N Selina Assignment 7 - Data 605

Noori Selina

1. Let \(X_1, X_2, \ldots, X_n\) 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 \(X_i\)’s. Find the distribution of \(Y\).

T Let \(X_1, X_2, \ldots, X_n\) 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 \(X_i\)’s. Find the distribution of \(Y\).

We want to find the probability distribution function of \(Y\), denoted as \(P(Y = y)\) for \(y = 1, 2, \ldots, k\).

For \(Y = 1\), this means that the minimum value among \(X_1, X_2, \ldots, X_n\) is 1. The probability \(P(Y = 1)\) is the ratio of the number of outcomes where the minimum is 1 to the total number of possible outcomes: \[ P(Y = 1) = \frac{k^n - (k-1)^n}{k^n} \]

For \(Y = m\) where \(2 \leq m \leq k\), the probability \(P(Y = m)\) is the ratio of the number of outcomes where the minimum is \(m\) to the total number of possible outcomes: \[ P(Y = m) = \frac{(k-m+1)^n - (k-m)^n}{k^n} \]

These expressions give the probability distribution function of \(Y\), the minimum of the \(X_i\)’s.

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.).

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..)

P(X = n) = (1 - p)^{n-1} p

prob_fail_after_8_years <- (.9)^7 * .1
cat("Probability of machine failure after 8 years:", prob_fail_after_8_years, "\n")

## Probability of machine failure after 8 years: 0.04782969

geo_EV <- 1 / (.1)
cat("Expected value for geometric distribution:", geo_EV, "\n")

## Expected value for geometric distribution: 10

sd <- sqrt((.9) / (.1^2))
cat("sd =", sd, "\n")

## sd = 9.486833

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

prob_greater_8_years <- exp(-8/10)
cat("Probability of machine failure after 8 years:", prob_greater_8_years, "\n")

## Probability of machine failure after 8 years: 0.449329

exp_EV <- 1 / (.1)
cat("Expected value for exponential distribution:", exp_EV, "\n")

## Expected value for exponential distribution: 10

sd <- sqrt((1) / (.1^2))
cat("Standard deviation for exponential distribution:", sd, "\n")

## Standard deviation for exponential distribution: 10

3. 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)

biprob_greater_8_years <- (.1)^(0) * (.9)^(8)
cat("Probability of machine failure after 8 years:", biprob_greater_8_years, "\n")

## Probability of machine failure after 8 years: 0.4304672

binomial_EV <- 8 * (.1)
cat("Expected value for binomial distribution:", binomial_EV, "failures\n")

## Expected value for binomial distribution: 0.8 failures

sd <- sqrt(8 * (.1) * (.9))
cat("Standard deviation for binomial distribution:", sd, "\n")

## Standard deviation for binomial distribution: 0.8485281

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

poisson_prob <- .8^(8) * exp(-.8/8)
cat("Probability of machine failure after 8 years:", poisson_prob, "\n")

## Probability of machine failure after 8 years: 0.1518065

Expected Value

E[X] = = 0.8

sd <- sqrt(.8)
cat("Standard deviation for Poisson distribution:", sd, "\n")

## Standard deviation for Poisson distribution: 0.8944272