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 .

Y is the minimum of the Xi’s. Each Xi had k possibilities (each Xi is uniformly distributed on the integers from 1 to k). Thus, all possible options for all Xi to have number from 1 to k is \[k^n\].

The minimum of X is 1, so if Y=1, the amount of options to get it is \[k^n-(k-1)^n\]. All possible options remove options when no Xi is 1.
If Y=2, the amount of options to get it is \[k^n-(k-2)^n-[k^n-(k-1)^n]\]. All the options remove options when no Xi is 2 and remove options when Y=1. simplified \[(k-1)^n-(k-2)^n\] If we continue until Y=j, the amount of ways to get minimum value j \[(k-j+1)^n-(k-j)^n\] The distribution of Y is \[1\le{j}\le{k},\frac{(k-j+1)^n-(k-j)^n}{k^n}\]. For example, we can illustrate the distribution for k=100, using R.

k = 100
sample <- runif(k, min = 1, max = k)
Y = min(sample)
hist(sample)

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..).

The probability that the machine will fail after 8 years is 43.05%. The expected value is 10, the standard deviation is 9.49.
The probability that it fails after 10 years is \[p=\frac{1}{10}\] For geometric distribution, after 8 years, \[p(8 years)=(1 - p)^{(x - 1)} * p\] Expected value, \[\frac{1}{p}=10\] Standard deviation, \[SD=\sqrt{\frac{1-p}{p}}=\sqrt{\frac{1-0.1}{0.1}}=9.49\] Using R:

p <- 1/10
n=8
round((1 - pgeom(n-1, p)),4)
## [1] 0.4305
expected <- 1/p
expected
## [1] 10
std <- round((sqrt((1-p)/p^2)),2)
std 
## [1] 9.49

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.

The probability that the machine will fail after 8 years is 44.93%. The expected value is 10, the standard deviation is 100.
The probability for an exponential model for 10 years is \[p=e^{-\lambda*n}=e^{-\frac{n}{10}}=e^{-\frac{10}{10}}=e^{-1}\] For 8 years, \[p(n=8)=e^{-\frac{n}{10}}=e^{-\frac{8}{10}}=0.4493\] Expected value is \[\frac{1}{\lambda}=10\] Standard deviation, \[SD=\frac{1}{\lambda^2}=100\] Using R:

x <- 8
lambda <- 1/10
round((1 - pexp(x, lambda)), 4)
## [1] 0.4493
 expected <- 1/lambda 
expected
## [1] 10
std <- sqrt(1/lambda^2)
round(std,3)
## [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).

The probability that the machine will fail after 8 years is 43%. The expected value is 0.8, the standard deviation is 0.849.
The probability for a binomial model with 0 success in 8 years is \[p=(^n_k)p^k*q^{n-k}=\frac{n!}{k!(n-k)!}p^k*q^{n-k}=\frac{8!}{0!(8-0)!}0.1^0*0.9^{8-0}=0.43\] Expected value, \[n*p=8*0.1=0.8\] Standard deviation, \[SD=\sqrt{npq}=\sqrt{8*0.1*0.9}=0.849\] Using R:

x <- 0
n <- 8
p <- 1/10
round((dbinom(x, n, p)), 2)
## [1] 0.43
expected <- n*p
expected
## [1] 0.8
std <- sqrt(n*p*(1-p))
round(std,3)
## [1] 0.849

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.

The probability that the machine will fail after 8 years is 44.9%. The expected value is 0.8, the standard deviation is 0.894.
The probability for a Poisson model is \[p=\frac{\lambda^k}{k!}e^{-\lambda}=\frac{0.8^0}{0!}e^{-0.8}=0.449\] Expected value, \[\lambda=\frac{n*p}{t}=\frac{8*0.1}{1}=0.8\] Standard deviation, \[SD=\sqrt{\lambda}=\sqrt{8*0.1*0.9}=0.894\] Using R:

x <- 0
n <- 8
p <- 1/10
t <- 1 
lambda <- n*p/t
round((dpois(x, lambda)), 3)
## [1] 0.449
expected <- lambda 
expected
## [1] 0.8
std <- sqrt(lambda)
round(std,3)
## [1] 0.894