Question

Let X=the distance (m) that an animal moves form its birthplace to the first territorial vacancy it encounters. Suppose that for a specific specie X has an exponential distribution with ?=0.01386.

a. If the animal has been walking for 150m, what is the probability that it will have to walk for another 100m?

b. If you observe 15 animals, what is the probability that less than 5 will have to walk more than 100m?

c. If you observe 75 animals, what is the probability that the average distance travel will be less than 72 m? What is the probability that the average distance will be more than 80 m?

Answer 1

a) probability that it will have to walk for another 100m given already walked 150 m =P(X>250|X>150) =P(X>250)/P(X>150)

=e^-0.01386*250/e^-0.01386*150 =0.25

b)Probability of walking more than 100 meter =P(X>100) =e^-0.01386*100 =0.25

hence for binomial distribution  probability that less than 5 will have to walk more than 100m

=P(X<5) =$\sum_{x=0}^{4}\binom{15}{x}(0.25)^{x}(0.75)^{11-x}$ =0.6865

c)

for 75 animals average distance =1/$\lambda$ =1/0.01386 =72.15

and std error of mean =(1/$\lambda$)/(n)^1/2 =72.15/(75)^1/2 =8.33

for normal approximation z score =(X-mean)/std errror

a)  probability that the average distance travel will be less than 72 m:

probability =

P(X<72)

=

P(Z<-0.02)=

0.4928

b)

probability that the average distance will be more than 80 m:

probability =

P(X>80)

=

P(Z>0.94)=

1-P(Z<0.94)=

1-0.827=

0.1730