Question

The number of earthquake tremors in a 12-month period appears to be distributed as a Poisson random variable with a mean of 6. Assume the number of tremors from one 12-month period is independent of the number in the next 12-month period. Round your answers to four decimal places (e.g 98.7654) (a) What is the probability that there are 10 tremors in 1 year? (b) What is the probability that there are 18 tremors in 2 years? (c) What is the probability that there are no tremors in a 1-month period? (d) What is the probability that there are more than 5 tremors in a 6-month period?

Answer 1

a)P(10 tremors in 1 year)=P(X=10)=e^-$\lambda$$\lambda$^x/x! =e^-6*6¹⁰/10! =0.0413

b)

expected tremor in 2 year =2*6=12 =$\lambda$

P(18 tremors in 2 year)=e^-12*12¹⁸/18!=0.0255

c)

expected tremor in 1 month =6/12=0.5 =$\lambda$

P(no tremors in 1 month)=P(X=0)=e^-0.5*0.5⁰/0! =0.6065

d)

expected tremor in 6 month =6*6/12=3 =$\lambda$

P(more then 5 tremors)=P(X>5)=1-P(X<=5)=1-$\sum_{x=0}^{5}e^{-3}3^{x}/x!$ =1-0.9161=0.0839