Respuesta :
Answer:
a) [tex] \lambda_1 = 2*2 = 4[/tex]
And let X our random variable who represent the "occurrence of structural loads over time" we know that:
[tex] X(2) \sim Poi (4)[/tex]
And the expected value is [tex] E(X) = \lambda =4[/tex]
So we expect 4 number of loads in the 2 year period.
b) [tex] P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)][/tex]
[tex] P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}][/tex]
And we got: [tex] P(X(2) >6) =1-0.889=0.111[/tex]
c) [tex] e^{-2t} \leq 2[/tex]
We can apply natural log in both sides and we got:
[tex] -2t \leq ln(0.2)[/tex]
If we multiply by -1 both sides of the inequality we have:
[tex] 2t \geq -ln(0.2)[/tex]
And if we divide both sides by 2 we got:
[tex] t \geq \frac{-ln(0.2)}{2}[/tex]
[tex] t \geq 0.8047[/tex]
And then we can conclude that the time period with any load would be 0.8047 years.
Step-by-step explanation:
Previous concepts
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"
Solution to the problem
Let X our random variable who represent the "occurrence of structural loads over time"
For this case we have the value for the mean given [tex]\mu = 0.5[/tex] and we can solve for the parameter [tex]\lambda[/tex] like this:
[tex] \frac{1}{\lambda} = 0.5[/tex]
[tex] \lambda =2 [/tex]
So then [tex] X(t) \sim Poi (\lambda t)[/tex]
X follows a Poisson process
Part a
For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:
[tex] \lambda_1 = 2*2 = 4[/tex]
And let X our random variable who represent the "occurrence of structural loads over time" we know that:
[tex] X(2) \sim Poi (4)[/tex]
And the expected value is [tex] E(X) = \lambda =4[/tex]
So we expect 4 number of loads in the 2 year period.
Part b
For this case we want the following probability:
[tex] P(X(2) >6)[/tex]
And we can use the complement rule like this
[tex] P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)][/tex]
And we can solve this like this using the masss function:
[tex] P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}][/tex]
And we got: [tex] P(X(2) >6) =1-0.889=0.111[/tex]
Part c
For this case we know that the arrival time follows an exponential distribution and let T the random variable:
[tex] T \sim Exp(\lambda=2)[/tex]
The probability of no arrival during a period of duration t is given by:
[tex] f(T) = e^{-\lambda t}[/tex]
And we want to find a value of t who satisfy this:
[tex] e^{-2t} \leq 2[/tex]
We can apply natural log in both sides and we got:
[tex] -2t \leq ln(0.2)[/tex]
If we multiply by -1 both sides of the inequality we have:
[tex] 2t \geq -ln(0.2)[/tex]
And if we divide both sides by 2 we got:
[tex] t \geq \frac{-ln(0.2)}{2}[/tex]
[tex] t \geq 0.8047[/tex]
And then we can conclude that the time period with any load would be 0.8047 years.