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.
https://math.dartmouth.edu/archive/m20f10/public_html/HW5Solutions.pdf
\[For\ 1 ≤ j ≤ k, m(j) = \frac{(k-j+1)^n-(k-j)^n}{k^n}\]
Since Y is the minimum value of Xi over all of the Xi’s, then in order to find the distribution function m(j) = P(Y=j), we will need to count the number of ways that we can assign X1, X2, …, Xn to values between j and k with at least one Xi being assigned to j and divide by the total number of possible ways to assign X1, X2, …, Xn to values between 1 and k.
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.).
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 = P(machine will fail after 10 years) = .1 p-1 = P(machine will not fail after 10 years) = .9
\[P(X=n) = (1-p)^{n-1}*p\]
prob_8_years = ((.9)^(7))*.1
paste("P(X=8) = ", prob_8_years)## [1] "P(X=8) = 0.04782969"Expected Value:
\[E[X] = \frac{1}{p}\]
geo_EV = 1/(.1)
paste("E(X) = ", geo_EV)## [1] "E(X) = 10"Expected value of 10 years for machine
\[Standard Deviation = \sqrt\frac{1-p}{p^2}\]
sd = sqrt((.9)/(.1^2))
paste("sd = ", sd)## [1] "sd = 9.48683298050514"What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
u = mean life of machine = 10 years
\[P(X >= 8) = e^{\frac{-k}{u}}\]
prob_greater_8_years = exp(-8/10)
paste("P(X>=8) = ", prob_greater_8_years)## [1] "P(X>=8) = 0.449328964117222"Expected Value:
\[E[X] = u = \frac{1}{λ} = 10\] \[λ = \frac{1}{10}\]
exp_EV = 1/(.1)
paste("E(X) = ", exp_EV)## [1] "E(X) = 10"Standard Deviation:
\[sd = \sqrt\frac{1}{λ^2}\]
sd = sqrt((1)/(.1^2))
paste("sd = ", sd)## [1] "sd = 10"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 = P(machine will fail after 10 years) = .1 p-1 = P(machine will not fail after 10 years) = .9
\[P(X>8) = 1 * p^x(1-p)^{n-x}\]
biprob_greater_8_years = (.1)^(0)*(.9)^(8)
paste("P(X>=8) = ", biprob_greater_8_years)## [1] "P(X>=8) = 0.43046721"Expected Value:
\[E[X] = np\]
binomial_EV = 8*(.1)
paste("E(X) = ", binomial_EV, "failures")## [1] "E(X) = 0.8 failures"Standard Deviation:
#sd = sqrt(n*p*q)
sd = sqrt(8*(.1)*(.9))
paste("sd = ", sd)## [1] "sd = 0.848528137423857"What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
\[P(X=8)= \frac{λ^xe^{-λ}}{x!}\]
\[λ = \frac{np}{t} = \frac{8*.1}{1} = .8\] \[x = 8\]
poisson_prob = .8^(8)*exp(-.8/8)
paste("P(X>=8) = ", poisson_prob)## [1] "P(X>=8) = 0.151806528072716"Expected Value
\[E[X] = λ = .8\]
Standard Deviation:
\[sd = \sqrtλ \]
sd = sqrt(.8)
paste("sd = ", sd)## [1] "sd = 0.894427190999916"