Respuesta :
Answer:
a) 0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.
b) 240.79 cfs.
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
In this problem, we have that:
[tex]m = 100, \mu = \frac{1}{100} = 0.01[/tex]
a. Find the probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.
This is [tex]P(X > 120)[/tex]
[tex]P(X > 120) = e^{-0.01*120} = 0.3012[/tex]
0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.
b. What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.09?
We want x for which
[tex]P(X > x) = 0.09[/tex]
So
[tex]e^{-0.01x} = 0.09[/tex]
[tex]\ln{e^{-0.01x}} = \ln{0.09}[/tex]
[tex]-0.01x = \ln{0.09}[/tex]
[tex]0.01x = -\ln{0.09}[/tex]
[tex]x = -\frac{\ln{0.09}}{0.01}[/tex]
[tex]x = 240.79[/tex]
So 240.79 cfs.
