/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A doctor has scheduled two appoi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A doctor has scheduled two appointments, one at 1 P.M. and the other at 1:30 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 amount of time that the \(1:30\) appointment spends at the doctor's office is \(30\) minutes.

Step by step solution

01

Define the random variables

Let \(X\) be the duration of the first appointment and \(Y\) be the duration of the second appointment. Since both durations are independent and identically distributed, we can write them as \(X \sim Exp(\lambda)\) and \(Y \sim Exp(\lambda)\), where \(\lambda\) is the rate parameter of the exponential distribution and its reciprocal, \(1/\lambda\), represents the mean duration of an appointment. From the exercise, we know that \(1/\lambda = 30\) minutes. So, we have \(\lambda = 1/30\).
02

Compute the conditional expectation

We are interested in the expected time that the second appointment spends at the doctor's office, which can be written as \(E[Y|X]\). To do this, we need to compute the conditional expectation \(E[Y|X=x]\) and integrate it over all possible values of \(x\). We start by finding \(E[Y|X=x]\): Since \(X\) and \(Y\) are independent, we can write \(E[Y|X=x] = E[Y]\). We know that for an exponential random variable, \(E[Y] = 1/\lambda = 30\).
03

Compute the probability

We need to find the probability that the second appointment starts before or exactly 30 minutes after the first appointment, in other words, the probability that \(X \le 30\). Since \(X\) follows an exponential distribution, we can compute this probability as: \(P(X \le 30) = 1 - P(X > 30) = 1 - e^{-\lambda x} = 1 - e^{-(1/30)(30)}=1-e^{-1}\).
04

Compute the expected time

Finally, we combine the conditional expectation and the probability to find the expected amount of time the second appointment spends at the doctor's office: \(E[\text{time the second appointment spends at the office}] = E[Y|X \le 30]P(X \le 30) + E[Y|X > 30]P(X > 30)\) \(= E[Y]P(X \le 30) + E[Y]P(X > 30)\) \(= 30(1 - e^{-1}) + 30e^{-1}\) \(= 30(1 - e^{-1} + e^{-1})\).
05

Simplify the expression

Now we can simplify the expression to find the final answer: \(E[\text{time the second appointment spends at the office}] = 30(1 - e^{-1} + e^{-1}) = 30(1)\) \(= 30\) minutes. So, the expected amount of time that the \(1:30\) appointment spends at the doctor's office is \(30\) minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Expectation
The concept of conditional expectation is fundamental in probability, especially when dealing with random variables. Imagine you already know some information that could affect the outcome of what you are predicting. Conditional expectation allows you to update your predictions based on that new piece of information. In the context of our exercise, the conditional expectation revolves around the time spent on the second appointment, given the information about the first appointment.

Even though we are interested in predicting the duration of the second appointment, knowing how long the first appointment lasts can potentially provide additional insight. However, due to the independent nature of our variables in this exercise, the conditional expectation of the second appointment's duration remains unaffected by the first appointment's time. The key takeaway is that the expected value of a random variable, given a condition, can be simplified if the variables involved are independent.
  • The expected duration of the second appointment, with or without information about the first one, remains 30 minutes.
  • For more complex scenarios where dependencies exist, conditional expectation significantly helps update predictions and refine outcomes based on additional data.
Independent Random Variables
Independent random variables are a particular type of random variables that do not influence each other's outcomes. This means the occurrence of one does not change the probability distribution of the other. In mathematical terms, two random variables, say \(X\) and \(Y\), are independent if the joint probability distribution equals the product of their individual distributions.

In our exercise, the durations of the two appointments, denoted as \(X\) and \(Y\), respectively, are independent. The significance of this independence is immense as it simplifies calculations such as conditional expectations and allows separate treatment of each variable. This makes it much more straightforward to determine the expectations and probabilities associated with the duration of the appointments.
  • Independence simplifies many statistical calculations by allowing separate analyses of each variable.
  • This property can be a powerful tool in scenarios involving multiple random events, providing an analytical handle on otherwise complex problems.
Probability Calculation
Probability calculation involves figuring out the likelihood of different outcomes given a specific scenario. In this exercise, the focus was on calculating the probability of the second appointment starting on time, considering the duration of the first appointment. Since both durations follow an exponential distribution, we utilized its properties for probability calculations.

The exponential distribution has a key feature: it is memoryless. This means that the probability of an event happening in the future is unaffected by past events. In mathematical terms, this is expressed by \(P(X > s + t | X > s) = P(X > t)\).

Calculating the probability that the first appointment ends within 30 minutes involves:
  • Using the complementary probability \(1 - e^{-\lambda x}\) to find \(P(X \le 30) = 1 - e^{-1}\).
  • Being comfortable with exponential functions and their properties is crucial, as they often appear in such real-life scenarios, helping us quantify uncertainty and timing.
Understanding and calculating probabilities help predict outcomes and allows individuals to make informed decisions based on those predictions.

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