/*! 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 69 A store is expecting \(n\) deliv... [FREE SOLUTION] | 91Ó°ÊÓ

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A store is expecting \(n\) deliveries between the hours of noon and 1 p.m. Suppose the arrival time of each delivery truck is uniformly distributed on this one-hour interval and that the times are independent of each other. What are the expected values of the ordered arrival times?

Short Answer

Expert verified
The expected arrival time for the i-th truck is \( E(X_{(i)}) = \frac{60i}{n+1} \) minutes after noon.

Step by step solution

01

Understand Uniform Distribution

The problem states that the arrival time of each delivery truck is uniformly distributed between noon and 1 p.m. This means that any minute within the one-hour interval is equally likely for the truck to arrive. The uniform distribution is defined over the interval [0, 60] minutes.
02

Determine Expected Values of Order Statistics

For a uniform distribution over the interval [a, b], if you have n independent samples, the expected value of the i-th order statistic can be calculated using \[ E(X_{(i)}) = a + \frac{i(b-a)}{n+1} \]In this problem, \(a = 0\) minutes and \(b = 60\) minutes.
03

Calculate Expected Value for First Arrival

Substituting i=1 into the formula, we find the expected time for the first truck to arrive as: \[ E(X_{(1)}) = 0 + \frac{1(60-0)}{n+1} = \frac{60}{n+1} \] This is the expected value of the earliest arriving truck.
04

Calculate Expected Value for Last Arrival

Substituting i=n for the last delivery truck, its expected arrival time is: \[ E(X_{(n)}) = 0 + \frac{n(60-0)}{n+1} = \frac{60n}{n+1} \] This is expected value of the latest arriving truck.
05

Calculate Expected Value for Any ith Arrival

For a general i-th order statistic truck arrival time, substitute i in the formula: \[ E(X_{(i)}) = \frac{60i}{n+1} \] This gives the time expected for the i-th truck arrival.

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Key Concepts

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

Uniform Distribution
Uniform distribution is a common concept in probability theory. It describes a situation where every outcome in a particular interval is equally likely. Imagine a clock from 12:00 to 1:00 PM where any minute is just as probable for a specific event to happen. For this exercise, this means that any delivery truck could arrive at any given minute between noon and 1 PM with equal chance.

You have an interval from 0 to 60 minutes, and the probability of arriving at minute 15 is just as probable as arriving at minute 30. The key takeaway is that, for any minute, there's no favor, bias, or predilection for the truck to arrive. This is the essence of a uniform distribution in such scenarios.
Expected Value
The expected value is a fundamental idea in statistics and probability used to predict the average outcome from a series of trials. In this scenario, you are considering the expected time a truck arrives. It sums up the concept of finding the mean of all possible outcomes.

When you deal with deliveries arriving uniformly between noon and 1 PM, the formula for the expected value of the \(i\)th order statistic becomes vital: \[ E(X_{(i)}) = a + \frac{i(b-a)}{n+1} \] where \(a=0\) and \(b=60\) for our problem.

This computation helps determine the expected arrival time of the first truck, the last truck, or any truck at the \(i\)th position. Simply put, it shows us the average time when a specific delivery truck arrives if this exercise is repeated over many days.
Independent Samples
When dealing with independent samples, it means that the outcome of one sample doesn't affect the others. For this problem, each truck's arrival is an independent event. The arrival time of one truck doesn’t influence when the other trucks will get there.

This independence is crucial when considering random experiments in probability. The computations for expected values of order statistics rely on this assumption. If they were not independent, the arrival time of one truck might skew the expected results for another, making predictions and calculations incorrect.
Probability Theory
Probability theory is the mathematical framework that deals with uncertainty and randomness. It is the core behind understanding events that happen randomly, like the arrival of trucks in this exercise.

Applying probability theory allows us to calculate the likelihood of events, such as what time trucks are expected to arrive. It involves concepts like uniform distribution and expected value to predict future occurrences.
  • Builds a structure to calculate probabilities
  • Helps in estimating expected outcomes
  • Analyzes random behaviors like truck arrivals
In our scenario, we explore how these trucks' arrival times are modeled using probability principles, determining expected values through statistical theory.

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Most popular questions from this chapter

Teresa and Allison each have arrival times uniformly distributed between 12:00 and 1:00. Their times do not influence each other. If \(Y\) is the first of the two times and \(X\) is the second, on a scale of \(0-1\), then the joint pdf of \(X\) and \(Y\) is \(f(x, y)=2\) for \(0

Let \(X\) be a random digit \((0,1,2, \ldots, 9\) are equally likely) and let \(Y\) be a random digit not equal to \(X\). That is, the nine digits other than \(X\) are equally likely for \(Y\). a. Determine \(p_{X}(x), p_{Y X X}(y \mid x), p_{X, Y}(x, y)\). b. Determine a formula for \(E(Y \mid X=x)\). Is this a linear function of \(x\) ?

A system consisting of two components will continue to operate only as long as both components function. Suppose the joint pdf of the lifetimes (months) of the two components in a system is given by \(f(x, y)=c[10-(x+y)]\) for \(x>0\), \(y>0, x+y<10\) a. If the first component functions for exactly 3 months, what is the probability that the second functions for more than 2 months? b. Suppose the system will continue to work only as long as both components function. Among 20 of these systems that operate independently of each other, what is the probability that at least half work for more than 3 months?

A stick is one foot long. You break it at a point \(X\) (measured from the left end) chosen randomly uniformly along its length. Then you break the left part at a point \(Y\) chosen randomly uniformly along its length. In other words, \(X\) is uniformly distributed between 0 and 1 and, given \(X=x, Y\) is uniformly distributed between 0 and \(x\). a. Determine \(E(Y \mid X=x)\) and then \(V(Y \mid X=x)\). Is \(E(Y \mid X=x)\) a linear function of \(x\) ? b. Determine \(f(x, y)\) using \(f_{X}(x)\) and \(f_{Y \mid X}(y \mid x)\). c. Determine \(f_{Y}(y)\). d. Use \(f_{Y}(y)\) from (c) to get \(E(Y)\) and \(V(Y)\). e. Use (a) and the theorem of this section to get \(E(Y)\) and \(V(Y)\).

Consider a sample of size \(n=3\) from the standard normal distribution, and obtain the expected value of the largest order statistic. What does this say about the expected value of the largest order statistic in a sample of this size from any normal distribution? [Hint: With \(\phi(x)\) denoting the standard normal pdf, use the fact that \((d / d x) \phi(x)=-x \phi(x)\) along with integration by parts.]

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