Respuesta :
Answer:
a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?
We want this probability"
[tex] P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48[/tex]
At most 3 hours?
[tex] P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667[/tex]
b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?
[tex] P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X<13.62) = 1-(1-e^{-0.367*13.62})=e^{-0.367*13.62}=0.00675[/tex]
What is the probability that it is less than the mean value by more than one standard deviation?
[tex] P(X<2.725-2.725) =P(X<0) = 1- e^{-0.367*0}= 1-1 =0[/tex]
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 cumulative distribution for this function is given by:
[tex] F(X) = 1- e^{-\lambda x}, x\ geq 0[/tex]
We know the value for the mean on this case we have that :
[tex] mean = \frac{1}{\lambda}[/tex]
[tex] \lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367[/tex]
Solution to the problem
Part a
What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?
We want this probability"
[tex] P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48[/tex]
At most 3 hours?
[tex] P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667[/tex]
Part b
What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?
The variance for the esponential distribution is given by: [tex] Var(X) =\frac{1}{\lambda^2}[/tex]
And the deviation would be:
[tex] Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725[/tex]
And the mean is given by [tex] Mean = 2.725[/tex]
Two deviations correspond to 5.540, so we want this probability:
[tex] P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X<13.62) = 1-(1-e^{-0.367*13.62})=e^{-0.367*13.62}=0.00675[/tex]
What is the probability that it is less than the mean value by more than one standard deviation?
For this case we want this probablity:
[tex] P(X<2.725-2.725) =P(X<0) = 1- e^{-0.367*0}= 1-1 =0[/tex]
