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 = 0.01372.

What is the probability that the distance is at most 100 m? At most 200 m? Between 100 and 200 m?

74.64% probability that the distance is at most 100 m

93.57% probability that the distance is at most 200 m

18.93% probability that the distance is between 100 and 200m.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

In which is the decay parameter.

The probability that x is lower or equal to a is given by:

Which has the following solution:

In this problem, we have that:

What is the probability that the distance is at most 100 m?

74.64% probability that the distance is at most 100 m

At most 200 m?

93.57% probability that the distance is at most 200 m

Between 100 and 200 m?

0.9357 - 0.7464 = 0.1893

18.93% probability that the distance is between 100 and 200m.

your answer is a

step-by-step explanation:

4*4*4*4=256

180

step-by-step explanation: