Respuesta :
Answer:
a) 0.202 = 20.2% probability that a customer has to wait more than 4 minutes.
b) 0.33 = 33% probability that a customer is served within the first minute.
c) 11.5 minutes.
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 question:
[tex]m = 2.5, \mu = \frac{1}{2.5} = 0.4[/tex]
(a) Find the probability that a customer has to wait more than 4 minutes.
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
[tex]P(X > 4) = e^{-0.4*4} = 0.202[/tex]
0.202 = 20.2% probability that a customer has to wait more than 4 minutes.
(b) Find the probability that a customer is served within the first minute.
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
[tex]P(X \leq 1) = 1 - e^{-0.4*1} = 0.33[/tex]
0.33 = 33% probability that a customer is served within the first minute.
(c) The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she doesn't want to give away free hamburgers to more than 1% of her customers. What number of minutes should the advertisement use
We have to find x for which:
[tex]P(X > x) = 0.01[/tex]
So
[tex]P(X > x) = e^{-0.4x}[/tex]
Then
[tex]e^{-0.4x} = 0.01[/tex]
[tex]\ln{e^{-0.4x}} = \ln{0.01}[/tex]
[tex]-0.4x = \ln{0.01}[/tex]
[tex]x = -\frac{\ln{0.01}}{0.4}[/tex]
[tex]x = 11.5[/tex]
So 11.5 minutes.
