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 = 0.01352. (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.)

Answer 1

Answer with Step-by-step explanation:

We are given that

X=Distance (m) that animal moves from its birth site to the first territorial vacancy it encounters.

Parameter=
`\lambda=0.01352`

Exponential distribution of probability is given by

`f(x)=\lambda e^(-\lambda x), x\geq 0`

`f(x)=0, x < 0`

Cumulative distribution is given by

`F(X)=P(X\leq x)=1-e^(-\lambda x ), x\geq 0`

Where
`\lambda=Parameter=0.01352`

a.We have to find the probability that the distance is at most 100m.

`P(X\leq 100)=1-e^(-0.01352* 100)=0.7413`

Hence, the probability that the distance is at most 100 m=0.7413

b.
`P(X\leq 200)=1-e^(-0.01352* 200)=0.9331`

Hence, the probability that the distance is at most 200m =0.9331

c.
`P(100\leq X\leq 200)=P(X\leq 200)-P(X\leq 100)`

`P(100\leq X\leq 200)=0.9931-0.7413=0.1918`

Hence, the probability that the distance between 100 m and 200 m=0.1918