Problem 11 - Calculus Concepts

Question

Waiting Time At a certain grocery checkout counter, the average waiting time is 2.5 minutes. Suppose the waiting times follow an exponential density function. a. Write the equation for the exponential distribution of waiting times. Graph the equation and locate the mean waiting time on the graph. b. What is the likelihood that a customer waits less than 2 minutes to check out? c. What is the probability of waiting between 2 and 4 minutes? d. What is the probability of waiting more than 5 minutes to check out?

Step-by-Step Solution

Answer

Exponential distribution: \( f(x) = 0.4e^{-0.4x} \). Probabilities: less than 2 mins \( \approx 0.55 \), 2-4 mins \( \approx 0.25 \), more than 5 mins \( \approx 0.14 \).

Step 1: Understand the Exponential Distribution

An exponential distribution with mean \( \mu \) is given by the density function \( f(x) = \lambda e^{-\lambda x} \), where \( \lambda = \frac{1}{\mu} \). Here, the mean waiting time is 2.5 minutes, so \( \lambda = \frac{1}{2.5} = 0.4 \). Thus, the equation becomes \( f(x) = 0.4 e^{-0.4x} \).

Step 2: Graph the Distribution and Indicate the Mean

To graph \( f(x) = 0.4 e^{-0.4x} \), plot \( f(x) \) against \( x \) for \( x \geq 0 \). The mean waiting time of 2.5 minutes is at the point where the graph reaches its average height, which helps indicate the distribution of probable waiting times around this mean.

Step 3: Calculate Probability for Less than 2 Minutes

The probability that a customer waits less than 2 minutes is the integral \( \int_0^2 0.4 e^{-0.4x} \, dx \). Evaluating this, \( P(X < 2) = 1 - e^{-0.8} \approx 0.5507 \).

Step 4: Calculate Probability for Between 2 and 4 Minutes

The probability of waiting between 2 and 4 minutes is \( P(2 < X < 4) = \int_2^4 0.4 e^{-0.4x} \, dx \), which evaluates to \( e^{-0.8} - e^{-1.6} \approx 0.2501 \).

Step 5: Calculate Probability for More than 5 Minutes

The probability of waiting more than 5 minutes is \( P(X > 5) = 1 - P(X < 5) = e^{-2} \approx 0.1353 \).

Key Concepts

Probability Calculation

In probability theory, calculating probabilities for a situation modeled by an exponential distribution involves determining the likelihood of different outcomes. This distribution is often used for modeling time intervals between events.
For a grocery checkout scenario, we use this distribution to find out various probabilities:

• The probability a customer waits less than 2 minutes is calculated using the expression \( P(X < 2) = 1 - e^{-0.8} \), which gives us approximately 0.5507.
• The probability of waiting between 2 and 4 minutes involves integrating the function over the interval, which results in \( P(2 < X < 4) = e^{-0.8} - e^{-1.6} \approx 0.2501 \).
• For waiting more than 5 minutes, we find \( P(X > 5) = e^{-2} \approx 0.1353 \), by subtracting the probability of being less or equal to 5 minutes from 1.

This approach allows us to pinpoint probable ranges for waiting times, offering insights into customer service efficiency.

Mean Waiting Time

The mean waiting time in an exponential distribution is a critical value that affects both the rate parameter \( \lambda \) and the probability calculations.
The mean, denoted by \( \mu \), is the expected or average time between events in a process. For our checkout example, the mean is given as 2.5 minutes.

• From this, we calculate the rate \( \lambda \) with the formula \( \lambda = \frac{1}{\mu} \).
• Thus, \( \lambda = \frac{1}{2.5} = 0.4 \), defining how often customers are expected to be served.
• This mean is marked on the graph as the point where the probability density function balances, representing where most events occur.

Understanding the mean helps predict waiting times and manage the checkout process better.

Probability Density Function

The probability density function (PDF) of an exponential distribution expresses the probability of a random variable falling within a particular range. In our checkout scenario, the PDF is essential for visualizing and calculating these probabilities.
The general form of the PDF for exponential distribution is \( f(x) = \lambda e^{-\lambda x} \). Given our mean of 2.5 minutes:

• The rate becomes \( \lambda = 0.4 \), so the function is \( f(x) = 0.4 e^{-0.4 x} \).
• This function characterizes the shape of the distribution, which is asymmetric and declines as \( x \) increases.
• The graph starts at the maximum value of \( 0.4 \) at \( x = 0 \) and falls off, showing higher probabilities for shorter waiting times.

This PDF gives a clear and quantitative means of understanding how the different waiting times are distributed, crucial for logistical planning and decision-making at the checkout. By integrating this function across a range, we determine the exact probability of different waiting time spans.