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.
Since the integers range from 1 to k, each of the n mutually
independent random variables has k possible values.
So, for the whole set of n random variables, the possible value
assignments is \(k^n\).
If we take out one assignment out for Y, i.e. Y=1, the rest of possible
assignments would look like:
\(k^n - (k-1)^n / k^n\)
In this case the integers from 1 to k so 1 should be the minimum of the
Xi’s otherwise we can sub in 1 for the new minimum.
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.).
For geometric, in R we use geom() calls.
pgeom(x,prob)
it would be easier to do 1 - prob of it failing in 8 or less years x = 8
years
prob = 1/10 or 1 failure in 10 years
prob = 1/10
failure_geometric <- 1 - pgeom(8, prob)
cat("The probability that your copier will fail after 8 years: ", failure_geometric*100, "%")
## The probability that your copier will fail after 8 years: 38.74205 %
expval_geometric <- 1/prob
cat("The expected value (1/prob): ", expval_geometric)
## The expected value (1/prob): 10
sd_geometric <- sqrt(1-prob)/prob
cat("The standard deviation (sqrt(1-prob)/prob): ", sd_geometric)
## The standard deviation (sqrt(1-prob)/prob): 9.486833
For exponential, in R we use exp() calls.
pexp(x,prob)
x = 8 years
prob = 1/10 or 1 failure in 10 years
prob = 1/10
failure_exponential <- pexp(8, prob)
cat("The probability that your copier will fail after 8 years: ", failure_exponential*100, "%")
## The probability that your copier will fail after 8 years: 55.0671 %
expval_exponential <- 1/prob
cat("The expected value (1/prob): ", expval_exponential)
## The expected value (1/prob): 10
sd_exponential <- 1/(prob^2)
cat("The standard deviation (1/prob^2): ", sd_exponential)
## The standard deviation (1/prob^2): 100
For exponential, in R we use binom() calls.
pbinom(y,x,prob) y = is the successes so 0 from the hint x = 8
years
prob = 1/10 or 1 failure in 10 years
prob = 1/10
failure_binomial <- pbinom(0, 8, prob)
cat("The probability that your copier will fail after 8 years: ", failure_binomial*100, "%")
## The probability that your copier will fail after 8 years: 43.04672 %
expval_binomial <- 8*prob
cat("The expected value (x*prob): ", expval_binomial)
## The expected value (x*prob): 0.8
sd_binomial <- sqrt(8*prob*(1-prob))
cat("The standard deviation (sqrt(x*prob*(1-prob))): ", sd_binomial)
## The standard deviation (sqrt(x*prob*(1-prob))): 0.8485281
For poisson, in R we use pois() calls.
ppois(y, lamba) y = is the successes so 0 from the hint lamba = 1/10 or
1 failure in 10 years * 8 years
lamba = 1/10*8
failure_poisson <- ppois(0, lamba)
cat("The probability that your copier will fail after 8 years: ", failure_poisson*100, "%")
## The probability that your copier will fail after 8 years: 44.9329 %
expval_poisson <- lamba
cat("The expected value (lamba): ", expval_poisson)
## The expected value (lamba): 0.8
sd_poisson <- sqrt(lamba)
cat("The standard deviation (sqrt(lamba)): ", sd_poisson)
## The standard deviation (sqrt(lamba)): 0.8944272