Problem 1

The CDF of y = the minimum of n Uniform(a,b) continuous random variables is:

\[ \begin{equation} F(y) = 1 - \Big(\dfrac{b-y}{b-a}\Big)^{n} \end{equation} \] We are interested in the discrete case of y = the minimum of n Uniform(1, k) discrete random variables. The CDF is:

\[ \begin{equation} F(y) = 1 - \Big(1 - \dfrac{y}{k}\Big)^{n} \end{equation} \]

By recognizing that the above formula is an accumulation function, we can use it to get the Probability Mass Function of the discrete case:

\[ \begin{equation} f(y) = \Big(1 - \dfrac{y-1}{k}\Big)^{n} - \Big(1 - \dfrac{y}{k}\Big)^{n} \end{equation} \]

Problem 2

“Your organization owns a machine that has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every 10 years. What is the probability that the machine will fail after 8 years?”


  1. Geometric

\[ \begin{equation} P(x>8) = \Big(1 - \dfrac{1}{10}\Big)^{8} \end{equation} \]

p <- 1/10
1-sum(dgeom(0:7, p))   #the R function starts at 0, so must do 0:7 instead of 1:8
## [1] 0.4304672
1/p   #mean
## [1] 10
sqrt((1-p)/p^2)   #sd
## [1] 9.486833


  1. Exponential

\[ \begin{equation} P(x>=8) = e^{-8/10} \end{equation} \]

lambda <- 1/10
pexp(8, lambda, lower.tail = F)
## [1] 0.449329
1/lambda   #mean
## [1] 10
1/lambda   #sd
## [1] 10


  1. Binomial

\[ \begin{equation} P(x=0) = \Big(\dfrac{9}{10}\Big)^{8} \end{equation} \]

p <- 1/10
n <- 8
dbinom(0, n, p)
## [1] 0.4304672
n*p   #mean
## [1] 0.8
sqrt(p*(1-p)*n)   #sd
## [1] 0.8485281


  1. Poisson

\[ \begin{equation} PMF = \dfrac{\lambda^{k}e^{-\lambda}}{k!} \end{equation} \]

p <- 1/10
n <- 8
lambda <- n*p
dpois(0, lambda)
## [1] 0.449329
lambda   #mean
## [1] 0.8
sqrt(lambda)   #sd
## [1] 0.8944272