Respuesta :
Answer:
(a) [tex]\frac{1}{4}[/tex]
(b) [tex]\frac{1}{2}[/tex]
Step-by-step explanation:
Let X denote the number of minutes starting at 7 A.M. that the passenger arrives at the stop.
Since X is a uniform random variable over the interval (0, 60), it follows that the passenger will have to wait less than 5 minutes if (and only if) he arrives between 7:15 and 7:20 or between 7:35 and 7:40 or between 7:55 and 8:00
Hence, probability is calculated as follows:
[tex]P\left ( 15<X<20 \right )+P\left ( 35<X<40 \right )+P\left ( 55<X<60 \right )\\=\int_{15}^{20}\frac{1}{60}\,dx+\int_{35}^{40}\frac{1}{60}\,dx+\int_{55}^{60}\frac{1}{60}\,dx\\=\frac{5}{60}+\frac{5}{60}+\frac{5}{60}\\=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\\=\frac{3}{12}\\=\frac{1}{4}[/tex]
Since X is a uniform random variable over the interval (0, 60), it follows that the passenger will have to wait for more than 10 minutes for a bus if (and only if) he arrives between 7:00 and 7:10 or between 7:20 and 7:30 or between 7:40 and 7:50
Hence, probability is calculated as follows:
[tex]P\left ( 0<X<10 \right )+P\left ( 20<X<30 \right )+P\left ( 40<X<50\right )\\=\int_{0}^{10}\frac{1}{60}\,dx+\int_{20}^{30}\frac{1}{60}\,dx+\int_{40}^{50}\frac{1}{60}\,dx\\=\frac{10}{60}+\frac{10}{60}+\frac{10}{60}\\=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\\=\frac{3}{6}\\=\frac{1}{2}[/tex]
