Question

The commute time for people in a city has an exponential distribution with an average of 0.66 hours. What is the probability that a randomly selected person in this city will have a commute time between 0.55 and 1.1 hours? Answer: (round to 3 decimal places)

Answer 1

`P(0.55 <X<1.1)= F(1.1) -F(0.55)`

And replacing we got:

`P(0.55 <X<1.1)= (1-e^{-(1)/(0.66) *1.1}) -(1-e^{-(1)/(0.66) *0.55})`

`P(0.55 <X<1.1)=e^{-(1)/(0.66) *0.55}- e^{-(1)/(0.66) *1.1}=0.2457`

And rounded the answer would be 0.246

Explanation:

For this case we can define the random variable X as "The commute time for people in a city" and for this case the distribution of X is given by:

`X \sim exp (\lambda = (1)/(0.66)= 1.515)`

And for this case we want to find the following probability:

`P(0.55 <X<1.1)`

And we can use the cumulative distribution function given by:

`F(x) =1- e^(-\lambda x)`

And using this formula we got:

`P(0.55 <X<1.1)= F(1.1) -F(0.55)`

And replacing we got:

`P(0.55 <X<1.1)= (1-e^{-(1)/(0.66) *1.1}) -(1-e^{-(1)/(0.66) *0.55})`

`P(0.55 <X<1.1)=e^{-(1)/(0.66) *0.55}- e^{-(1)/(0.66) *1.1}=0.2457`

And rounded the answer would be 0.246