(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.

To get an idea of the distribution, the code below generates 1 million rows of 5 random variables from 0 to 1 (uniform). Then for each row, the minimum is selected. Then a histogram is generated. As you can see, the distribution looks like the exponential distribution.

vec <- c(runif(1, 0, 1), runif(1, 0, 1), runif(1, 0, 1), runif(1, 0, 1),runif(1, 0, 1))
datalist <- list(vec)
count <- 2 
while (count < 1000001)
{
    vec <- c(runif(1, 0, 1), runif(1, 0, 1), runif(1, 0, 1), runif(1, 0, 1),runif(1, 0, 1))
    datalist[[count]] <- vec
    count = count + 1
}
minvals = c()
for(i in 1:length(datalist))
{
  minvals[i] <-  datalist[[i]][which.min(datalist[[i]])]
}
hist(minvals)

(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.).

(a) 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).

Geometric

P = 1/10 = 0.10 –> probability of machine failing in one year

Expected Value = 1/p = 1/.10 = 10 –> expected number of years to see first failure

Standard Deviation = sqrt((1-p)/p^2) = 9.49

p <- .10
round(sqrt((1-p)/p^2),2)
## [1] 9.49

Probability that machine will fail after 8 years –> machine does not fail during the first 8 years.

1 - probability machine fail within the first 8 years = .43

p_fail_year_1 = p
p_fail_year_2 = (1-p)^1 * p
p_fail_year_3 = (1-p)^2 * p 
p_fail_year_4 = (1-p)^3 * p
p_fail_year_5 = (1-p)^4 * p
p_fail_year_6 = (1-p)^5 * p 
P_fail_year_7 = (1-p)^6 * p 
p_fail_year_8 = (1-p)^7 * p 

P_fail_after_8_years <- 
1 - (p_fail_year_1 + p_fail_year_2 + p_fail_year_3 + p_fail_year_4 + p_fail_year_5 + p_fail_year_6 + P_fail_year_7     + p_fail_year_8)

round(P_fail_after_8_years,2)
## [1] 0.43

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

Exponential

lambda = 1 failure every ten years = 1/10 = .10 failure every year

lambda <- .10

Expected value = 1/lambda = 1/.10 = 10

expected_value <- 1/lambda
expected_value
## [1] 10

Variance = (lambda)^-2

Standard deviation - sqrt(variance)

standard_deviation <- sqrt(lambda^-2)
standard_deviation
## [1] 10

P(X >(x=8)) = e^((-lambda)x) = e^((-.10)8) = .45

e <- exp(1)
round(e^(-.10*8),2)
## [1] 0.45

(c) 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).

p = .10 –> probability of machine failing for the year n is not given in this case

Expected value = np = n(.10) Standard deviation sqrt(np(1-p)) = sqrt(n.10.90)

Probability machine fails after 8 years –> same as machine does not fail in first 8 years

p <- .10
round((1-p)^8, 2)
## [1] 0.43

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

1 failure in 10 years = .10 failure in a year

8 year period –> 8 * .10 = .8 occurrence lambda = .8 occurrence in 8 year period

0 occurrence of the event within 8 years

P(X=0) = ((lambda^x) * (e^-lambda))/ x! = .45

lambda <- .8
e <- exp(1)
round((lambda^0 * e^(-1*lambda)) / factorial(0), 2)
## [1] 0.45

Expected value is lambda = .8

Standard deviation = sqrt(lambda) = sqrt(.8) = 0.8944272

sqrt(.8)
## [1] 0.8944272