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 .

Y = min(X1, X2,…Xn)

Since all variables are randomly independent with uniformly distrubtion, no variable is overlapped and repeated, the distribution of Y is always count as 1 or probablity of 1/n

Sample:

 k = 20
 sample <- runif(k, min = 1, max = k)
 
 summary(sample)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.255   8.201  10.700  10.640  14.020  19.830
 Y = min(sample)
 
 (Y)
## [1] 1.254658
 plot(sample)

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..)

P(T = n) = q^(n−1) * p

the probablity of not failing during the first 8 years = fail at 9th year

n = 9 p = 0.5 for chance that machine fails and q = 1 - 0.5 that machine doesn’t not fail

  n=9
  p = 0.5
  P_fail_9th_Yr = ((1-p)^(n-1)) * p
  
  (P_fail_9th_Yr)
## [1] 0.001953125

Expected value for Geometric Distribution: E(X) = 1/p

  p = 0.5
  Expected_Value =  1/ p
  
  (Expected_Value)
## [1] 2

Standard Deviation for Geometric Distribution:

Var(X) = (1-p) / p^2

 p = 0.5
 Standard_dev_geom = sqrt((1-p)/p^2)
 (Standard_dev_geom)
## [1] 1.414214

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.

The machine has expected’s lifetime = 10 yrs

λ = probablity that the machine will fail = 1/10 = 0.1

PDF function : f(x) = λ e^(−λx) CDF function : F(x) = 1 - e^(−λx)

For probablity that the machine will fail after 8 years

P (T > 8) = 1 - F(8) = 1 - (1 - e^(−(0.1)*8)) = e^(-0.8)

 (exp(-0.8)) 
## [1] 0.449329

Expected Value: E(x) = 1/λ, E(x) = 1/0.1 = 10

Standard Deviation: Var(x) = 1/λ^2, Standard deviation = sqrt(1/λ^2) = 1/λ = 1/0.1 = 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)

p = 0.5 that is not fail x = number of sucuess not to fail n = 10

The probablity that the machine is not failed in first 8 yers P(x) = (nCx) * p^x * (1 - p) ^ (n - x) and P(8)

The probablity that the machine will fail after 8 years = 1 - P(8)

 p = 0.5
 x = 8
 n = 10
 
 prob_that_the_machine__not_failed = choose(n,10) * (p^x) * (1-p)^(n-x)
 
 (1 - prob_that_the_machine__not_failed)
## [1] 0.9990234

Expected VAlue E(x) = np

 (n*p)
## [1] 5

Standard deviation:

Var(x) = np(1-p), sqrt(Var(x))

 (sqrt(n*p*(1-p)))
## [1] 1.581139

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.

expected life time of machine is 10 yr, λ = 10, assume it is constant

we are interested in the machine fail, a success is machine fail, x = 8

P(x) = ( λ^(x) * e^(-λ) )/x!

Probablity that we fail happen after 8 years = P(x=8)

 (  (( 10^(8) * exp(-10) )/factorial(8))  )
## [1] 0.112599

Expected Value:

E(x) = λ = 10

Standard Deviation: Var(x) = λ, standard deviation = sqrt(VAr(x)) = sqrt(10)

(sqrt(10))
## [1] 3.162278