October 7th, 2019
Answer:
Let’s define Success as getting a minimum of Xi’s and failure as getting something other than a minimum.
The p of getting a mimimum is 1/k.
This is a geometric distribution: P(T = n) = q^(n−1)*p
Y = (1-1/k)^(n-1) * (1/k)
P(T = n) = q^(n−1)*p
#Let's calculate the probability of failing in the 1st 8 years
n <- 9
p <- 1/10
q <- 1-p
Pfail <- 1 - (1-p)^n
P <- 1 - Pfail
P
## [1] 0.3874205
# Method 2: Using pgeom
P <- pgeom(8, p, lower.tail = F)
P
## [1] 0.3874205
#Calculating Expected Value
EV <- 1/p
EV
## [1] 10
#Calculating standard deviation
Var <- (1-p)/(p^2)
sqrt(Var)
## [1] 9.486833
P(T<=x) = 1−e−λx.
p <- 1/10
P<-pexp(8, rate = p, lower.tail = FALSE)
P
## [1] 0.449329
EX=1/λ
EV<-1/p
EV
## [1] 10
Var(X)=1/(λ^2)
#standard deviation
Var <- 1/(p^2)
sqrt(Var)
## [1] 10
n<-8
p <- 1/10
P<-pbinom(0, n, p)
P
## [1] 0.4304672
#Expected value
EV<-n*p
EV
## [1] 0.8
#standard deviation
Var <- n * p * ( 1 - p )
sqrt(Var)
## [1] 0.8485281
λ = n*p/t
lam<-10
P<- ppois(8, 10)
P
## [1] 0.3328197
Expected value
#expected value of a Poisson distribution with parameter λ also has expectation equal to λ
EV<-lam
EV
## [1] 10
Variance
#standard deviation
#Given a Poisson distribution with parameter λ, we should guess that its variance is λ.
Var <- lam
sqrt(Var)
## [1] 3.162278