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:
![f(x) = \mu e^{-\mu x}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cmu%20e%5E%7B-%5Cmu%20x%7D)
In which
is the decay parameter.
The probability that x is lower or equal to a is given by:
![P(X \leq x) = \int\limits^a_0 {f(x)} \, dx](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%20%5Cint%5Climits%5Ea_0%20%7Bf%28x%29%7D%20%5C%2C%20dx)
Which has the following solution:
![P(X \leq x) = 1 - e^{-\mu x}](https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%20%3D%201%20-%20e%5E%7B-%5Cmu%20x%7D)
During lunch hour, customers arrive at a fast food drive-through window, on average, every 3.8 minutes.
This means that ![m = 3.8, \mu = \frac{1}{3.8} = 0.2632](https://tex.z-dn.net/?f=m%20%3D%203.8%2C%20%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B3.8%7D%20%3D%200.2632)
a) What is the probability that the next customer will arrive within the next 3 minutes?
![P(X \leq 3) = 1 - e^{-0.2638*3} = 0.5468](https://tex.z-dn.net/?f=P%28X%20%5Cleq%203%29%20%3D%201%20-%20e%5E%7B-0.2638%2A3%7D%20%3D%200.5468)
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
![P(X \leq 7) + P(X > 7) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%207%29%20%2B%20P%28X%20%3E%207%29%20%3D%201)
We want P(X > 7). So
![P(X > 7) = 1 - P(X \leq 7)](https://tex.z-dn.net/?f=P%28X%20%3E%207%29%20%3D%201%20-%20P%28X%20%5Cleq%207%29)
![P(X \leq 7) = 1 - e^{-0.2638*7} = 0.8422](https://tex.z-dn.net/?f=P%28X%20%5Cleq%207%29%20%3D%201%20-%20e%5E%7B-0.2638%2A7%7D%20%3D%200.8422)
![P(X > 7) = 1 - P(X \leq 7) = 1 - 0.8422 = 0.1578](https://tex.z-dn.net/?f=P%28X%20%3E%207%29%20%3D%201%20-%20P%28X%20%5Cleq%207%29%20%3D%201%20-%200.8422%20%3D%200.1578)
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?
![P(1 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1)](https://tex.z-dn.net/?f=P%281%20%5Cleq%20X%20%5Cleq%206%29%20%3D%20P%28X%20%5Cleq%206%29%20-%20P%28X%20%5Cleq%201%29)
![P(X \leq 6) = 1 - e^{-0.2638*6} = 0.7946](https://tex.z-dn.net/?f=P%28X%20%5Cleq%206%29%20%3D%201%20-%20e%5E%7B-0.2638%2A6%7D%20%3D%200.7946)
![P(X \leq 1) = 1 - e^{-0.2638*1} = 0.2319](https://tex.z-dn.net/?f=P%28X%20%5Cleq%201%29%20%3D%201%20-%20e%5E%7B-0.2638%2A1%7D%20%3D%200.2319)
![P(1 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) = 0.7946 - 0.2319 = 0.5627](https://tex.z-dn.net/?f=P%281%20%5Cleq%20X%20%5Cleq%206%29%20%3D%20P%28X%20%5Cleq%206%29%20-%20P%28X%20%5Cleq%201%29%20%3D%200.7946%20-%200.2319%20%3D%200.5627)
56.27% probability that the next customer will arrive between 1 and 6 minutes