1 Question 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:

\(\because x_{i}\) is uniformly distrubted on the intergers from 1 to k

\(\therefore\) the pdf of $x_{i} $ is \(P(X=x_{i}) = \frac{1}{k}\)

\(\because x_{i}'s\) are mutually independtent from each other

\(\therefore\) the joint pdf of \(P(All(X_{i}) > x) = P(x_{1} > x,x_{2} > x,...,x_{n} > x \;for\;x\;in\;range\;(1,k)) = (1-\frac{x}{k})^n\)

\(\therefore P(Min(x_{i})=1) = P(Y = 1) = 1-P(All(X_{i}) > 1)=1-(1-\frac{1}{k})^n=(1-\frac{0}{k})^n-(1-\frac{1}{k})^n\)

\(\therefore P(Y = 2) = 1-P(All(X_{i}) > 2) - P(Y = 1)=(1-\frac{1}{k})^n-(1-\frac{2}{k})^n\)

\(\therefore P(Y = 3) = 1-P(All(X_{i}) > 3) - P(Y = 1) - P(Y = 2)=(1-\frac{2}{k})^n-(1-\frac{3}{k})^n\)

\(...\)

\(\therefore P(Y = y) = 1-P(All(X_{i}) > y) - P(Y = j\;for\;j < y)=(1-\frac{y-1}{k})^n-(1-\frac{y}{k})^n\)

\(\therefore P(Y = y) = (\frac{k-y+1}{k})^n-(\frac{k-y}{k})^n\)

2 Question 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.).

2.1 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..)

Answer:

Modeling using geometric distirbution, the probability that the machine will fail after 8 years is equivalent to it will not fail in the first 8 year, therefore

\(P(X = n) = (1-p)^{n-1}p\) where \(p = \frac{1}{10}\)

and \(P(X > 8) = 1 - P(X \leq 8)\)

compute in R

  1. \(P(x > 8)\)
## [1] 0.4304672
  1. \(E[x] = \frac{1}{p}\)
## [1] 10
  1. \(SD(x)=\sqrt{\frac{1-p}{p^2}}\)
## [1] 9.486833

2.2 Part 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.

Answer:

Modeling using exponential distirbution, the probability that the machine will fail after 8 years is equivalent to it will not fail in the first 8 year, therefore

\(P(X <= n) = 1-e^{-\lambda x}\) where \(\lambda = 1/10\)

and \(P(X > 8) = 1 - P(X \leq 8)\)

compute in R

  1. \(P(x > 8)\)
## [1] 0.449329
  1. \(E[x] = \frac{1}{\lambda}\)
## [1] 10
  1. \(SD(x) = \sqrt{\frac{1}{\lambda^2}}\)
## [1] 10

2.3 Part 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)

Modeling using binomial distirbution, the probability that the machine will fail after 8 years is equivalent to it fail 0 times in the first 8 year, therefore

\(P(X = n) = \begin{pmatrix}n\\ x\end{pmatrix}p^x(1-p)^{n-x}\) where \(p = \frac{1}{10}\;and\;n=8\)

and \(P(x) = \begin{pmatrix}8\\ 0\end{pmatrix}(\frac{1}{10})^0(\frac{9}{10})^{8}\)

compute in R

1.\(P(x=0)\;given\;p=\frac{1}{10}\;and\;n=8\)

## [1] 0.4304672
  1. \(E[x] = np\)
## [1] 0.8
  1. \(SD[x]=\sqrt{np(1-p)}\)
## [1] 0.8485281

2.4 Part 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.

Modeling using Poisson distirbution, the probability that the machine will fail after 8 years is equivalent to it will fail zero times in the first 8 year, therefore

for 1 year, \(P(X = n) = \frac{\lambda^{x}e^{-\lambda}}{x!}\) where \(\lambda = 1/10\), and

for 8 year, \(P(X = n) = \frac{\lambda^{x}e^{-\lambda}}{x!}\) where \(\lambda = 8/10\)

compute in R

1.\(P(x=0)\;given\;\lambda=\frac{8}{10}\)

## [1] 0.449329

2.\(E[x] = \lambda\)

## [1] 0.8

3.\(SD[x] = \sqrt{\lambda}\)

## [1] 0.8944272