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, has an exponential distribution with parameter λ = 0.01392 (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.) at most 100 m at most 200 m between 100 and 200 m (b) What is the probability that distance exceeds the mean distance by more than 2 standard deviations? (Round your answer to four decimal places.) (c) What is the value of the median distance? (Round your answer to two decimal places.)

Answer 1

Solution:

Given information

μ = 0.01392
mean = 1/μ = 71.84
standard deviation = 1/μ = 71.84

CDF, C(x) = 1 - e^(-μx)

a.
probability that the distance is at most 100 m
P(X<=100)
= C(100)
= 1 - e^{(-0.01392*100)}
= 0.7514

probability that the distance is at most 200 m
P(X<=200)
= C(200)
= 1 - e^{(-0.01392*200)}
= 0.9382

probability that the distance is between 100 and 200 m
= 0.9382 - 0.7514
= 0.1868

b.
P(X> mean + 2*standard deviation)
= P(X > 215.52)
= 1 - P(X<=215.52)
= 1 - C(215.52)
= 1 - (1 - e^{(-0.01392*215.52)})
= e^{(-0.01392*215.52)}
= 0.0498

c.
Calculate the value of median distance.
P(X ≤ M_d)= 0.5
F_x(M_d) = 0.5
e^{-λ x M_d} = 0.5
Take log on both sides and solve.
-λ x M_d = ln(0.5)
M_d - (ln(0.5)/λ)
Substitute 0.01392 for λ.
M_d = - [(ln(0.5)/0.01392] = 49.7951 m