Suppose you are at a train station in Connecticut. Trains headed to New York city arrive at the station at 15 minute intervals starting at 7:00 am, whereas trains headed to Boston arrive at 15 minute intervals starting at 7:05 am. Required:
a. If you arrive at a time uniformly distributed between 7 and 8 am, and then get on the first train that arrives (you love both cities!), what proportion of time do you end up in New York city?
b. What if you arrive at a time uniformly distributed between 7:10 and 8:10 am?

The answer is below

Step-by-step explanation:

a) For the person to get to New York city, the person must arrive at a time after the train to Boston has left and before the train to New York leaves. Between 7 and 8 am, there is a total of 60 minutes. For the person end up in New York city he has to arrive at the train station between the following intervals:

7:05 to 7:15 or 7:20 to 7:30 or 7:35 to 7:45 or 7:50 to 8:00

Let X represent the minutes the passenger arrives between 7 am to 8 am, hence:

P(goes to New York) = P(5 < X < 15) + P(20 < X < 30) + P(35 < X < 45) + P(50 < X < 60) = 10/60 + 10/60 + 10/60 + 10/60 = 40/60 = 2/3

b) As explained in part A, Let X represent the minutes the passenger arrives between 7 am to 8:10 am. But since time uniformly distributed between 7:10 and 8:10 am, this means 10 < X < 70:

For the person end up in New York city he has to arrive at the train station between the following intervals:

7:10 to 7:15 or 7:20 to 7:30 or 7:35 to 7:45 or 7:50 to 8:00 or 8:05 to 8:10

P(goes to New York) = P(10 < X < 15) + P(20 < X < 30) + P(35 < X < 45) + P(50 < X < 60) + P(65< X <70)= 5/60 + 10/60 + 10/60 + 10/60 + 10/60 + 5/60= 40/60 = 2/3

probably the last option

step-by-step explanation: