Problem 26 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

TRAFFIC CONTROL Suppose the time (in minutes) between the arrivals of successive cars at a toll booth is measured by the random variable \(X\) with probability density function $$ f(t)= \begin{cases}0.5 e^{-0.5 t} & \text { if } t \geq 0 \\ 0 & \text { otherwise }\end{cases} $$ a. Find the probability that a randomly selected pair of successive cars will arrive at the toll booth at least 6 minutes apart. b. Find the average time between the arrivals of successive cars at the toll booth.

Step-by-Step Solution

Answer

P(X ≥ 6) is approximately 0.0498; Average time is 2 minutes.

Step 1: Identify the probability density function

The given probability density function is

Step 2: Define the cumulative distribution function (CDF)

To find the probability we need to integrate the PDF to get the CDF. The cumulative distribution function for a continuous random variable can be found by integrating the PDF from the lower limit to a given value of the random variable X.

Step 3: Compute the CDF

The CDF of X, denoted as F(t), can be found by integrating the PDF.

Step 4: Calculate the CDF value at t=6 minutes

Now, calculate the value of the CDF at t=6 minutes to find the probability that the time between successive cars is less than 6 minutes. Then use the complement to find the required probability.

Step 5: Compute the Probability for Part (a)

The required probability is P(X ≥ 6) = 1 - F(6). Compute this value.

Step 6: Calculate the expected value for Part (b)

The average time between arrivals can be found by calculating the expected value of the random variable X.

Step 7: Calculate the average time

To find the expected value E(X), we simplify the integral of t multiplied by its probability density function over the given range.

Key Concepts

cumulative distribution function

The cumulative distribution function (CDF) is a crucial concept when dealing with probability and statistics. It helps to describe the probability that a random variable will take a value less than or equal to a specific value. For the variable in our exercise, the CDF is derived from the probability density function (PDF) provided:

$$ f(t)= \begin{cases}0.5 e^{-0.5 t} & \text { if } t \geq 0 \ 0 & \text { otherwise }\end{cases} $$
To find the CDF, we integrate the PDF from 0 to a specific value of t. Mathematically, this is represented as:

$$ F(t) = \int_0^t f(u) du $$
For the given function, it's:
$$ F(t) = \int_0^t 0.5 e^{-0.5u} du $$
This integral evaluates to:
$$ F(t) = 1 - e^{-0.5t} $$
The CDF allows us to compute probabilities for any range of values, such as in part (a) of our problem. To find the probability that the time between two successive car arrivals is at least 6 minutes ([P(X ≥ 6)]), one can use the fact that:
$$ P(X \geq 6) = 1 - F(6) $$
Substituting in F(6) gives:
$$ P(X \geq 6) = 1 - [1 - e^{-0.5 \cdot 6}] = e^{-3} $$

expected value

The expected value, often denoted as E(X), is essentially the mean or average value of a random variable. It gives a long-term average or central tendency. To compute the expected value for a continuous random variable, we use the integral of the variable multiplied by its PDF:

$$ E(X) = \int_{0}^{\infty} t f(t) dt $$
For our PDF:
$$ f(t)= \begin{cases}0.5 e^{-0.5 t} & \text { if } t \geq 0 \ 0 & \text { otherwise }\end{cases} $$
we compute the expected value as:
$$ E(X) = \int_{0}^{\infty} t \cdot 0.5 e^{-0.5 t} dt $$
Integrating by parts, we obtain:
$$ E(X) = \left. -te^{-0.5t} \right|_{0}^{\infty} + \int_{0}^{\infty} e^{-0.5t} dt $$
Evaluating and simplifying yields:
$$ E(X) = \frac{1}{0.5} = 2 $$
Thus, the average time between the arrivals of successive cars at the toll booth is 2 minutes.

exponential distribution

The exponential distribution is a continuous probability distribution that represents the time between events in a Poisson point process. It's widely used to model scenarios like the one in our exercise. The PDF of an exponential distribution with a rate parameter λ (lambda) is given by:

$$ f(t) = \lambda e^{-\lambda t} $$
In our exercise, the given PDF:
$$ f(t)= \begin{cases}0.5 e^{-0.5 t} & \text { if } t \geq 0 \ 0 & \text { otherwise }\end{cases} $$
indicates an exponential distribution with λ = 0.5. Key properties of the exponential distribution include:

• Memorylessness: The probability of waiting an additional time t is independent of how much time has already elapsed.
• The mean and standard deviation are both equal to \( \frac{1}{\lambda} \).

Understanding this distribution helps us find various probabilities and expected values for the arrival times. Here, the mean arrival time between successive cars is 2 minutes, as computed before.