Question

Part 3: The Uniform Distribution Suppose that you need to take a bus that comes every 30 minutes. Assume that the amount of time you have to wait for this bus has a uniform distribution between 0 and 30 minutes. The probability density curve for this distribution is given below. Distribution of Bus Wait Time Wait Time (min.) 1) 2) 3) 4) 5) 6) Is waiting time a discrete or continuous random variable? What is the area of this entire rectangle? What numbers are represented by a, b and c (note: c is the height of the rectangle)? What is the probability of waiting between 10 and 15 minutes? What is the probability of waiting between 5 and 10 minutes or between 20 and 30 minutes? What is the probability of waiting exactly 12 minutes?

Part 3: The Uniform Distribution

Suppose that you need to take a bus that comes every 30 minutes. Assume that the amount of time you have to wait for this bus has a uniform distribution between 0 and 30 minutes. The probability density curve for this distribution is given below.

1) Is waiting time a discrete or continuous random variable?

2) What is the area of this entire rectangle?

3) What numbers are represented by a, b and c (note: c is the height of the rectangle)?

4) What is the probability of waiting between 10 and 15 minutes?

5) What is the probability of waiting between 5 and 10 minutes or between 20 and 30 minutes?

6) What is the probability of waiting exactly 12 minutes?

Answer 1

Let X denote the waiting time. By, the question, X ~ U(0, 30). The probability density function of X is given by,

f(x) = (1/30).I(x) where I(x) = 1 if 0<x<30, I(x) = 0, otherwise.

1. The waiting time X is continuously distributed.

2. The area of this rectangle is actually the area under the curve of f(x) and x-axis bounded by x=0 and x=30. From the properties of a probability density function, this area is 1 (unit minute^2).

3. a=0, b=30, c=1/30.

4. P(X<x) = $\int_{0}^{x}f(u)du=\int_{0}^{x}1/30 du$ = x/30 if 0<x<30.

P(10<X<15)=P(X<15)-P(X<10)=(15/30)-(10/30)=1/6.

5. Similarly, P(5<X<10)=5/30=1/6. P(20<X<30)=10/30=1/3.

6. Required probability = P(X=12) = 0 since X is continuous.