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 .

Answer: Y denotes the minimum of \(X_{i}\)s. In order to count the distribution of m(j) = P(X = j) we need to find two things.

We need to find two things. 1. Count the number of ways that we can assign \(X_{1}\),\(X_{2}\),…,\(X_{n}\) to values between j and k at least one \(X_{i}\) being assigned to j Number of ways of getting = 1 is \(k^{n} - (k-1)^{n}\), since \(k^{n}\) represents the total number of options and \((k-1)^n\) represents all of the options where non of the \(X_{i}\)’s are equal to 1. For Y = 2, \(k^{n} - (k-1)^{n} -[k^n-(k-1)^n]\)

Generalizing for Y = j, \((k- j + 1)^n - (k-j)^n\) ways to assign \(X_{1}\),\(X_{2}\),…,\(X_{n}\) so that the minimum value is j. 2. Total number of possible ways to assign \(X_{1}\),\(X_{2}\),…,\(X_{n}\) to values between 1 and k.

\(X_{i}\) has k possibilities: 1,2,..,k. Total possible number of assignment is \(k^{n}\).

Divide 1 by 2 to we find the distribution

\(\frac{(k-j+1)^n-(k-j)^n-(k-j)^n}{k^n}\)

  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. (Include the probability statements and R Code for each part.).
  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. (Hint: the probability is equivalent to not failing during the first 8 years..)

Answer:

p = 1/10
q = 1-p
n <- 8
pgeom(8,p,lower.tail = F)
## [1] 0.3874205

The probability that the machine will fail after 8 years is 0.3874205.

1/p
## [1] 10
sqrt(q/p^2)
## [1] 9.486833

Expected value 10 and standard deviation 9.486833

  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.

Answer:

lambda <- 1/10
k = 8
exp(-lambda*k)
## [1] 0.449329

The probability following an exponential distribution is 0.449329.

1/lambda
## [1] 10
sqrt(1/lambda^2)
## [1] 10

The mean and standard deviation is 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)

Answer:

n <- 8
p <- 1/10
q <- 1-p
k <- 0

dbinom(k, n, p)
## [1] 0.4304672

The probability following a binomial distribution is 0.4304672.

n*p
## [1] 0.8
sqrt(n*p*q)
## [1] 0.8485281

Following binomial distribution mean is 0.8 and standard deviation is 0.8485281.

  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.

Answer:

lambda <- 8/10
ppois(0, lambda = 0.8)
## [1] 0.449329

The probability following a poison distribution is 0.449329.

sqrt(0.8)
## [1] 0.8944272

Following Poisson distribution mean is 0.8 and standard deviation is 0.8944272.