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.)

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 

yo answer should be 6 if im not mistaken

step-by-step explanation: