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2 years ago

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An article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the me

an time between occurrences of loads is 0.4 year.
(a) How many loads can be expected to occur during a 4-year period?
(b) What is the probability that more than eleven loads occur during a 4-year period?
(c) How long must a time period be so that the probability of no loads occurring during that period is at most 0.3?

1 answer:

pashok25 [27]2 years ago

5 0

Answer:

A. 10 loads

B. 0.428335

C. 3.2 years

Step-by-step explanation:

Given

Mean time between occurrence = 0.4 year

A Number of loads expected to occur during a 4 year period

Period = 4

Mean time = 0.4

Formula = Period/Mean Time

Formula =

So, the number of loads expected to occur during a 4-year period = 4/0.4 =10 loads

B. Probability that more than 11 loads occur during 4 year period

Since, the expected number of loads during 4-year period = 10 (from A above)

Then the mean is 10

Using Poisson distribution,

P(k events in interval)= (λ^k * e^-k)/k!

where: k = 0, 1, 2,3,4, . . ., 11 and λ = 10.

P(k=0)= (10^0 * e^-10)/0! = 0.000045

P(k=1)= (10^1 * e^-10)/1! = 0.000454

P(k=2)= (10^2 * e^-10)/2! = 0.00227

P(k=3)= (10^3 * e^-10)/3! = 0.007567

P(k=4)= (10^4 * e^-10)/4! = 0.018917

P(k=5)= (10^5 * e^-10)/5! = 0.037833

P(k=6)= (10^6 * e^-10)/6! = 0.063055

P(k=7)= (10^7 * e^-10)/7! = 0.090079

P(k=8)= (10^8 * e^-10)/8! = 0.112599

P(k=9)= (10^9 * e^-10)/9! = 0.12511

P(k=10)= (10^10 * e^-10)/10! = 0.12511

P(k=11)= (10^11 * e^-10)/11! = 0.113736

The probability that more than 11 loads occur during a 4-year period is then given by the following Express

1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]

= 1 - [0.000045 + 0.000454 + 0.00227 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.12511+ 0.12511 + 0.113736]

= 1 - 0.571665 = 0.428335

C. How long must a time period be so that the probability of no loads occurring during that period is at most 0.3

This is

(λ^0 * e^-λ)/0! <= 0.3

e^-λ/1 <= 0.3

e^-λ <= 0.3

-λ <= ln 0.3

-λ <= -1.204

λ >= 1.204

The time period = 4 years / 1.204

Time period = 3.2 years

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