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 .

We know there are N total observations, that there are k unique values for each of these observations and that Y is the minimum.

(k-Y+1)^n represents the total number of permutations of X values, where the X values are between and include Y and k. This alone does not guarantee that any value of X is equal to Y.

(k-Y)^n represents the total number of permutations of X values where all X values are greater than Y.

By subtracting the first term above, from the second, we can calculate the number of permutations where at least one value of X is equal to Y. We then can divide this by our total number of permutations X to calculate the PDF.

P(min=Y) = ((k-Y+1)^n - (k-Y)^n )/k^n

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..)

If we expect one failure every 10 years, using a geometric distribution, we know that the probability of failure in any give year is (1/10). We also know that the probability of the machine not failing in a given year is (9/10).

As the question does not specify in what year the machine fails, we effectively want to calculate the likelihood of 8 consecutive years where the machine does not fail.

(9/10)^8
## [1] 0.4304672

Expected Value

The expected value is equal to 1/p:

1/(1/10)
## [1] 10

Standard Deviation

The Standard Deviation is equal to [(1-p)/p^2]^(1/2)

((1-.1)/(.1^2))^.5
## [1] 9.486833

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

We can calculate the probability the machine will fail after 8 years by taking e to the ratio of the years before failure * -1, over the total expected lifespan of the machine.

e^(-YearsBeforeFailure/Expected Lifespan)

exp(-8/10)
## [1] 0.449329

Expected Value

1/(1/10)
## [1] 10

Standard Deviation

(1/((1/10)^2))^.5
## [1] 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)

This will look a similar to our geometric distribution. The probability of a “success” is .1, and a failure ,9. 8 consecutive failures can be calculated by:

.9^8
## [1] 0.4304672

Expected Value

e(x) = np

8*.1
## [1] 0.8

Standard Deviation

SD = (p(1-p)n)^.5

(.1*(1-.1)*8)^.5
## [1] 0.8485281

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

For this one we will use Rs Poisson functions

ppois(0,lambda = (8/10))
## [1] 0.449329

Expected Value

e(x)= lambda

8/10

Standard Deviation

SD = lambda^.5

(8/10)^.5
## [1] 0.8944272