According to NOAAs report a major flooding in Amherst in a g

According to NOAA’s report, a major flooding in Amherst in a given year has a Poisson distribution with a mean occurrence of 2.5. a) What is the probability that there will be at least one major flooding in the next one year?

b) Using the same parameter in a), what is the probability that it will be at least 6 months until the next major flooding?

c) If one year has passed with no flooding, what is the probability that there will be flooding within the next six months?

d) Let T be the time until the occurrence of fifth flooding. Find the probability of T is between 2 and 4 years?

Solution

Let X be the random variable that,  a major flooding in Amherst in a given year.

X ~ Poison(mean = 2.5)

The parameter of the Poison distribution is mean ( ).

The probability mass function of the Poison disribution is,

P(X = x) = ( e^- * ^x ) / x!

where = 2.5

a) What is the probability that there will be at least one major flooding in the next one year?

that is we have to find here P(X 1).

P(X 1) = 1 - P(X=0)

P(X=0) = ( e^-2.5 * 2.5⁰ ) / 0! = 0.0821

P(X 1) = 1 - 0.0821 = 0.9179

a) What is the probability that there will be at least one major flooding in the next one year?

that is we have to calculate here P(X 6)

P(X 6) = 1 - P(X < 6)

= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) ]

P(X=0) = ( e^-2.5 * 2.5⁰ ) / 0! = 0.0821

P(X=1) = ( e^-2.5 * 2.5¹ ) / 1! = 0.2052

P(X=2) = ( e^-2.5 * 2.5² ) / 2! = 0.2565

P(X=3) = ( e^-2.5 * 2.5³ ) / 3! = 0.2138

P(X=4) = ( e^-2.5 * 2.5⁴ ) / 4! = 0.1336

P(X=5) = ( e^-2.5 * 2.5⁵ ) / 5! = 0.0668

P(X 6) = 1- [ 0.0821 + 0.2052 + 0.2565 + 0.2138 + 0.1336 + 0.0668 ]

P(X 6) = 1 - 0.9580

P(X 6) = 0.0420

c) If one year has passed with no flooding, what is the probability that there will be flooding within the next six months?

that is here we have to find P(X 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)

P(X 6) = 0.0821 + 0.2052 + 0.2565 + 0.2138 + 0.1336 + 0.0668 + 0.0278 = 0.9858

d) Let T be the time until the occurrence of fifth flooding. Find the probability of T is between 2 and 4 years?

Here T follows exponential distribution with 1 / = 1 / 2.5 = 0.4

The probability distribution function of T is,

P(T=t) = * e^-t

P(2 < T < 4) = * e^-t dt

= *   e^-t dt

=   * [ e^- t / - ]

= - [ e^-0.4*4 - e^-0.4*2 ]

= - [ 0.2019 - 0.4493 ]

= 0.2474