Question

When crossing the Golden Gate Bridge traveling into San Francisco, all drivers must pay a toll. Suppose the amount of time (in minutes) drivers wait in line to pay the toll follows an exponential distribution with a probability density function of f(x) = 0.2e−0.2x.

a. What is the mean waiting time that drivers face when entering San Francisco via the Golden Gate Bridge?

b. What is the probability that a driver spends more than the average time to pay the toll? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

c. What is the probability that a driver spends more than 10 minutes to pay the toll? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.) d. What is the probability that a driver spends between 4 and 6 minutes to pay the toll? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

Answer 1

Answer:

Given,

f(x)  = 0.2^e − 0.2x

.a)

To determine the mean waiting time that drivers face when entering San Francisco via the Golden Gate Bridge

Let us consider the exponential probability density function

i.e.,

f(x) = e^-x

CDF of exponential distribution is F(x) = 1 - e^-x

Mean = 1/

From the given function

= 0.2

So mean = 1/0.2

Mean = 5

b)

To give the probability that a driver spends more than the average time to pay the toll

P(x > 5) = e^-5

= e^-5*0.2

= e^-1

P(x > 5) = 0.3679

c)

To give the probability that a driver spends more than 10 minutes to pay the toll

P(X > 10) = e^-10

= e^-10*0.2

= e^-2

P(X > 10) = 0.1353

d)

To give the probability that a driver spends between 4 and 6 minutes to pay the toll

P(4 < X < 6) = P(X < 6) - P(X < 4)

= (1 - e^-6) - (1 - e^-4)

= e^-4 - e^-6

= e^-4*0.2 - e^-6*0.2

= e^-0.8 - e^-1.2

= 0.4493 - 0.3012

P(4 < X < 6) = 0.1481