Ans.
\[ 1 ≤ j ≤ k, m(j) = \tfrac{(k-j+1)^n - (k-j)^n}{k^n} \]
Ans.
The probability that the machine will fail after 8 years
## Probability machine will fail after 8 years.
P_of_failure <- 1/10
P_of_no_failure <- 1 - P_of_failure
n <- 8
P_10 <- 1-pgeom(n,P_of_failure)
print(P_10)
## [1] 0.3874205
The expected value
## Expected Value
expected_value_A <- 1/ P_of_failure
print(expected_value_A)
## [1] 10
The Standard Deviation.
## Standard Deviation
SD_A<- sqrt(P_of_no_failure/(P_of_failure^2))
print(SD_A)
## [1] 9.486833
Probability the machine will fail after 8 years.
## Probability will fail after 8 years
n <- 8
Pro8 <- 1/10
Pro_exponent <- pexp(n, Pro8, lower.tail = FALSE)
print(Pro_exponent)
## [1] 0.449329
The Expected value
## Expected Value
expected_value_B <- 1/Pro8
print(expected_value_B)
## [1] 10
Standard Deviation
## Standard Deviation
SD_B <- sqrt(1/(Pro8^2))
print(SD_B)
## [1] 10
The probability that the machine will fail after 8 years.
## Probability the machine will fail after 8 years
n <- 8
P_of_failure <- 1/10
S_8years <- 0
P_of_no_failure <- (1 - P_of_failure)
Binom_Prob <- dbinom(S_8years,n, P_of_failure)
print(Binom_Prob)
## [1] 0.4304672
The Expected Value
## The expected value
n <- 8
P_of_failure <- 1/10
expected_value_C <- (n*P_of_failure)
print(expected_value_C)
## [1] 0.8
The Standard Deviation
## Standard Deviation
n <- 8
P_of_failure <- 1/10
P_of_no_failure <- (1 - P_of_failure)
SD_C <- sqrt(n*P_of_failure*P_of_no_failure)
print(SD_C)
## [1] 0.8485281
The probability that the machine will fail after 8 years
n <- 8
P_of_failure <- 1/10
S_8years <- 0
lambda <- n*P_of_failure
pois_prob <- ppois(0,lambda,lower.tail = TRUE, log.p = FALSE)
print(pois_prob)
## [1] 0.449329
The Expected value
## expected value
expected_value_D <- (lambda)
print(expected_value_D)
## [1] 0.8
\[EV = \lambda \]
The Standard Deviation
## Standard Deviation
SD_D <- sqrt(lambda)
print(SD_D)
## [1] 0.8944272
\[ SD = \sqrt{{\lambda}} \]