Buses arrive at a specified stop at 15-minute intervals starting at 7 a. m. that is, they arrive at 7, 7: 15, 7: 30, 7: 45, and so on. if a passenger arrives at the stop at a time that is uniformly distributed between 7 and 7: 30, find the probability that he waits

(a) less than 5 minutes for a bus
(b) more than 10 minutes for a bus.

The answers are a) 1/3 and b) 1/3

Step-by-step explanation:

we will consider t to be the arrival time variable in minutes. We will have it run from 0 min to 30 min omitting the 7 hours , which won't change the results since the PDF we are about to calculate is a constant.

So the PDF of the arrival time is a constant and the since the area under this PDF distribution should be equal to 1 so, Let the height of the constant distribution equal to c, so c*30 (which is equal to the total probability) would be the area under the distribution, but this area should be equal to 1, So that gives us c=1/30 which is the value of the constant PDF for all corresponding arrival times.

part a) of the question asks for the probability that the passengers wait less than 5 minute. The passengers would have to wait for the bus 5 min or late if they arrive between the times (10 - 15 )min and (25  - 30) min. So we will have to integrate the PDF corresponding to these times and then we will will just have to add the probabilities calculated as give below,

Now part b) of the question asks for the probability that the passenger waits for more than 10 minutes. Which can be calculated by noting that that can only happen if the passenger arrives between the times (0 - 5) min and (15 - 20) min. So we will have to integrate the PDF corresponding to these times and then we will will just have to add the probabilities calculated as give below,

The answer to part b) is 1/3.

cut 138 centimeters off

step-by-step explanation:

250-112 = 138

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation: