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 .
Per the problem, each Xi can take on values between 1 to k, let’s assume that these random variable values are independent. Hence the population (denominator) k ^ n Xi’s.
Per the problem, Y is the minimum of Xi’s, so let’s assume Y = 1. If k^n represents all options and (k-1)^n is all of the options where none of the Xi’s are 1. Then [k^n - (k - 1)^n] is the collection of all options where Y = 1.
Similarly for Y = 2, we get k^n - (k - 2)^n - [k^n - (k - 1)^n]. If we simplify this expression, we get below:
Extrapolating this to higher-order effects, we get ((k-j+1)^n - (k-j)^n) / k^n which is the distribution function.
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.).
p <- 1/10 #probability of the event
q <- (1 - p)
k <- 8 #no failures in first 8 trials
calc1 <- dgeom(k, p)
sprintf("The probability that the machine will fail after 8 years is: %f", calc1)## [1] "The probability that the machine will fail after 8 years is: 0.043047"sprintf("The expected value is: %d", 1/p)## [1] "The expected value is: 10"sprintf("The standard deviation is: %f", (sqrt((1 - p) / p^2)))## [1] "The standard deviation is: 9.486833"lambda <- 1/10
calc2 <- dexp(k, p)
sprintf("The probability that the machine will fail after 8 years is: %f", calc2)## [1] "The probability that the machine will fail after 8 years is: 0.044933"sprintf("The expected value is: %d", 1/lambda)## [1] "The expected value is: 10"sprintf("The standard deviation is: %d", (sqrt(1 / lambda ^ 2)))## [1] "The standard deviation is: 10"n <- 8 #Number of trials
p <- 1/10 #Probability of success
q <- (1 - p) #probability of failure
k <- 0
calc3 <- dbinom(k, n, p)
sprintf("The probability that the machine will fail after 8 years is: %f", calc3)## [1] "The probability that the machine will fail after 8 years is: 0.430467"sprintf("The expected value is: %f", (n * p)) ## [1] "The expected value is: 0.800000"sprintf("The standard deviation is: %f", (sqrt(n * p * q)))## [1] "The standard deviation is: 0.848528"## [1] "The probability that the machine will fail after 8 years is: 0.449329"## [1] "The expected value is: 0.800000"## [1] "The standard deviation is: 0.894427"