buses arrive at the bus stop at 15min
interval starting from 7am in the morning. If a
passenger arrived at the bus-stop at a time that
is uniformly distributed between 7 and
7:30am.
what is the probability that the passenger waits less than 5minutes for a bus
The pdf of a uniform distribution is 1/(b-a) where a≤x≤b.
Thus mean,μ=(1/2)(a+b).
To wait less than (or equal to) five minutes means he has to arrive between 7:10 and 7:15, or 7:25 and 7:30, out of the 30 minute interval, which makes P(≤5)=(5+5)/30=1/3.
how is it 5+5?
There are two buses after 7 and on or before 7:30.
The two corresponding periods are between 7:10 and 7:15, or 7:25 and 7:30, which makes 5+5 = 10 minutes.
Equivalently, we can calculate for just one bus/period, which makes
P(<5)=5/15=1/3...again.
To find the probability that the passenger waits less than 5 minutes for a bus, we need to determine the arrival times of the buses and compare them to the passenger's arrival time.
Given that buses arrive at the bus stop at 15-minute intervals starting from 7 am, we need to determine the number of 15-minute intervals that have passed by the time the passenger arrives.
1. Determine the number of 15-minute intervals before the passenger arrives:
- The time window between 7 am and 7:30 am is 30 minutes.
- Each interval is 15 minutes long.
- So, the number of intervals is 30 / 15 = 2 intervals.
2. Calculate the probability of waiting less than 5 minutes:
- For the passenger to wait less than 5 minutes, the bus must arrive either during the 0th interval or the 1st interval (since there are only 2 intervals when the passenger arrives).
- The probability of the bus arriving during the 0th interval is 1 / 2 (50% chance).
- The probability of the bus arriving during the 1st interval is also 1 / 2 (50% chance).
- Therefore, the total probability of waiting less than 5 minutes is 1 / 2 + 1 / 2 = 1 (or 100%).
So, the probability that the passenger waits less than 5 minutes for a bus is 1 (or 100%).