Question 1

Hi Dr. Larry, I really struggled with this quesiton and look forward to the walk through in class, see you then!

Solution: Since Y is the minimum of the \(Xi\) and each \(Xi\) is uniformly distributed from 1 to \(k\), then each possibility for \(Xi\) is represented by \(k^n\).

Question 2

Part A:

The question states that there is one expected failure one every 10 years, so the probability of a failure happening in 10 years is \(\dfrac{1}{10}\).

prob <- 1/10
n <-8
p_not <- 1-prob
fail_chance <- pgeom(n-1, prob, lower.tail = F)
cat("The probability of failing during the after 8 years:", fail_chance, "\n")
## The probability of failing during the after 8 years: 0.4304672

Expected Value:

EV <- p_not/prob
cat("The expected value is:", EV)
## The expected value is: 9

Standard Deviation:

sd_not <- sqrt((p_not)/(prob^2))
cat("The sd is:", sd_not)
## The sd is: 9.486833

Part B:

p <- 1/10
n <- 8
exp_prob <- pexp(n, p, lower.tail = F)
cat("The probability of failing during the after 8 years using exponential:", exp_prob, "\n")
## The probability of failing during the after 8 years using exponential: 0.449329

Expected Value:

EX_EV <- 1/p
cat("The expected value is:", EX_EV)
## The expected value is: 10

Standard Deviation:

EX_sd <- sqrt((1)/(p^2))
cat("The sd is:", EX_sd)
## The sd is: 10

Part C:

x <- 0
p <- 1/10
p_not <- 1-p
n <- 8
binom_prob <- dbinom(x, n, p)
cat("The probablitity using the binomial probability function:", binom_prob, "\n")
## The probablitity using the binomial probability function: 0.4304672
BIN_EV <- n*p
cat("The expected value is:", BIN_EV)
## The expected value is: 0.8

Standard Deviation:

BIN_sd <- sqrt(n*p*(p_not))
cat("The sd is:", BIN_sd)
## The sd is: 0.8485281

Part D:

p <- 8/10
x <- 0
ppo_prob <- ppois(x, lambda = p)
cat("The probablitity using the binomial probability function:", ppo_prob, "\n")
## The probablitity using the binomial probability function: 0.449329
PPO_EV <- p
cat("The expected value is:", PPO_EV)
## The expected value is: 0.8

Standard Deviation:

PPO_sd <- sqrt(p)
cat("The sd is:", PPO_sd)
## The sd is: 0.8944272