If \({ K }^{ n }\) represents the total sum of the variables then \({ (k-1) }^{ n }\) can represent the varibles wheren none of Xi is equal to 1.
\[P(X=1)=\frac { { k }^{ n }-{ (k-1) }^{ n } }{ { k }^{ n } }\] \[P(X=2)=\frac { { (k-2+1) }^{ n }-{ (k-2) }^{ n } }{ { k }^{ n } }\] \[P(X=y)=\frac { { (k-y+1) }^{ n }-{ (k-y) }^{ n } }{ { k }^{ n } }\]
If the life expenticy for a failure is every 10 years, then we have an increasing presentage of 10% per year until we get to the 10th year.
p <- .10
initialsd <- round(sqrt((1-p)/p^2),4)
initialsd
## [1] 9.4868
q2a <- pgeom(8,p,lower.tail = FALSE)
q2a
## [1] 0.3874205
asd <- round(sqrt((1-p)/p^2),4)
asd
## [1] 9.4868
q2b <- pexp(8, p, lower.tail = FALSE)
q2b
## [1] 0.449329
bsd<- sqrt(p^-2)
bsd
## [1] 10
pbinom(0,8,p)
## [1] 0.4304672
e <- exp(1)
csd <- round(e^(-p*8),4)
csd
## [1] 0.4493
dpois(8,10)
## [1] 0.112599
round(sqrt(.8),4)
## [1] 0.8944