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 .

Answer: Let us first show the minimum value from a single random uniform distribution, with 50 random values on a uniform distribution between 1 and 100

k = 100
n = 100

min(runif(50, min = 1, max = k))
## [1] 1.517051
#Now replicating this n (100) times
y <- replicate(n, min(floor(runif(50, min = 1, max = k))))

hist(y)

As we see above, it is skewed on the right.

  1. 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..) Answer: This is a Geometric distribution. In this case, the success of the event signifies the failure of a machine. As a machine does not fail for 10 years, that means it fails in the 11th year. Hence, probability of success, p = 1/11

In this question, the machine should fail anytime after 8 years.

Using the geometric probability function: p(machine fails anytime after 8 successes) = 1 - P(machine fails in any of the first 8 successes) = 1 - P(x<=8) = 1 - (0.1 + (0.9)(0.1) + (0.9)^2(0.1) + (0.9)^3(0.1) + .. + (0.9)^7(0.1))

we have to calculate the cumulative in this case, i.e. failure happens anytime after 8 years.

Now using R:

#dgeom will be used as we need to get the probability of the first success event
# no. of failures --> first parameter will be 8

p_2_a <- 1 - pgeom(8, 1/11)
p_2_a
## [1] 0.4240976

Expected Value = 1/p

1/(1/11)
## [1] 11

Variance = (1-p)/p^2 SD = sqrt(V)

variance = (1 - 1/11)/(1/11)^2
variance
## [1] 110
SD = sqrt(variance)
SD
## [1] 10.48809
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

lambda = 1/10

Variance = 1/lambda^2= 1/100

SD = sqrt(Variance) = 1/10

Applying the formula for exponential probability:

P(x>k) = e^(-k/µ) = e^(-0.1*8) = 0.4493

  1. 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)

np = 1

p = 1/10

Variance = np(1-p) = 10 x 1/10 x (1 - 1/10) = 0.9

sd = sqrt(variance) = 0.9487

  1. 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 = (lambda^x * e^(-lambda)) / x!

= (10^8 * exp(-10))/factorial(8)

= 0.112599

Variance = lambda = 10

SD = sqrt(variance) = 3.1623