Problem 8 - Applied Calculus

Question

Let \(p(t)=0.1 e^{-0.1 t}\) be the density function for the waiting time at a subway stop, with \(t\) in minutes, \(0 \leq t \leq 60\). (a) Graph \(p(t)\). Use the graph to estimate visually the median and the mean. (b) Calculate the median and the mean. Plot both on the graph of \(p(t)\). (c) Interpret the median and mean in terms of waiting time.

Step-by-Step Solution

Answer

The median waiting time is 6.93 minutes, and the mean is 10 minutes.

Step 1: Understanding the Problem

We are given a probability density function (pdf) for the waiting time at a subway stop: \( p(t) = 0.1 e^{-0.1 t} \) for \( 0 \leq t \leq 60 \). We need to graph this function and find both the median and mean waiting times, then interpret these values.

Step 2: Graphing the Density Function

Plot the function \( p(t) = 0.1 e^{-0.1 t} \) on a graph over the interval \([0, 60]\). This is an exponential decay function, starting at \( p(0) = 0.1 \) and decreasing towards \( p(60) \approx 0 \). Visually, the curve slopes downward, and we can estimate that the median is somewhere around the middle of the first half of the interval, and the mean will depend on the tail.

Step 3: Calculating the Median

The median of a continuous probability distribution is the value of \( t \) such that the integral of the probability density function from 0 to \( t \) is 0.5. Solve for \( t_m \) from \( \int_{0}^{t_m} 0.1 e^{-0.1t} \, dt = 0.5 \). This integral calculates to \( [-e^{-0.1t}]_{0}^{t_m} = 0.5 \), which simplifies to \( 1 - e^{-0.1t_m} = 0.5 \). Solving gives \( e^{-0.1t_m} = 0.5 \), so \( t_m = -10 \ln(0.5) \approx 6.93 \) minutes.

Step 4: Calculating the Mean

The mean of a continuous probability distribution is given by \( \int_{0}^{60} t \cdot 0.1 e^{-0.1t} \, dt \). This requires integration by parts, where \( u = t \) and \( dv = 0.1 e^{-0.1t} \, dt \). The mean is found to be 10 minutes after evaluating the integral from 0 to 60.

Step 5: Plotting and Interpreting

Mark \( t_m = 6.93 \) for the median and the mean \( \mu = 10 \) minutes on the graph. The median of 6.93 minutes represents the point where 50% of people wait less. The mean of 10 minutes indicates the average wait time, showing a slightly right-skewed distribution due to the exponential decay.

Key Concepts

Exponential Decay

An exponential decay function is a mathematical concept that describes a process which decreases rapidly at first, and then slows over time. This behavior is visualized in curves that slope downward as you move from left to right across a graph. The function given in the exercise, \( p(t) = 0.1 e^{-0.1 t} \), serves as an example of an exponential decay. Here, the parameter inside the exponent, \(-0.1\), determines how quickly the function decays. It's important to note that the starting value at \( t = 0 \) is 0.1, and it declines over the 60-minute range. By the time \( t = 60 \), the value of \( p(t) \) is very close to zero, but never actually reaches it. This characteristic is typical in real-world processes, such as radioactive decay or cooling of items. These functions are handy in predicting how fast a process will diminish from its initial state.

Continuous Probability Distribution

A continuous probability distribution is a function that describes the probabilities of all possible outcomes of a continuous random variable. Unlike discrete distributions, which work with specific values, continuous distributions concern ranges of values. The density function provided in this exercise, \( p(t) = 0.1 e^{-0.1 t} \), represents such a distribution for the time waiting at a subway stop. Here, the function defines the likelihood of any given waiting time within the interval \([0, 60]\) minutes.

• Continuous distributions are usually represented by a smooth curve rather than a set of individual probabilities.
• The area under the curve signifies the total probability, which always equals 1.

In practical terms, this means if you integrated the function from \( t = 0 \) to \( t = 60 \), it would result in a total probability of 100%. This coverage of all possible waiting times allows the density function to describe all scenarios in that range.

Median Calculation

The median of a continuous probability distribution is the value that splits the area under the density function into two equal parts. Practically, this means that 50% of waiting times are less than the median, whereas the other 50% are more. In the given problem, the median \( t_m \) is calculated using the integral \( \int_{0}^{t_m} 0.1 e^{-0.1t} \, dt = 0.5 \). Solving this integral, we conclude that the median waiting time \( t_m \) is approximately 6.93 minutes.

• Using integration to find the median is typical in continuous distributions.
• This value is vital as it provides a clear threshold of where the middle of the distribution lies.

Knowing the median helps understand the distribution of waiting times, indicating that almost half of the people waited less than about 7 minutes.

Mean Calculation

The mean of a continuous probability distribution is the average value it takes. To find the mean waiting time for this subway problem, we integrate to compute \( \int_{0}^{60} t \cdot 0.1 e^{-0.1t} \, dt \). The calculation results in a mean of 10 minutes. This tells us the average time a person would expect to wait for a subway.

• Mean involves integrating the product of the variable and its density function across the given interval.
• This results in a number that signifies a typical expected outcome.

Both mean and median provide insights into the distribution's shape. The mean being slightly higher than the median indicates a typical right-skewed distribution, which is common in exponential decay, where most values pile up at the lower end, with fewer occurrences at longer waiting times.