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Magazine Chapter 5: Problem 31
A doctor has scheduled two appointments, one at \(1 \mathrm{P} . \mathrm{M}\). and the other at \(1: 30 \mathrm{P.M}\). The amounts of time that appointments last are independent exponential random variables with mean 30 minutes. Assuming that both patients are on time, find the expected amount of time that the \(1: 30\) appointment spends at the doctor's office.
Short Answer
Expert verified
The expected time the second patient spends at the doctor's office is given by:
\(E[T] = \int_{0}^{\infty} w \frac{1}{30} e^{-\frac{1}{30}(w + 30)} dw + 30\)
Step by step solution
01
Define random variables
Let X and Y be the durations of the first and second appointments, respectively, both of which follow exponential distributions with mean 30 minutes. Let W be the waiting time for the second patient.
02
Calculate the probability density functions for X and Y
Since the appointments are independent and follow exponential distributions with mean 30 minutes, we have the probability density functions (PDFs) for each appointment as follows:
\(f_X(x) = \frac{1}{30} e^{-\frac{1}{30}x}\) for \(x \geq 0\)
\(f_Y(y) = \frac{1}{30} e^{-\frac{1}{30}y}\) for \(y \geq 0\)
03
Find the expected waiting time for the second appointment
To find the expected waiting time W for the second patient, we can determine the expected value of the random variable \(W = max(0, X - 30)\), since the second patient has to wait for the first appointment to finish if it takes longer than 30 minutes.
So, we first need to find the probability density function of W. To do this, let's first find the cumulative distribution function (CDF) of W. We can express the CDF of W as:
\(F_W(w) = P(W \leq w) = P(max(0, X - 30) \leq w)\)
As W is non-negative, if \(w < 0\), we have \(F_W(w) = 0\). And, for \(w \geq 0\), we have:
\(F_W(w) = P(max(0, X - 30) \leq w) = P(0 \leq w \leq X - 30)\)
Using the CDF of X, we get:
\(F_W(w) = F_X(w + 30) = 1 - e^{-\frac{1}{30}(w + 30)}\)
Now, we can find the PDF of W by differentiating the CDF of W with respect to w:
\(f_W(w) = \frac{dF_W(w)}{dw} = \frac{1}{30} e^{-\frac{1}{30}(w + 30)}\)
Finally, we find the expected waiting time E[W]:
\(E[W] = \int_{0}^{\infty} w f_W(w) dw = \int_{0}^{\infty} w \frac{1}{30} e^{-\frac{1}{30}(w + 30)} dw\)
04
Find the expected time the second patient spends at the doctor's office
Since the second patient's waiting time and appointment duration are independent, the expected time they spend at the doctor's office is given by the sum of the expected waiting time and the expected appointment duration:
\(E[T] = E[W] + E[Y]\)
We know that \(E[Y] = 30\) as the second appointment's duration is an exponential random variable with mean 30 minutes. We already have the integral expression to find \(E[W]\) in Step 3. Therefore:
\(E[T] = \int_{0}^{\infty} w \frac{1}{30} e^{-\frac{1}{30}(w + 30)} dw + 30\)
The expected time the second patient spends at the doctor's office is given by the above integral calculation added to 30 minutes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Distribution
The exponential distribution is an important concept in statistics and probability theory, especially when it comes to modeling the time until an event occurs. Imagine waiting for a bus to arrive, and the time it takes can be described by the exponential distribution if the bus comes on average at fixed intervals, but with some variation.
For the case of the doctor's appointments described in the exercise, each appointment can be seen as an event that has a certain expected duration 鈥� in this case, 30 minutes. The exponential distribution has the unique property that it is 'memoryless,' meaning that the probability of the event happening in the next instant is the same, regardless of how long we have already waited. This mirrors the nature of the patients' waiting time, as the wait only depends on the current occurrence, rather than any previous events.
Mathematically, the exponential distribution is characterized by its mean, which is the average time until the event occurs. The exponential random variables in the problem have a mean of 30 minutes, indicating that, on average, each appointment is expected to last for 30 minutes. However, the waiting time for the second patient, calculated from these variables, follows its own distribution, which is derived from that of the first appointment.
For the case of the doctor's appointments described in the exercise, each appointment can be seen as an event that has a certain expected duration 鈥� in this case, 30 minutes. The exponential distribution has the unique property that it is 'memoryless,' meaning that the probability of the event happening in the next instant is the same, regardless of how long we have already waited. This mirrors the nature of the patients' waiting time, as the wait only depends on the current occurrence, rather than any previous events.
Mathematically, the exponential distribution is characterized by its mean, which is the average time until the event occurs. The exponential random variables in the problem have a mean of 30 minutes, indicating that, on average, each appointment is expected to last for 30 minutes. However, the waiting time for the second patient, calculated from these variables, follows its own distribution, which is derived from that of the first appointment.
Random Variables
In the world of probability and statistics, a random variable is a crucial notion. It essentially assigns numerical values to each outcome of a random phenomenon, thereby quantifying uncertainty. For our example about the doctor's appointments, we're particularly interested in two types of random variables: the durations of appointments (X and Y) and the waiting time (W) for the second patient.
The durations of the appointments X and Y are expressed as independent exponential random variables with a mean of 30 minutes. The independence here implies that the length of one appointment is not affected by the length of the other. This aligns with the scenario of the doctor's office where the second patient's appointment is unaffected by the first one. Meanwhile, W represents a more complex random variable since it depends on the outcome of X (the duration of the first appointment).
The task at hand is to understand how these random variables interact with one another and to find out what the expected waiting time, and ultimately the total time spent at the doctor's office, looks like for the second patient. To accomplish this, we use the mathematical descriptions for these variables provided by the exponential distribution and the probability density function.
The durations of the appointments X and Y are expressed as independent exponential random variables with a mean of 30 minutes. The independence here implies that the length of one appointment is not affected by the length of the other. This aligns with the scenario of the doctor's office where the second patient's appointment is unaffected by the first one. Meanwhile, W represents a more complex random variable since it depends on the outcome of X (the duration of the first appointment).
The task at hand is to understand how these random variables interact with one another and to find out what the expected waiting time, and ultimately the total time spent at the doctor's office, looks like for the second patient. To accomplish this, we use the mathematical descriptions for these variables provided by the exponential distribution and the probability density function.
Probability Density Function
Capturing the likelihood of a random variable taking on a particular value, the probability density function (PDF) is a fundamental concept in statistical analysis of continuous variables. For a continuous random variable like the duration of a doctor's appointment, the PDF gives us the probability that the variable will fall within a particular range of values.
In the exercise we're considering, we calculate the PDFs for two exponential random variables X and Y, which represent the durations of the first and second appointments respectively. The PDF tells us the likelihood of these appointments lasting any given amount of time, assuming that they will on average last for 30 minutes.
Furthermore, for our second patient, who potentially needs to wait for the first to finish, we derive the PDF for their waiting time, W, based on the PDF for the first appointment X. This derivation involves understanding that the waiting time starts only if the first appointment exceeds 30 minutes 鈥� highlighting the relationship between the two variables. The final step, integrating this new PDF, allows us to calculate the expected waiting time 鈥� an integral feature for the patient's overall time spent in the doctor's office.
In the exercise we're considering, we calculate the PDFs for two exponential random variables X and Y, which represent the durations of the first and second appointments respectively. The PDF tells us the likelihood of these appointments lasting any given amount of time, assuming that they will on average last for 30 minutes.
Furthermore, for our second patient, who potentially needs to wait for the first to finish, we derive the PDF for their waiting time, W, based on the PDF for the first appointment X. This derivation involves understanding that the waiting time starts only if the first appointment exceeds 30 minutes 鈥� highlighting the relationship between the two variables. The final step, integrating this new PDF, allows us to calculate the expected waiting time 鈥� an integral feature for the patient's overall time spent in the doctor's office.
