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 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=[tex]\lambda=0.01352[/tex]Exponential distribution of probability is given by [tex]f(x)=\lambda e^{-\lambda x}, x\geq 0[/tex][tex]f(x)=0, x < 0[/tex]Cumulative distribution is given by [tex]F(X)=P(X\leq x)=1-e^{-\lambda x }, x\geq 0[/tex]Where [tex]\lambda=Parameter=0.01352[/tex]a.We have to find the probability that the distance is at most 100m.[tex]P(X\leq 100)=1-e^{-0.01352\times 100}=0.7413[/tex]Hence, the probability that the distance is at most 100 m=0.7413b. [tex]P(X\leq 200)=1-e^{-0.01352\times 200}=0.9331[/tex]Hence, the probability that the distance is at most 200m =0.9331c. [tex]P(100\leq X\leq 200)=P(X\leq 200)-P(X\leq 100)[/tex][tex]P(100\leq X\leq 200)=0.9931-0.7413=0.1918[/tex]Hence, the probability that the distance between 100 m and 200 m=0.1918