The uniformly distributed on the integers from 1 to k, the miniumum value of Y is one Xi from Xi’s:
Assume the distribution function P(Y=z) = min{X1,X2,X3….Xi}
Case 1: Assumption: Y=1, which is minimum sample Xi (Select one sample from 10 samples) x = {1,2,3,4,5} Size of all sample x = {1,2,3,4,5} : 510 Size of non-sample x = {1,2,3,4} : \((5−1)^{10} P(Y=1)=((5−0)^{10}−(5−1)^{10})/5^{10}\)
Case 2: Assumption: (Select 2 sample from 10 samples) x = {1,2,3,4,5} Size of all sample: \(5^10\) Size of non-sample = \((5^{10}−4^{10})+3^{10} P(Y=1)=(5^{10}−(5^{10}−4^{10})−3^{10})/5^{10}\)
Using variables: \(k^n−(k^n−(k−1)^n)−(k−2)^n\) \(=(k−1)^n−(k−2)^n\)
When m=2; \(=((k+m−1)^n−(k−m)^n)/k^n\)
We see here p=0.1, and q=1−p=0.9 machine fails after 8 years
r # Calculating P(X>8) using geometric distribution pgeom(8, 0.1, lower.tail=FALSE)
## [1] 0.3874205 Expected number of years before the first machine failure is \(E(X)=q/p= 0.9/0.1 = 9\).
Standard deviation \(\sigma^2 = \sqrt{q/p^2} = \sqrt{0.9/0.1^2} \approx 9.4868\)
Using Exponential Distribution \(p(n)=e^{−λx}\), probability that copier will be working or fail after 8 years is Standard deviation = \(sqrt(V(X))\)
# Calculating P(X>8) using exponential distribution
pexp(8, 0.1, lower.tail=FALSE)## [1] 0.449329
Expected value is \(E(X) = 1/\lambda = 1/0.1 = 10\).
Standard deviation \(\sigma^2 = \sqrt{1/\lambda^2} = 1/\lambda = 10\).
# Calculating P(X=0) for n=8 using binomial distribution
pbinom(0,8,0.1,lower.tail=TRUE)## [1] 0.4304672
Expected value \(E(X)=np= 8 \times 0.1 = 0.8\).
Standard deviation \(\sigma^2 = \sqrt{npq} = \sqrt{8 \times 0.1 \times 0.9} \approx 0.8485\).
# Calculating P(X=0) for 8 intervals using Poisson distribution
ppois(0,0.1,lower.tail=TRUE)^8## [1] 0.449329
For Poisson distribution, \(E(X) = \sigma^2 = \lambda = 0.1\).