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 7.5 minutes. What is the probability that a car can complete the transaction in less than 5 minutes?

Answer 1

Answer: 0.49

Explanation:

The mean= E[x]=1/l= 7.5 min

l=1/7.5 =10/75=2/15

The exponential distribution's probability density function is looking like that:

F(x)= l*e^(-l*x) = 2/15*e^(-2*x/15)

P(0<x<5)= e^(-0*2/15)- e^(-5*2/15)=1- e^(-2/3)= approx 0.49