Answer:
a) P=1/3
b) P=1/3
Step-by-step explanation:
We know that buses arrive at a specified stop at 15-minute intervals starting at 7A.M. Therefore, ,they arrive at 7, 7:15 ,7:30 ,7:45,and so on.
a) We calculate the probability that a passenger waits less than 5 minutes for a bus between 7 and 7:30. We know that a passenger arrives at the stopat a time that is uniformly distributed between 7 and 7:30, we have:
[tex]f(x)=\left \{ {{\frac{1}{30}\, ,\, \, \, 0\leq x \leq 30} \atop {0,\, \, \, \, \, otherwise}} \right.[/tex]
We conclude that the passenger will wait less than 5 minutes for a bus if he arrives between 7:10 and 7:15, or between 7:25 and 7:30. We get:
[tex]P=\int\limits^{15}_{10} {\frac{1}{30}} \, dx + \int\limits^{30}_{25} {\frac{1}{30}} \, dx\\\\ P=\frac{1}{30} [x]^{15}_{10}+\frac{1}{30} [x]^{30}_{25}\\\\P=\frac{1}{30}(15-10)+\frac{1}{30}(30-25)\\\\P=\frac{1}{3}[/tex]
Therefore, the probability is P=1/3.
b) We conclude that the passenger will wait more than 10 minutes for a bus if he arrives between 7:00 and 7:05, or between 7:15 and 7:20. We get:
[tex]P=\int\limits^{5}_{0} {\frac{1}{30}} \, dx + \int\limits^{20}_{15} {\frac{1}{30}} \, dx\\\\ P=\frac{1}{30} [x]^{5}_{0}+\frac{1}{30} [x]^{20}_{15}\\\\P=\frac{1}{30}(5-0)+\frac{1}{30}(20-15)\\\\P=\frac{1}{3}[/tex]
Therefore, the probability is P=1/3.
