Question

Let x denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. suppose that for banner-tailed kangaroo rats, x has an exponential distribution with parameter λ = 0.01382. (a) what is the probability that the distance is at most 100 m? at most 200 m? between 100 and 200 m? (round your answers to four decimal places.)

Answer 1

To solve this problem, we have to use the formula:

P = 1 - e^(-λx)

where λ = 0.01382, and x is the distance

A. when x = 100

(P < or equal to 100) = 1 - e^(-0.01382 * 100)

(P < or equal to 100) = 0.75

B. when x = 200
(P < or equal to 200) = 1 - e^(-0.01382 * 200)
(P < or equal to 200) = 0.94

C. when 100 < x < 200
P (100 < x < 200) = P ((x < 100) + P(x > 200))
P (100 < x < 200) = 0.75 + (1 – 0.94)

P (100 < x < 200) = 0.81