I didn’t figure this one out.
The probability the machine will fail any given year is P(X=1) = \(\frac{1}{10}\)
P(X = n) = \((1-p)^n(p)\)
n = 8 as we have 8 failures before one success. The probability the machine will fail any given year is p = \(\frac{1}{10}\)
P(X = 8) = \((1-\frac{1}{10})^{8}(\frac{1}{10})\) = dgeom(8,.1) = .4304
Expected value = E(X) = \(\frac{1}{p}\) = \(\frac{1}{\frac{1}{10}}\) = 10
Standard Deviation = \(\sqrt{\frac{1-p}{p^2}}\) = \(\sqrt{\frac{1-\frac{1}{10}}{\frac{1}{10}^2}}\) = 9.487
P(X \(\ge\) k) = \(e^{\frac{-k}{\mu}}\) where \(\mu\) = $
The probability that the machine will fail as a exponential distribution is given by the mean which is 10 and would like to know when k = 8 and \(\lambda\) = 10.
P(X \(\ge\) k) = \(e^{\frac{-8}{10}}\) = .449 * .1 = .0449
Standard Deviation = npq = (8)(.1)(.9)= .72
The probability formula for a binomial distribution is:
P(X=x) = \(\binom{n}{k}p^k(q^{n-k})\)
Expected Value = (.1)(8) = .8
Standard Deviation = npq = (8)(.1)(.9)= .72
To find the probability that the machine will fail after 8 years is 8 failures followed by one success. In this case the probability we will have 0 successes in 8 years and the probability of a success being .1.
P(X=0) = \(\binom{8}{0}p^k(q^{n-k})\) = \(1\cdot(.1^0)(.9^8)\) = .4304
Provided that the trials are independent, if we have a failure on the 9th year we multiply that probability with .1 giving us .04304
Poisson distribution for X occurences in a fixed amount of time is P(X = x) = \(\frac{\lambda^{x}e^{-\lambda}}{x!}\)
In this case, we will look at the probability of 0 occurences in 8 years where lambda is .8 multiplied by the probability it will happen in one year.
P(X=0) = \(\frac{.1^{0}e^{-.8}}{0!}\) = .4493 probability it will happen any given year = .1
.4493\(\cdot\).1= .0493
Expected Value = .1
SD = \(\sqrt{.1}\) = .01