Respuesta :
Answer:
The advertisement should use 16 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]
The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.
This means that [tex]m = 3.5, \mu = \frac{1}{3.5} = 0.2857[/tex]
What number of minutes should the advertisement use?
The values of x for which:
[tex]P(X > x) = 0.01[/tex]
So
[tex]e^{-\mu x} = 0.01[/tex]
[tex]e^{-0.2857x} = 0.01[/tex]
[tex]\ln{e^{-0.2857x}} = \ln{0.01}[/tex]
[tex]-0.2857x = \ln{0.01}[/tex]
[tex]x = -\frac{\ln{0.01}}{0.2857}[/tex]
[tex]x = 16.12[/tex]
Rounding to the nearest number, the advertisement should use 16 minutes.
