Question

The length of time it takes to drive from Rexburg to CEI (assuming no construction delays!) is uniformly distributed over the interval of 30.0 minutes and 37.0 minutes.

(a) If one of the drives is randomly selected, find the probability that the drive takes between between 31.3 and 34.9 minutes.

(b) What is the average time the drive takes?

(c) Based on the uniform distribution described in this question, what is the probability that a drive will take longer than 37 minutes?

Answer 1

Solution :

Let X be the length of time it takes to drive from Rexburg to CEI (assuming no construction delays!), so it is given that X is uniformly distributed over the interval of 30.0 minutes and 37.0 minutes.

The pdf of X is given by,

f(x) = 1 / (37 - 30) = 1 / 7 ; for 30 <= x <= 37 and

f(x) = 0 ; otherwise

(a). For an Uniformly distributed random variable X the probability that X lies between points "a" and "b" is given by,

P(a <= X <= b) = (b - a) / 7 ; for 30 <= x <= 37.

If one of the drives is randomly selected, then the probability that the drive takes between between 31.3 and 34.9 minutes is, P(31.3 <= X <= 34.9) = (34.9 - 31.3) / 7 = 3.6 / 7 = 0.5143.

(b). The average for the uniform distribution is given by, (37 + 30) / 2 = 67 / 2 = 33.5 minutes

(c). Since the pdf of uniform distribution vanishes outside the interval (30, 37), the f(x) = 0 ; for x < 30 and x > 37.

So the probability that a drive will take longer than 37 minutes is given by, P(X > 37) = 0.