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 = .01386

a. What is the probability that the distance is 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?
c. What is the value of the median distance?

Answer 1

To find probabilities for different distances in an exponential distribution, use the exponential distribution formula P(X ≤ x) = 1 - e^(-λ * x) with the given parameter λ. Adjust the limits of x accordingly to find the desired probabilities.

Step-by-step explanation:

The distance that an animal moves from its birth site to the first territorial vacancy it encounters, denoted by X, follows an exponential distribution with parameter lambda (λ) = 0.01386. To find the probability that the distance is at most 100 m, we need to calculate P(X ≤ 100). Using the exponential distribution formula, we have:

P(X ≤ 100) = 1 - e^(-λ * 100) = 1 - e^(-0.01386 * 100)

By evaluating the expression, we can find the probability. Similarly, we can find the probabilities for the other questions by adjusting the limits accordingly and using the same formula.