failure

expected_failure2
## [1] 0.8
# standard deviation
stand_dev2 <- sqrt(num_year * prob_failure * no_issue_prob)

stand_dev2
## [1] 0.8485281
success_rate <- 0

no_issue_prob_8_bin <- dbinom(success_rate, num_year, prob_failure)

no_issue_prob_8_bin
## [1] 0.4304672

Part D

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

# expected value
expected_failure3 <- num_year * prob_failure

expected_failure3
## [1] 0.8
# standard deviation
lambda <- num_year/10
stand_dev3 <- sqrt(lambda)

stand_dev3
## [1] 0.8944272
no_issue_prob_8_pois <- ppois(success_rate, lambda)

no_issue_prob_8_pois
## [1] 0.449329