/*! 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 4 A surveyor wishes to lay out a s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A surveyor wishes to lay out a square region with each side having length \(L\). However, because of a measurement error, he instead lays out a rectangle in which the north-south sides both have length \(X\) and the east-west sides both have length \(Y\). Suppose that \(X\) and \(Y\) are independent and that each is uniformly distributed on the interval \([L-A, L+A]\) (where \(0

Short Answer

Expert verified
The expected area of the rectangle is \(L^2\).

Step by step solution

01

Understand the Problem

We are given a rectangle mistakenly laid with sides \(X\) and \(Y\), which are independent and uniformly distributed between \([L-A, L+A]\). We need to find the expected area of this rectangle. The area of a rectangle is calculated as \(A = X \times Y\).
02

Define Expected Value of X and Y

Since \(X\) is uniformly distributed over \([L-A, L+A]\), the expected value \(E[X]\) is the midpoint of the interval, which is \(L\). Similarly, \(E[Y] = L\) because \(Y\) is also uniformly distributed on the same interval.
03

Use Independence Property

The random variables \(X\) and \(Y\) are independent. This means the expected value of their product is the product of their expected values: \(E[XY] = E[X] \times E[Y]\).
04

Calculate Expected Area

By substituting the expected values from Step 2 into the expression from Step 3, we have \(E[XY] = L \times L = L^2\). Therefore, the expected area of the rectangle is \(L^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Uniform Distribution
In probability theory, a uniform distribution is one where all outcomes are equally likely within a certain range. For our rectangle problem, we have the north-south side lengths and the east-west side lengths distributed uniformly over the interval \[ [L-A, L+A] \]. This means that any length within this interval has an equal chance of being the actual side length. The key characteristic here is that the probability density function is constant across the interval.

For a uniform distribution over the interval \[ [a, b] \], the expected value can be simply found as the midpoint, given by \[ E[X] = \frac{a + b}{2} \]. For our variables \(X\) and \(Y\), this translates to \[ E[X] = E[Y] = L \] since they are uniformly distributed around \(L\). This is essential for determining the expected area of the rectangle.
The Nature of Independent Random Variables
Independent random variables have a particular importance in probability theory because their outcomes do not affect each other. That is, the probability distribution of one does not change when we have information about the other.

In our scenario, the lengths \(X\) and \(Y\) are independent random variables. This independence is crucial for simplifying the calculation of the expected area. When variables are independent, the expected value of their product is the product of their expected values: \[ E[XY] = E[X] \times E[Y] \]. In this problem, given that \[ E[X] = L \] and \[ E[Y] = L \], the expected area calculates simply as \[ L^2 \].

Independence makes complex problems manageable by allowing us to use neat mathematical properties.
Demystifying Area Calculation
Calculating the area of a rectangle is straightforward: you multiply the length of one side by the length of the adjacent side. For our rectangle with uncertain side lengths due to measurement error, the formula becomes \ A = X \times Y \. With exact lengths, the area is just these two multiplied together, but with random lengths, we seek the expected value, i.e., the average area covered.

For uniform distributions, since we have calculated \[ E[X] = L \] and \[ E[Y] = L \], the expected area \[ E[XY] = L^2 \] matches the area computed for a perfectly measured square. Thus, even with slight errors in measurement, on average, the rectangle will approximate the size of the intended square.
Basics of Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random phenomena. It provides the framework for dealing with uncertainties and measuring the likelihood of different outcomes.

Within this framework, expected value is a dominant concept, predicting the average result over many trials. It’s a way of quantifying the center of a probability distribution. When dealing with uniformly distributed variables, we can calculate expected values easily because outcomes are symmetrically spread around the midpoint.
  • The expected value gives a single number summarizing a random variable's distribution. This can simplify predictions and guide decision-making under uncertainty.
  • Probability theory allows us to leverage the independence of variables to simplify calculations, like deriving \[ E[XY] = E[X] \times E[Y] \] when \(X\) and \(Y\) are independent.
Probability theory equips us with the tools to systematically approach randomness and make informed predictions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a certain region is normally distributed with mean \(2.65\) and standard deviation .85 (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants," Water Research, 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most \(3.00\) ? Between \(2.65\) and \(3.00\) ? b. How large a sample size would be required to ensure that the first probability in part (a) is at least 99 ?

There are 40 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of \(6 \mathrm{~min}\) and a standard deviation of \(6 \mathrm{~min}\). a. If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins? b. If the sports report begins at \(11: 10\), what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?

Suppose your waiting time for a bus in the morning is uniformly distributed on \([0,8]\), whereas waiting time in the evening is uniformly distributed on \([0,10]\) independent of morning waiting time. a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define rv's \(X_{1}, \ldots, X_{10}\) and use a rule of expected value.] b. What is the variance of your total waiting time? c. What are the expected value and variance of the difference between morning and evening waiting times on a given day? d. What are the expected value and variance of the difference between total morning waiting time and total evening waiting time for a particular week?

Six individuals, including \(\mathrm{A}\) and \(\mathrm{B}\), take seats around \(\mathrm{a}\) circular table in a completely random fashion. Suppose the seats are numbered \(1, \ldots, 6\). Let \(X=\) A's seat number and \(Y=\mathrm{B}\) 's seat number. If A sends a written message around the table to \(B\) in the direction in which they are closest, how many individuals (including \(A\) and \(B\) ) would you expect to handle the message?

Suppose the amount of liquid dispensed by a certain machine is uniformly distributed with lower limit \(A=8 \mathrm{oz}\) and upper limit \(B=10 \mathrm{oz}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes \(n=5,10,20\), and \(30 .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.