On the average, a category 4 (on the Stafford/Simpson scale) or stronger hurrica

Question

On the average, a category 4 (on the Stafford/Simpson scale) or stronger hurricane strikes the United States once every 6 years. A hurricane of this strength has winds of at least 131 miles per hour and can cause extreme damage. An insurance agency is considering whether it might want to stop insuring oceanfront homes and wants to assess the risk involved. The president of the company wants to know how long (in years) before the next hurricane that is category 4 or stronger hits the US.

d) What is the expected length of time (in years) between now and when the next hurricane that is category 4 or stronger?

e) What is the variance in this length of time?

f) What is the probability density function for the length of time before the next hurricane that is category 4 or stronger? Write your answer in function form and show a graph.

g) What is the CDF for the length of time before the next hurricane that is category 4 or stronger? Write your answer in function form and show a graph.

h) What is the probability that there will be a hurricane that is category 4 or stronger within the next 5 years?

i) What is the probability that there will not be any hurricane that is category 4 or stronger, during the next 3 years?

j) What is the probability that there will be a hurricane that is category 4 or stronger, during the period that is between 5 to 10 years from now?

k) Given that there are no hurricanes that are category 4 or stronger during the next 3 years, what is the probability that there will not be any during the next 10 years?

l) How long a waiting time do we need, if we want to be 75% sure that there is a hurricane that is category 4 or stronger during the waiting time?

Explanation / Answer

d)

expected length of time =6 year

e)

variance = 6^2 = 36 year^2

f)

f(t) = 1/6 * e^(-t/6)

g) P(T < t) = 1 -e^(-t/6)

h)

P(T < 5) = 1 -e^(-5/6) = 0.565401

i)

P(X > 3) = e^(-3/6) = e^(-1/2) = 0.606530

j)

P(5<X<10) =e-5/6-e-10/6 =0.4346-0.1889 =0.2457

l)

let waiting time be x:

therefore P(X<x) =0.75

1-e-x/6 =0.75

e-x/6 =0.25

taking log and solving

x =-6*ln(0.25)=8.318 years

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