Answer:
P(X>5) = 0.857
Step-by-step explanation:
Let X [tex]\sim[/tex] uniform(3.17)
[tex]f(x) = \dfrac{1}{17-3} ; \ \ \ 3 \le x \le 17[/tex]
The required probability that it will take Isabella more than 5 minutes to wait for the bus can be computed as:
[tex]P(X > 5) = \int ^{17}_{5} f(x) \ dx[/tex]
[tex]P(X > 5) = \int ^{17}_{5} \dfrac{1}{17-3} \ dx[/tex]
[tex]P(X > 5) =\dfrac{1}{14} \Big [x \Big ] ^{17}_{5}[/tex]
[tex]P(X > 5) =\dfrac{1}{14} [17-5][/tex]
[tex]P(X > 5) =\dfrac{12}{14}[/tex]
P(X>5) = 0.857
