Continuous Random Variables we are working with commonly use

Continuous Random Variables

we are working with commonly used discrete and continuous distribution

You are running a 5km race. The time limit for the race is 50 minutes, after which you are disqualified. Assume that the random variable T for the minutes it takes a runner to finish the race is uniformly distributed over [25, 50]. E(X) and V(X); What is the probability that a runner takes less than 45 minutes to finish, given that he took more than 30 minutes?

Solution

a)

Note that in a uniform distribution from [a,b],

E(x) = (a+b)/2 = (25+50)/2 = 37.5 [ANSWER]

var(x) = (b-a)^2/12 = (50-25)^2/12 = 52.08333333 [ANSWER]

**********************
B)

P(x<45|x>30) = P(30<x<45)/P(x>30)

Here, as the distribution is only up to 50,

P(x>30) = P(30<x<50) = (50-30)/(50-25) = 0.8

also,

P(30<x<45) = (45-30)/(50-25) = 0.6

Thus,

P(x<45|x>30) = P(30<x<45)/P(x>30) = 0.6/0.8 = 0.75 [ANSWER]