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.

$P(X\leq 100)=1-e^{-0.01352\times 100}=0.7413$

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

b. $P(X\leq 200)=1-e^{-0.01352\times 200}=0.9331$

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