1. Minimum Uniform Distribution
Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y.
Admission: Struggled a bit on #1. Looked at some previous semester soltuions.
\(X_i\) has \(k\) possible values, so denominator is \(k^n\). \(k^{n-1}\) captures all the values of \(X_i\) for which \(X_i\) does not equal 1. Thus, the number of ways of getting \(Y = 1\) is \(k^n - (k - 1)^n\). We’re, however, looking for \(Y = j\). To get this for the minimum, we have:
\(\frac{(k - j + 1)^n - (k - j)^n}{k^n}\)
2. Failure Distributions
Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).
a. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years.)
\(P(x) = .1\) \(P(x=n) = q^{(n-1)}p\) \(P(x\geq9) = 1 - P(x\leq8) = 1 - (.1 + .9(.1) + .9^2(1)...+.9^8(.1))\)
Using R:
1 - pgeom(8,.1)## [1] 0.3874205
So there’s a 38.7% chance the machine will fail after 8 years.
Expected value is \(1/p\):
1/.1## [1] 10
Standard deviation ofr geometric distribution is \(\sqrt{q/p^2}\):
sqrt(.9/(.1^2))## [1] 9.486833
b. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
\(P(x\geq9) = 1 - (1-e^{-k/u})\) \(P(x\geq9) = e^{-8/10}\)
Using R:
1-pexp(8,.1)## [1] 0.449329
For expontential distribution, expected value will again be:
1/.1## [1] 10
Standard deviation is:
sqrt(1/.1^2)## [1] 10
c. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)
\((P\leq8) = .1^0.9^8\)
Using R:
pbinom(0,8,.1)## [1] 0.4304672
For binomail distribution, expected value will be number of failures in the 8 years:
8 * .1## [1] 0.8
Standard deviation is:
sqrt(8 * .1 * .9)## [1] 0.8485281
d. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
\(P(X=0) = e^{-(8*.1)}\)
Using R:
ppois(0,.8)## [1] 0.449329
Expected value:
8/10## [1] 0.8
Standard deviation of Poisson:
sqrt(.8)## [1] 0.8944272