Question

The manager of a gas station has observed that the times required by drivers to fill their car's tank and pay are quite variable. In fact, the times are exponentially distributed with a mean of 7 minutes. What is the probability that a car can complete the transaction in less than 4 minutes?

Answer 1

Answer:

0.4353 = 43.53%

Step-by-step explanation:

The times are exponentially distributed, so let's call n the mean, and x the time of transaction. The times of the transaction occurs in a certain time is:

$e^{-x/n}$

So, the probability that a car complete the transaction in less than 4 minutes, is 1 less the times it does it at 4 minutes:

P = 1 - $e^{-4/7}$

P = 0.4353

P = 43.53%