Let X represent the number of years until the machine fails.
For geometric distribution:
\(f(x) = P(X=x) = (1 - p)^{x-1}p\)
In this case: \(P(X=8) = (\frac{9}{10})^7 * \frac{1}{10}\)
Ans <- dgeom(8,0.1)
Ans
## [1] 0.04304672
Expected value = \(\frac{1}{p}\)
Ans = 1/(1/10)
Ans
## [1] 10
standard deviation = \(\sqrt(1-p)/p^2\)
Ans = sqrt((1 - (1/10))/((1/10)^2))
Ans
## [1] 9.486833
For exponential distribution
Rate = 1/10
Failure after 8 years: 1 - P(X <= 8)
Ans <- 1 - pexp(8,0.1)
Ans
## [1] 0.449329
Expected Value - 1/0.1
Ans = 1/0.1
Ans
## [1] 10
Standard Deviation = 1/(0.1)^2
Ans = 1/((0.1)^2)
Ans
## [1] 100
For Binomial Distribution
X ~ B(8, 0.1) P(X = 0)
Ans = dbinom(0, size=8, prob=0.1)
Ans
## [1] 0.4304672
Expected Value = n * p
Ans = 8 * 0.1
Ans
## [1] 0.8
Standard Deviation = \(\sqrt(np * (1-p)\)
Ans = sqrt((8 * 0.1) * (1 - 0.1))
Ans
## [1] 0.8485281
X ~ P(0.8)
P(X=0)
Ans = ppois(0,0.8)
Ans
## [1] 0.449329
Expected Value = Variance = 0.8 Standard Deviation:
Ans = sqrt(0.8)
Ans
## [1] 0.8944272