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 = 0.01387.

(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.)
m

Answer 1

According to the question,

Here, X follows an exponential distribution with λ = 0.01387.

(a)

at most 100m   0.7502
at most 200m 0.9376
between 100 and 200m 0.1874
The answers are obtained by the R program:

> pexp(100,rate=0.01387)
[1] 0.7501763
> pexp(200,rate=0.01387)
[1] 0.9375881
> pexp(200,rate=0.01387)-pexp(100,rate=0.01387)
[1] 0.1874118

(b) The mean and standard deviation of an exponential distribution are same, 1/\lambda. Thus, the probability that distance exceeds the mean distance by more than 2 standard deviations is P(X>3/\lambda) = 1 - P() = 0.0498 .

The R program is

> 1-pexp(3/0.01387,rate=0.01387)
[1] 0.04978707

(c) The value of the median distance is 49.9746.

> qexp(0.5,rate=0.01387)
[1] 49.97456