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 .
We have X1, X2, … to Xn, then total number of possibilities = kn The possibilities none of Xi equals to 1 is (k-1)^n
When X = 1 \[P(X=1) = (k^n-(k-1)^n)/k^n\] When X = 2 \[P(X=2) = ((k-2+1)^n)-(k-2)^n)/k^n\] When X = 3 \[P(X=2) = ((k-3+1)^n)-(k-3)^n)/k^n\] Therefore, we can get the Y distribution as \[P(X=y) = ((k-y+1)^n)-(k-n)^n)/k^n\]
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.).
Probability machine fails after 8 years.
prob_f<- 1/10
y<-8
prob_nf<- 1-prob_f
prob_f8y_geom<- 1-pgeom(y-1, prob_f)
prob_f8y_geom
## [1] 0.4304672
Expected value
ev<-1/prob_f
ev
## [1] 10
Standard deviation
sd<-sqrt(prob_nf/prob_f^2)
sd
## [1] 9.486833
Probability machine fails after 8 years.
y<- 8
prob_f<- 1/10
prob_f8y_exp<- pexp(y, prob_f, lower.tail = F)
prob_f8y_exp
## [1] 0.449329
Expected value
ev<-1/prob_f
ev
## [1] 10
Standard deviation
sd<- sqrt(1/prob_f^2)
sd
## [1] 10
y<- 8
prob_f<- 1/10
prob_nf<- 1-prob_f
k<- 0
prob_f8y_bino<- dbinom(k, y, prob_f)
prob_f8y_bino
## [1] 0.4304672
Expected value
ev<- y * prob_f
ev
## [1] 0.8
Standard deviation
sd<-sqrt(y * prob_f * prob_nf)
sd
## [1] 0.8485281
prob_f<- 1/10
y<- 8
k<- 0
lambda<- prob_f * y
lambda
## [1] 0.8
ppois(0, lambda )
## [1] 0.449329
Expected value
ev<-y * prob_f
ev
## [1] 0.8
Standard deviation
sd<- sqrt(ev)
sd
## [1] 0.8944272