Two friends (A and B) agree to meet on 4:00 PM. A usually arrives between 5 minutes early and 5 minutes late. B usually arrives between 5 minutes early and 15 minutes late. Their times of arrival are independent from each other.
a) \(P(B > 5) = \dfrac{5 - (-5)}{15 - (-5)} = 1/2\)
b) \(E[B] - E[A] = 5 - 0 = 5\)
c) \(P(B < 0, A < 0) = P(B < 0) P(A < 0) = 5/20 * 5/10 = 1/8\)
\[P(X < x) = 1 - e^{-\lambda x}\]
\[P(X > x) = e^{-\lambda x}\]
The solution is 1 - no machine provides an answer within the hour \(1 - P(X_1 > 1, X_2 > 1, X_3 > 1)\).
\[1 - P(X_1 > 1, X_2 > 1, X_3 > 1) = 1 - e^{-\lambda_1 - \lambda_2 - \lambda_3}\]
1 - pexp(1,rate=3+4+6)
## [1] 2.260329e-06
\(P(X>15) = 1 - P(X < 15) = 1 - (1 - e^{-\lambda x}) = e^{-15/6} = 0.08\)
\(P(X > t) = e^{-t/6} = 0.1\), then \(t = 13.81\).
\(\lambda * t = 1/6 * 60 = 10\)
\(P(X=10) = \dfrac{e^{-\lambda *t}(\lambda t)^10}{10!} = 0.125\).
\(P(X < 15 | X > 10) = 1 - P(X > 15 | X > 10) = 1 - P(X > 5) = 1 - e^{-5/6} = 0.565\) (memoryless property)
Hint: Check the relationship between Poisson and Exponential distributions.
A pack of flour contains 1 kg of flour. Though a flour pouring machine has a standard deviation of 50 gr.
We have \(\mu = 1000,\ \sigma = 50\). Define \(\Phi(.)\) as the cdf of the standard normal distribution.
#a
pnorm(1075,mean=1000,sd=50)-pnorm(925,mean=1000,sd=50)
## [1] 0.8663856
#b
qnorm(0.9,sd=50)
## [1] 64.07758
#c
1000-qnorm(0.05,sd=50)
## [1] 1082.243
There are two different roads to get to Sarıyer. Road A takes 35 minutes on average with standard deviation 5 minutes. Road B takes 32 minutes on average with standard deviation 8 minutes.
a) For Road A it will take 42 minutes or less with probability \(\Phi(\dfrac{42-35}{5}) = \Phi(1.4) = 0.92\). For Road B, \(\Phi(\dfrac{42-32}{8}) = \Phi(1.4) = 0.89\). Take Road A. b) It is the quantile value of \(\alpha = 0.9\). Then we need to find \(\Phi^{-1}(\alpha) = 1.28 = \dfrac{x-\mu}{\sigma}\). For Road A \(1.28*5 + 35 = 41.4\), B \(1.28*8 + 32 = 42.2\).
Note: b is phrased weakly. It is also possible to understand it as a probability interval as the probability of an event occurring in normal distribution happens in intervals (i.e. originates from mu). In that case the quantile max value is \(\Phi^{-1}(0.95)=1.64\) as there should be 5% slack at each side of the distribution.