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.
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
3 answers
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.
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.
1 answer