Question

When a bus service reduces fares, a particular trip from New York City to Albany, New York, is very popular. A small bus can carry four passengers. The time between calls for tickets is exponentially distributed with a mean of 30 minutes. Assume that each caller orders one ticket. What is the probability that the bus is filled in less than 3 hours from the time of the fare reduction

Answer 1

Answer:

The probability that the bus is filled in less than 3 hours from the time of the fare reduction is 0.9975.

Step-by-step explanation:

We are given that the time between calls for tickets is exponentially distributed with a mean of 30 minutes. Assume that each caller orders one ticket.

Let X = the time between calls for tickets

The probability distribution for the exponential distribution is given by;

$f(x) = \lambda e^{-\lambda x}$ ,  where x > 0.

Here, $\lambda$ = exponential parameter.

As we are given in the question that the mean of the exponential distribution is 30 minutes, i.e;

Mean of an exponential distribution =  $\frac{1}{\lambda}$

                        $30 = \frac{1}{\lambda}$

                         $\lambda=\frac{1}{30}$

For calculating the required probability, we have to use the cumulative distribution function of an exponential distribution which is given by;

$F(x) = P(X \leq x) = 1-e^{-\lambda x}$ ;  where x > 0

Now, the probability that the bus is filled in less than 3 hours from the time of the fare reduction is given by = P(X < 180 minutes)

          P(X < 180 min) =  $1 - e^{-\frac{1}{30} \times 180 }$

                                   =  $1 - e^{-6 }$

                                   =  0.9975