Question 8.8: A fishermen catches, on average, three fish in an hour. Let ......
A fishermen catches, on average, three fish in an hour. Let Y be a random variable denoting the number of fish caught in one hour and let X be the time interval between catching two fishes.We assume that X follows an exponential distribution.
(a) What is the distribution of Y ?
(b) Determine E(Y ) and E(X).
(c) Calculate P(Y = 5) and P(Y < 1).
(a) The random variable Y follows a Poisson distribution, see Theorem 8.2.1 for more details.
(b) The fisherman catches, on average, 3 fish an hour. We can thus assume that the rate λ is 3 and thus E(Y ) = λ = 3. Similarly, E(X) = \frac{1}{\lambda } =\frac{1}{3} which means that it takes, on average, 20 min to catch another fish.
(c) Using the PDF of the Poisson distribution, we get:
P(Y = 5) = \frac{3^{5}}{5 ! } exp(−3) = 0.1 = 10%
P(Y < 1) = P(Y = 0) = \frac{3^{0}}{0 ! } exp(−3) ≈ 0.0498 ≈ 5%.
We would have obtained the same results in R using the dpois(5,3) and dpois (0,3) commands.