DATA605 HW7

1. 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 .

For 1≤j≤k,m(j)=((k−j+1)^n−(k−j)n)/kn

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.

2. 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.).

1. 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]=1/p

geo_EV = 1/(.1)
paste("E(X)=", geo_EV)

## [1] "E(X)= 10"

Expected value of 10 years for machine

StandardDeviation=√(1−p/p^2)

sd=sqrt((.9)/(.1^2))
paste("sd=", sd)

## [1] "sd= 9.48683298050514"

2. 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^−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=1/λ=10
                         λ=1/10

exp_EV = 1/(.1)
paste("E(X)=",exp_EV)

## [1] "E(X)= 10"

Standard Deviation:

sd=√(1/λ^2)

sd = sqrt((1)/(.1^2))
paste("sd = ", sd)

## [1] "sd =  10"

3. 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"

4. 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)=(λxe−λ)/x!

λ=np/t=(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=√λ

sd = sqrt(.8)
paste("sd = ", sd)

## [1] "sd =  0.894427190999916"