Respuesta :
Answer:
a) 54.68% probability that the next customer will arrive within the next 3 minutes
b) 15.78% probability that the next customer will arrive in more than 7 minutes
c) 56.27% probability that the next customer will arrive between 1 and 6 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]
During lunch hour, customers arrive at a fast food drive-through window, on average, every 3.8 minutes.
This means that [tex]m = 3.8, \mu = \frac{1}{3.8} = 0.2632[/tex]
a) What is the probability that the next customer will arrive within the next 3 minutes?
[tex]P(X \leq 3) = 1 - e^{-0.2638*3} = 0.5468[/tex]
54.68% probability that the next customer will arrive within the next 3 minutes
b) What is the probability that the next customer will arrive in more than 7 minutes?
Either it will arrive in 7 minutes or less, or it will arrive in more than 7 minutes. The sum of the probabilities of these outcomes is decimal 1. So
[tex]P(X \leq 7) + P(X > 7) = 1[/tex]
We want P(X > 7). So
[tex]P(X > 7) = 1 - P(X \leq 7)[/tex]
[tex]P(X \leq 7) = 1 - e^{-0.2638*7} = 0.8422[/tex]
[tex]P(X > 7) = 1 - P(X \leq 7) = 1 - 0.8422 = 0.1578[/tex]
15.78% probability that the next customer will arrive in more than 7 minutes
c) What is the probability that the next customer will arrive between 1 and 6 minutes?
[tex]P(1 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1)[/tex]
[tex]P(X \leq 6) = 1 - e^{-0.2638*6} = 0.7946[/tex]
[tex]P(X \leq 1) = 1 - e^{-0.2638*1} = 0.2319[/tex]
[tex]P(1 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) = 0.7946 - 0.2319 = 0.5627[/tex]
56.27% probability that the next customer will arrive between 1 and 6 minutes
