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.
Solution:
Given, \(X1, X2, . . . , Xn\), n mutually independent random variables uniformly distributed on the integers from 1 to k.
Total number of possibilities = \(k^{n}\)
Possibilities where none of the Xi’s are equal to 1 is \((k-1)^{n}\)
For X=1
\(P(X=1)=\frac{k^{n}-(k-1)^{n}}{k^{n}}\)
Similarly for X=2
\(P(X=2)=\frac{(k-2+1)^{n}-(k-2)^{n}}{k^{n}}\)
X=3
\(P(X=3)=\frac{(k-3+1)^{n}-(k-3)^{n}}{k^{n}}\)
We can generalize this as below
\(P(X=y)=\frac{(k-y+1)^{n}-(k-y)^{n}}{k^{n}}\)
2. 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. (Include the probability statements and R Code for each 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..)
Solution:
Probability that the machine will fail after 8 years
n<-8
prob_failure <- 1/10
prob_nonfailure <- 1-prob_failure
prob_geomet <-1-pgeom(n-1,prob_failure)
prob_geomet
## [1] 0.4304672
Expected value
e_val <- 1/prob_failure
e_val
## [1] 10
Standard deviation
sd<-sqrt(prob_nonfailure/(prob_failure^2))
sd
## [1] 9.486833
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.
Solution:
Probability that the machine will fail after 8 years
n <- 8
lambda <- 1/10
p_expo <- pexp(n, lambda, lower.tail=FALSE)
p_expo
## [1] 0.449329
Expected value
e_val <- 1/lambda
e_val
## [1] 10
Standard deviation
sd <- sqrt(1/lambda^2)
sd
## [1] 10
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)
Solution
Probability that the machine will fail after 8 years
n <- 8
p <- 1/10
q <- (1-p)
k <- 0
p_binomial <- dbinom(k, n, p)
p_binomial
## [1] 0.4304672
Expected value
e_val <- n * p
e_val
## [1] 0.8
Standard deviation
sd <- sqrt(n*p*q)
sd
## [1] 0.8485281
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.
Solution
Probability that the machine will fail after 8 years
lambda <- 8/10
k <- 0
ppois(0,lambda = .8 )
## [1] 0.449329
Expected value
e_val <- 8/10
e_val
## [1] 0.8
Standard deviation
sd <- sqrt(8/10)
sd
## [1] 0.8944272