Answer:
P(X > 50.3) = 0.85
Step-by-step explanation:
Uniform probability distribution:
An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
The probability of finding a value between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
The probability of finding a value above x is:
[tex]P(X > x) = \frac{b - x}{b - a}[/tex]
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.
This means that [tex]a = 50, b = 52[/tex]
If one such class is randomly selected, find the probability that the class length is more than 50.3 min.
[tex]P(X > 50.3) = \frac{52 - 50.3}{52 - 50} = 0.85[/tex]
So
P(X > 50.3) = 0.85
