Answer:
a) 0.4
b) 0.6
Step-by-step explanation:
We are given the following information in the question:
The time students wait for a bus can be described by a uniform random variable X, where X is between 5 minutes and 75 minutes.
a = 5, b = 75
Then,
[tex]f(x) = \displaystyle\frac{1}{b-a} = \frac{1}{75-5} = \frac{1}{70}[/tex]
We are given a uniform distribution.
a) P(student will wait between 5 and 33 minutes)
P( 5 < x < 33)
[tex]=\displaystyle\int_{33}^{5} f(x) dx\\\\=\displaystyle\int_{33}^{5} \frac{1}{70}dx\\\\=\frac{1}{70}[x]_{33}^{5} = \frac{1}{70}(33-5) = 0.4[/tex]
b) P(a student will have to wait at least 33 minutes)
P(33 < x < 75)
[tex]=\displaystyle\int_{75}^{33} f(x) dx\\\\=\displaystyle\int_{75}^{33} \frac{1}{70}dx\\\\=\frac{1}{70}[x]_{75}^{33} = \frac{1}{70}(75-33) = 0.6[/tex]