Let \(X_1\), \(X_2\),…,\(X_n\) 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 \(X_{i}'s\). Find the distribution of \(Y\).
Solution: The distribution of \(Y\) can be calculated using the distribution function:
\[ m(j)=P(Y=j) \]
We know that \(Y\) is the minimum value of \(X_i\) over all of the \(X_i\)’s. Thus, we can solve this equation by counting the total ways in which \(X_1\), \(X_2\),…,\(X_n\) can be assigned between values \(j\) and \(k\) that are greater or equal to one.
Using the refence below, we can define \(m(j)\) as follows:
\[ \text{For }1 \leq j \leq k, m(j)= \frac{(k-j+1)^{n}-(k-j)^{n}}{k^{n}} \]
Reference: Dartmouth University, Public Solutions
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).
Solution:
The geometric distribution equation is:
\[ P(X>n)=(1???p)^{n???1}*p \]
Geometric distribution calculations (using the pgeom function from the stats package):
q <- 8
p <- .1
pgeom <- pgeom(q, p, lower.tail = TRUE)
round(pgeom,2)## [1] 0.61Expected value calculations:
\[ E[X]=\frac{1}{p} \]
ev <- 1/p
round(ev, 2)## [1] 10Standard deviation calculations:
\[ \sigma = \sqrt \frac{1-p}{p^2} \]
sd <- sqrt((1-p)/p^2)
round(sd, 2)## [1] 9.49What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
Solution:
The exponential distribution equation is:
\[ P(X>n)=e^{\frac{-k}{u}} \]
Exponential distribution calculations (using the pexp function from the stats package):
q <- 8
rate <- .1 #lambda
pexp <- pexp(q, rate, lower.tail = TRUE)
round(pexp, 2)## [1] 0.55Expected value calculations:
\[ E[X]=\frac{1}{\lambda} \]
ev <- 1/rate
round(ev, 2)## [1] 10Standard deviation calculations:
\[ \sigma = \sqrt \frac{1}{\lambda^2} \]
sd <- sqrt((1)/rate^2)
round(sd, 2)## [1] 10What 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)
Solution:
The binomial distribution equation is:
\[ P(X=n)=1*p^x(1-p)^{n-x} \]
Binomial distribution calculations (using the pbinom function from the stats package):
q <- 0
n <- 8
p <- .1
pbinom <- pbinom(q, n, p, lower.tail = TRUE)
round(pbinom, 2)## [1] 0.43Expected value calculations:
\[ E[X]=n*p \]
ev <- n*p
round(ev, 2)## [1] 0.8Standard deviation calculations:
\[ \sigma = \sqrt{np(1-p)} \]
sd <- sqrt(n*p*(1-p))
round(sd, 2)## [1] 0.85What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Solution:
The poisson distribution equation is:
\[ P(X>n)=\frac{\lambda^{k}e^{-\lambda}}{k!} \]
Poisson distribution calculations (using the ppois function from the stats package):
lambda <- 8/10
q <- 0
ppois <- ppois(q, lambda, lower.tail = TRUE)
round(ppois, 2)## [1] 0.45Expected value calculations:
\[ E[X]=\lambda \]
ev <- lambda
round(ev, 2)## [1] 0.8Standard deviation calculations:
\[ \sigma = \sqrt{\lambda} \]
sd <- sqrt(lambda)
round(sd, 2)## [1] 0.89