Problem 9 - Calculus Concepts

Question

Waiting Time A traffic light on campus remains red for 30 seconds at a time. A car arrives at that light and finds it red. Assume that the waiting time \(t\) seconds at the light follows a uniform density function \(f\). a. Calculate the car's chances of waiting at least 10 seconds at the red light. b. Calculate the probability of waiting no more than 20 seconds at the red light. c. What is the average expected wait time?

Step-by-Step Solution

Answer

a. Probability = \(\frac{2}{3}\). b. Probability = \(\frac{2}{3}\). c. Average wait time = 15 seconds.

Step 1: Define the Uniform Distribution

Since the waiting time at the red light is uniformly distributed between 0 and 30 seconds, we define it as \( f(t) = \frac{1}{{b-a}} \) where \( a = 0 \) and \( b = 30 \). Therefore, the probability density function (pdf) \( f(t) = \frac{1}{30} \) for \( 0 \leq t \leq 30 \).

Step 2: Calculate Probability of Waiting at Least 10 Seconds

We want the probability that the waiting time is at least 10 seconds, i.e., \( P(t \geq 10) \). Using the uniform distribution, this probability can be calculated as the area under the pdf from 10 to 30. This is \( \frac{{b - 10}}{{b-a}} = \frac{{30 - 10}}{{30 - 0}} = \frac{20}{30} = \frac{2}{3} \).

Step 3: Calculate Probability of Waiting No More than 20 Seconds

The probability that the waiting time is no more than 20 seconds is \( P(t \leq 20) \). This is the area under the pdf from 0 to 20, which is \( \frac{{20 - 0}}{{30 - 0}} = \frac{20}{30} = \frac{2}{3} \).

Step 4: Calculate Average Expected Wait Time

The average expected wait time is the mean of the uniform distribution, given by \( \frac{{a+b}}{2} \). Substituting the values, we get \( \frac{{0+30}}{2} = 15 \) seconds.

Key Concepts

Probability Density Function

In a uniform distribution, every outcome within a defined range is equally likely to occur. In the context of the given traffic light problem, we rely on the Probability Density Function (PDF) to understand these probabilities. The waiting time at the red light is distributed between 0 to 30 seconds as a uniform distribution.
The PDF is mathematically expressed as: \( f(t) = \frac{1}{b-a} \).
Here, \( a \) and \( b \) are the minimum and maximum bounds of the distribution, respectively. Substituting the values from the exercise:

• \( a = 0 \)
• \( b = 30 \)

Thus, the PDF becomes \( f(t) = \frac{1}{30} \) for \( 0 \leq t \leq 30 \).
This constant value signifies that the probability of waiting any specific number of seconds is evenly distributed across the 30-second interval.

Expected Value

The expected value in statistics is what you would anticipate as the average or "mean" outcome given a certain probability distribution. In the scenario of uniform distribution, it serves as a useful metric to predict central tendency over numerous trials.
For the traffic light problem, the expected value, which represents average waiting time, is computed using the formula for the mean of a uniform distribution: \( \frac{a+b}{2} \).
Applying the specific values of our problem:

• \( a = 0 \)
• \( b = 30 \)

The calculation reveals that the expected waiting time is \( \frac{0+30}{2} = 15 \) seconds.
This average indicates that, over many repeated instances of this problem, the car would spend about 15 seconds waiting at the light.

Waiting Time Problem

The waiting time problem in probability contexts typically examines how long an event is expected to last under certain conditions. Our specific dilemma revolves around a traffic light, where the focus is on the time a car has to wait when it encounters a red signal. Utilizing uniform distribution helps in assessing the probability of various waiting times.
In this exercise, we calculate two different probabilities:

• The likelihood of waiting at least 10 seconds, which involves calculating the area under the PDF from 10 to 30 seconds. The result is \( P(t \geq 10) = \frac{2}{3} \).
• The chance of waiting no more than 20 seconds, which requires evaluating the area from 0 to 20 seconds, yielding \( P(t \leq 20) = \frac{2}{3} \).

Uniform distributions simplify these calculations, making them straightforward by focusing on the range's endpoints.
Understanding these concepts helps grasp how uniform distribution can address practical scenarios like gauging waiting periods at traffic lights.