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 .
The probability that \(Y=y\) is the probability that none of the variables is less than \(y\) AND that the variables are NOT all greater than \(y\) given that they are not less than Y. Since these events are independent, the probability can be calculated using multiplication.
The probability that none of the variables is less than \(y\) is \(P_1=(\frac{k-y+1}{k})^n\).
The probability that the variables are NOT all greater than \(y\) given that they are not less than \(y\) is one minus the probability that they are all greater than \(y\) given that none of them are less than \(y\), which can be calculated as follows:
\(P_2=1-(\frac{k-y}{k-y+1})^n\)
Therefore:
\(P(Y=y)=(\frac{k-y+1}{k})^n\times (1-(\frac{k-y}{k-y+1})^n)=(\frac{k-y+1}{k})^n-(\frac{k-y}{k})^n=\frac{(k-y+1)^n}{k^n}-\frac{(k-y)^n}{k^n}=\frac{(k-y+1)^n-(k-y)^n}{k^n}\)
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..)
The probability that the machine fails after 8 years is equivalent to the probability that the machine doesn’t fail for the first 8 years. Each year there is a probability of 0.1 that the machine fails, so the probability that it does not fail in the first 8 years is: \(P(X>8)=1-P(X\leq8)=1-(1-(0.9)^8)=0.4304672\)
So there is approximately a 43% chance that the machine will fail after 8 years.
\(E(X)=\frac{1}{.1}=10\)
The expected value of the number of years until failure is 10 years (this is also information given by the problem).
\(\text{Standard Deviation}=\sqrt{\frac{1-.1}{.1^2}}=\sqrt{90}=9.486833\)
Using R:
1-pgeom(7,0.1)## [1] 0.4304672What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
\(P(X>8)=1-P(X\leq8)=1-(1-e^{-0.1\times8})=1-(1-e^{-0.8})=0.449329\)
To there is approximately a 44.9% chance that the machine will fail after 8 years.
\(E(X)=\frac{1}{.1}=10\)
The expected value of the number of years until failure is 10 years.
\(\text{Standard Deviation}=\sqrt{\frac{1}{.1^2}}=\sqrt{100}=10\)
Using R:
1-pexp(8, rate = 0.1)## [1] 0.449329What 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)
\(P(X>8)=1-P(X\leq8)=1-(1-(0.9)^8)=0.4304672\)
So there is approximately a 43% chance that the machine will fail after 8 years.
\(E(X)=8(0.1)=0.8\) so we would expect 0.8 failures in 8 years.
\(\text{Standard Deviation}=\sqrt{(8)(.1)(.9)}=\sqrt{0.72}=0.84585281\)
Using R:
pbinom(0,8,0.1)## [1] 0.4304672What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
The poisson distribution cannot compute the expected number of failures in 8 years given the expected failure rate of once every ten years. In order to use the poisson distribution, we need the expected number of failures in 8 years, which was calculated in the solution to part C:
\(E(X)=8(0.1)=0.8\) so we would expect 0.8 failures in 8 years.
Therefore, the probability that the machine will fail after 8 years can be determined by calculating the probability of 0 events with an expected value of 0.8 failures in 8 years, as follows:
\(P(X=0)=\frac{0.8^0e^{-0.8}}{0!}=e^{-0.8}=0.449329\)
So there is approximately a 44.9% chance that the machine will fail after 8 years.
The expected value was calculated above.
\(\text{Standard Deviation}=\sqrt{0.8}=0.8944272\)
Using R:
dpois(0,0.8)## [1] 0.449329