We know that the probability being between two values, [latex]x_1[/latex] and [latex]x_2[/latex] is:
\[P(x_1 \le x \le x_2) = (x_2 – x_1) \times \frac{1}{b-a} = \frac{x_2 – x_1}{b-a} \]

Where the following is true:

• [latex]a[/latex] = minimum (lowest) value
• [latex]b[/latex] = maximum (highest) value
• [latex]x_1[/latex] = lowest [latex]x[/latex] value
• [latex]x_2[/latex] = highest [latex]x[/latex] value

Bus Arrival Times (Example)

Let us look at some examples below that go through calculating the probabilities of being between [latex]x[/latex]-values.

Example 34.1.1

Problem Setup: During peak periods, the times between 130 busses arriving at BCIT follow a continuous uniform distribution with a minimum of 2 minutes and a maximum of 12 minutes.

Question: What is the probability that it takes between 5 to 10 minutes for the next bus to arrive?

Solution: This gives the following values:

• [latex]a=2[/latex]
• [latex]b=12[/latex]
• [latex]x_1 = 5[/latex]
• [latex]x_2 = 10[/latex]

Plugging this all into the formula gives: \[ P(5\le x \le 10) = \frac{x_2-x_1}{b-a} = \frac{10-5}{12-2} = 0.5 \]

Conclusion: There is a 50% chance that the 130 bus will take between 5 to 10 minutes to arrive.

Bus Arrival Times Example Continued (Video)

Let us now change up the problem slightly and show the solution in a video (see below).

Example 34.1.2

Problem Setup: Your friend was waiting for you at the bus stop and texted that you JUST missed the bus. You are hoping to join them to watch a movie. If you can catch a bus in the next 10 minutes, you will make it the movie on time.

• It takes you 2 minutes to pack up your stuff.
• It takes 1 minute to run to the bus stop.
• The times between busses are between 2 and 12 minutes
• The times follow a continuous uniform distribution

Question: What is the probability of making it to the movie on time?

Solution: Click here to download the written solutions. Also, see the video below:

Key Takeaways (EXERCISE)

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