Answer :
Answer:
23.97% probability that a customer has to wait more than 5 minutes.
Step-by-step explanation:
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 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[/tex]. So
[tex]\mu = \frac{1}{m} = \frac{1}{3.5} = 0.2857[/tex]
Find the probability that a customer has to wait more than 5 minutes.
Either the customer has to wait for 5 minutes or less, or he has to wait for more than 5 minutes. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 5) + P(X > 5) = 1[/tex]
We want [tex]P(X > 5)[/tex]. So
[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]
In which
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
[tex]P(X \leq 5) = 1 - e^{-0.2857*5} = 0.7603[/tex]
[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.7603 = 0.2397[/tex]
23.97% probability that a customer has to wait more than 5 minutes.