Data 605 Homework 7

library(tidyverse)
library(openintro)

Exercise 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.

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.

The probability that \(Y\) is greater than or equal to \(y\), denoted as \(P(Y \geq y)\), is the probability that all of the \(X_i\)’s are greater than or equal to \(y\). Since the random variables are independent and uniformly distributed, we can calculate this probability.

\[ P(Y \geq y) = P(X_1 \geq y) \cdot P(X_2 \geq y) \cdot \ldots \cdot P(X_n \geq y) \]

Given that each \(X_i\) is uniformly distributed on the integers from 1 to \(k\), the probability \(P(X_i \geq y)\) is equal to \(\frac{{k - y + 1}}{k}\), as there are \(k - y + 1\) integers greater than or equal to \(y\) in the range [1, \(k\)].

Now, we can express the distribution function of \(Y\):

\[ P(Y \geq y) = \left( \frac{{k - y + 1}}{k} \right)^n \]

The probability mass function (PMF) of \(Y\), denoted as \(P(Y = y)\), is then given by:

\[ P(Y = y) = P(Y \geq y) - P(Y \geq y + 1) \]

\[ P(Y = y) = \left( \frac{{k - y + 1}}{k} \right)^n - \left( \frac{{k - y}}{k} \right)^n \]

This formula provides the distribution of \(Y\) for each value of \(y\) in the range [1, \(k\)].

Exercise 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.).

Let X be the number of failures in a 10-year period. If failures occur with an average rate of one failure every ten years, we can model X with a Poisson distribution.

X∼Poisson(λ), where λ is the average rate of failures per 10 years.

The probability of having no failures in a 10-year period is given by the Poisson probability mass function.

P(X=0)=e −λ, where e is the base of the natural logarithm.

#Average rate of failures per 10 years
lambda <- 1

#Probability of no failures
prob_no_failures <- exp(-lambda)
prob_no_failures

## [1] 0.3678794

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..)

Probability Statements:

1. Geometric Distribution: Let \(X\) be the number of trials until the first success (failure in this case) in a sequence of independent Bernoulli trials, where the probability of success (failure) on each trial is \(p\). The probability mass function for a geometric distribution is given by: \[ P(X = k) = (1 - p)^{k-1} \cdot p \]

2. Probability of Failing after 8 Years: The probability of the machine failing after 8 years is equivalent to the probability of not failing during the first 8 years. Thus, \[ P(X > 8) = 1 - P(X \leq 8) = 1 - \sum_{k=1}^{8} P(X = k) \]

R Code for Calculations:

Assuming \(p = \frac{1}{10}\) (one failure every ten years), the R code would be:

#Probability of failure in a given year
p_failure <- 1 / 10

#Probability of not failing during the first 8 years
prob_not_failing_8_years <- sum((1 - p_failure)^(1:8 - 1) * p_failure)

#Probability of failing after 8 years
prob_failing_after_8_years <- 1 - prob_not_failing_8_years
prob_failing_after_8_years

## [1] 0.4304672

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.

Statements:

1. Exponential Distribution: Let \(X\) be the time until the first failure of the machine, modeled as an exponential distribution with a rate parameter \(\lambda\). The probability density function (PDF) is: \[ f(t) = \lambda e^{-\lambda t} \] The cumulative distribution function (CDF) is: \[ F(t) = 1 - e^{-\lambda t} \]

2. Probability of Failing after 8 Years: The probability that the machine fails after 8 years is \(1 - F(8)\).

R Code for Calculations:

Assuming \(\lambda = \frac{1}{10}\) (one failure every ten years), the R code would be:

#Rating parameter for the exponential distribution
lambda <- 1 / 10

#Probability of failing after 8 years
prob_failing_after_8_years <- 1 - exp(-lambda * 8)
prob_failing_after_8_years

## [1] 0.550671

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)

Statements:

1. Binomial Distribution: Let \(X\) be the number of failures in a fixed number of trials (8 years), where each year is a Bernoulli trial with a probability of failure (\(p\)). The probability mass function (PMF) is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(n\) is the number of trials and \(k\) is the number of successes.

2. Probability of Failing after 8 Years: The probability of the machine failing after 8 years is \(P(X = 0)\).

R Code for Calculations:

Assuming \(p = \frac{1}{10}\) (one failure every ten years), the R code would be:

#Probability of failure in a given year
p_failure <- 1 / 10

#Number of trials (years)
n_trials <- 8

#Probability of failing after 8 years
prob_failing_after_8_years <- dbinom(0, n_trials, p_failure)
prob_failing_after_8_years

## [1] 0.4304672

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.

#Average rate of failures in a given year
lambda <- 1 / 10

#Number of years
n_years <- 8

#Probability of failing after 8 years
prob_failing_after_8_years <- 1 - ppois(0, lambda * n_years)
prob_failing_after_8_years

## [1] 0.550671

#Expected value and standard deviation
expected_value <- lambda
standard_deviation <- sqrt(lambda)

expected_value

## [1] 0.1

standard_deviation

## [1] 0.3162278