Untitled

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 .

Solution: When X = 1: P(X=1) = k^n - (k-1)^n / k^n

When X = 2: P(X=2) = (k-2+1)^n - (k-2)^n / k^n

when X = 3: P(X=3) = (k-3+1)^n - (k-3)^n / k^n

So, when X = m:P(X=m) = (k-m+1)^n - (k-m)^n * k^n

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.).

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..)

geometric model: P(X=n)=[(1???p)^(n???1)]???p

p = P(machine will fail after 10 years) = 0.1 p-1 = P(machine will not fail after 10 years) = 0.9

pro_8years = ((.9)^(7))*.1
pro_8years

## [1] 0.04782969

Expected value = E[X]=1/p

Ex_value = 1/(0.1)
Ex_value

## [1] 10

Standard deviation = sqrt[(1???p)/p^2]

Standard_deviation = sqrt((0.9)/(0.1^2))
Standard_deviation

## [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.

exponential model: P(X>=8) = e^(-k/u)

u = mean life of machine = 10 years

pro_grater_8yrs = exp(-8/10)
pro_grater_8yrs

## [1] 0.449329

Expected value = E[X] = u = 1/?? = 10

?? = 1/10

Ex_value_b = 1/(1/10)
Ex_value_b

## [1] 10

Standard deviation = sqrt[1/(??^2)]

Standard_deviation_b = sqrt(1/(0.1^2))
Standard_deviation_b

## [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)

binomial model: P(X>8) = [1 x (p^x)] x (1???p)^(n???x)

p = P(machine will fail after 10 years) = 0.1 p-1 = P(machine will not fail after 10 years) = 0.9

binomial_greater_8_years = (0.1)^(0)*(1-0.1)^(8-0)
binomial_greater_8_years

## [1] 0.4304672

Expected value = E[X] = np

Ex_value_c = 8 * 0.1
Ex_value_c

## [1] 0.8

Standard deviation = sqrt(npq)

Standard_deviation_c = sqrt(8 * 0.1 * (1 - 0.1))
Standard_deviation_c

## [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.

Poisson model: P(X=8) = [[??^(x)] *[e^(?????)]] / x!

?? = np/t = (8???0.1)/1 = 0.8

x = 8

poisson_grater_8_years = 0.8^(8)*exp(-0.8/8)
poisson_grater_8_years

## [1] 0.1518065

Expected value = E[X] = ?? = 0.8

Standard deviation = sqrt(??)

Standard_deviation_d = sqrt(0.8)
Standard_deviation_d

## [1] 0.8944272