1 votes 1 votes Question A subway train in a certain line runs after every half hour, between every midnight and 6 in the morning. What is the probability that a man entering the station at random will have to wait at least 20 minutes? I'm stuck here... It can be solved using uniform distribution method . But 1) I'm bit confused about the limits ? X denote the RV waiting time of man for the next train given that man arrives at random time So is probability has to be computed for P(X>20 2) What happens if man arrives of a day time after 6 in the morning ? in this case what is the upper limit of the integral ? URL : https://www.assignmentexpert.com/homework-answers/Math-Answer-40654.pdf Probability probability uniform-distribution + – pC asked Jun 17, 2016 pC 6.8k views answer comment Share Follow See 1 comment See all 1 1 comment reply Pranav Kant Gaur commented Jun 20, 2016 reply Follow Share Applying simple logic and assuming resolution of time to be in minutes, shouldn't the answer be 0.3?....because say between 12:00 - 12:30 AM, there are 9 time instants which result in at least 20 minutes of waiting time...this results in total of 18 * 6 such time instants between 12:00 AM to 6:00 AM. Therefore, probability of waiting time exceeding 20 minutes = (18 * 6)/(6 * 60) = 3 / 10 = 0.3. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes It is a Uniform distribution on [0,30] a=0, b= 30, f(x) = 1/(b-a) = 1/30 here, if a person comes between 0 to 10, then he has to wait for at least 20 minutes. so, P(X<=10) = $\int_{0}^{10}$(1/30)dx = 1/30(10-0) = 1/3 Ami Ladani answered Feb 4, 2017 • edited Feb 4, 2017 by Ami Ladani Ami Ladani comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes (1). I think it should be $P(30 > X \geq 20)$, because questions asks for atleast 20 minutes and train arrives every 30 minutes, so $X < 30$. (2). I suspect this possibility is out of scope of this problem because train arrives only between 12:00 AM to 6:00 AM, but maybe I am missing your point here. As described in the referred solution, this problem is reduced to the problem in 30 minutes(instead of its original duration between 12:00 - 6:00) because of its periodic nature(train arriving every 30 minutes). Pranav Kant Gaur answered Jun 20, 2016 Pranav Kant Gaur comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes As here it is mentioned between midnight and 6 am in the morning, we only consider that period Now man entering the station have to wait minimum 20 mins. So, the man have to enter maximum between 0 to 10 mins. after leaving the first train Probability of it have to wait atleast 20 mins (302 -1/2*2*(202)) / 302 =5/9 srestha answered Jul 9, 2016 srestha comment Share Follow See 1 comment See all 1 1 comment reply ankit commented Jul 10, 2016 reply Follow Share how u got (302 -1/2*2*(202)) / 302 ??? 0 votes 0 votes Please log in or register to add a comment.