/*! 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 6 Annie and Alvie have agreed to m... [FREE SOLUTION] | 91Ó°ÊÓ

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Annie and Alvie have agreed to meet for lunch between noon (0:00 P.M.) and 1:00 P.M. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cc} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{X}(y)=\left\\{\begin{array}{cl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: \(h(X, Y)=|X-Y| .]\)

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Step by step solution

01

Define the Problem

We want to find the expected waiting time for the person who arrives first, given that Annie arrives at time \(X\) and Alvie at time \(Y\). This waiting time is represented by the function \(h(X,Y) = |X-Y|\) since it measures the absolute difference in their arrival times.
02

Determine the Joint PDF

The random variables \(X\) and \(Y\) are independent, so the joint pdf is the product of their individual pdf's:\[ f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y) = 3x^2 \cdot 2y = 6x^2y \quad \text{for} \quad 0 \leq x, y \leq 1 \]},
03

Set Up the Expected Value

The expected value of the waiting time \(E[|X-Y|]\) can be expressed using the double integral:\[ E[|X-Y|] = \int_0^1 \int_0^1 |x-y| 6x^2y \: dy \: dx \]
04

Consider Symmetry

Notice that \(|x-y|\) can be split based on whether \(x > y\) or \(x < y\). Due to symmetry, consider the integration over the region \(x > y\), multiply by 2 for symmetry, and simplify the integration range:\[ E[|X-Y|] = 2\int_0^1 \int_0^x (x-y) 6x^2y \: dy \: dx \]
05

Integrate with Respect to y

Compute the integral with respect to \(y\) first:\[ \int_0^x (x-y) 6x^2y \: dy = 6x^2 \left(\int_0^x xy \: dy - \int_0^x y^2 \: dy \right) \]

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Key Concepts

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

Joint Probability Density Function
The concept of the Joint Probability Density Function (PDF) is key in probability when dealing with more than one random variable. In the case of Annie and Alvie, who arrive at their meeting place at different times, we have two random variables: - Annie's arrival time, denoted by the random variable \(X\)- Alvie's arrival time, denoted by the random variable \(Y\)If these variables are independent, as is the case here, the joint PDF helps us calculate the likelihood of both events occurring together at specific times. The joint PDF, denoted \(f_{X,Y}(x,y)\), is simply the product of the two individual PDFs when variables are independent.For Annie and Alvie's arrival, we have:- \(f_X(x) = 3x^2\) for \(0 \leq x \leq 1\)- \(f_Y(y) = 2y\) for \(0 \leq y \leq 1\)Thus, their joint PDF is given by:\[f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y) = 6x^2y\]This equation holds only within the defined range \(0 \leq x, y \leq 1\). Understanding the joint PDF allows us to explore the combined behavior of the two independent variables.
Integration in Probability
Integration plays a vital role in probability, especially when calculating expected values such as the expected waiting time for Annie or Alvie. When we want to find the expected value of a function of two continuous random variables, like \(|X-Y|\), integration helps us.To calculate the expected waiting time \(E[|X-Y|]\), we use a double integral over the range of possible values that \(X\) and \(Y\) can take. The expression is:\[E[|X-Y|] = \int_0^1 \int_0^1 |x-y| \cdot 6x^2y \, dy \, dx\]Here, the integrals go from 0 to 1, corresponding to the time between noon and 1:00 PM. The function \(|x-y|\) measures the absolute time difference in their arrivals.The tricky part is considering the regions where \(x > y\) and \(x < y\). Due to symmetry, we only need to compute for \(x > y\) and then double the result. This involves setting up separate integrals and then simplifying them, often requiring substitution and partial fraction techniques.
Independent Random Variables
The concept of independent random variables is crucial in probability. Two random variables are independent if the occurrence of one event does not affect the occurrence of another. For Annie and Alvie's arrival times:- Independence implies that Annie's time \(X\) does not affect Alvie's time \(Y\) and vice versa.Mathematically, this means that their joint PDF is the product of their respective PDFs, \(f_X(x)\) and \(f_Y(y)\). This simplifies calculations immensely:\[f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)\]In our case, Annie's arrival time is modeled by \(3x^2\) and Alvie's by \(2y\).Understanding independence allows us to create models where more complex interactions would otherwise be too complicated or impractical to evaluate. Independence can simplify the distribution and expected value calculations, making them feasible for integration. As with Annie and Alvie, it can help solve real-world problems like determining the expected waiting time without overly complicated math.

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Most popular questions from this chapter

A binary communication channel transmits a sequence of "bits" (0s and 1s). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000 -bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

A shipping company handles containers in three different sizes: (1) \(27 \mathrm{ft}^{3}(3 \times 3 \times 3)\), (2) \(125 \mathrm{ft}^{3}\), and (3) \(512 \mathrm{ft}^{3}\). Let \(X_{i}(i=1,2,3)\) denote the number of type \(i\) containers shipped during a given week. With \(\mu_{i}=E\left(X_{i}\right)\) and \(\sigma_{i}^{2}=V\left(X_{i}\right)\), suppose that the mean values and standard deviations are as follows: $$ \begin{array}{lll} \mu_{1}=200 & \mu_{2}=250 & \mu_{3}=100 \\ \sigma_{1}=10 & \sigma_{2}=12 & \sigma_{3}=8 \end{array} $$ a. Assuming that \(X_{1}, X_{2}, X_{3}\) are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume \(=27 X_{1}+125 X_{2}+512 X_{3}\).] b. Would your calculations necessarily be correct if the \(X_{i}^{\prime}\) s were not independent? Explain.

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter \(\lambda\). a. What is the probability that all ten bulbs fail before time \(f\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(f\) ? c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \(\lambda\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \(\theta\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time \(t\) ?

A restaurant serves three fixed-price dinners costing \(\$ 12\), \(\$ 15\), and \(\$ 20\). For a randomly selected couple dining at this restaurant, let \(X=\) the cost of the man's dinner and \(Y=\) the cost of the woman's dinner. The joint pmf of \(X\) and \(Y\) is given in the following table: \begin{tabular}{cc|ccc} \(p(x, y)\) & & 12 & 15 & 20 \\ \hline & 12 & \(.05\) & \(.05\) & \(.10\) \\ \multirow{x}{*}{} & 15 & \(.05\) & \(.10\) & \(.35\) \\ & 20 & 0 & \(.20\) & \(.10\) \end{tabular} a. Compute the marginal pmf's of \(X\) and \(Y\). b. What is the probability that the man's and the woman's dinner cost at most \(\$ 15\) each? c. Are \(X\) and \(Y\) independent? Justify your answer. d. What is the expected total cost of the dinner for the two people? e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message "You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen." How much would the restaurant expect to refund?

The National Health Statistics Reports dated Oct. 22, 2008, stated that for a sample size of 277 18-year-old American males, the sample mean waist circumference was \(86.3 \mathrm{~cm}\). A somewhat complicated method was used to estimate various population percentiles, resulting in the following values: \(\begin{array}{lllllll}5^{\text {th }} & 10^{\text {th }} & 25^{\text {th }} & 50^{\text {th }} & 75^{\text {th }} & 90^{\text {th }} & 95^{\text {th }} \\\ 69.6 & 70.9 & 75.2 & 81.3 & 95.4 & 107.1 & 116.4\end{array}\) a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution. b. Suppose that the population mean waist size is \(85 \mathrm{~cm}\) and that the population standard deviation is \(15 \mathrm{~cm}\). How likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least \(86.3 \mathrm{~cm}\) ? c. Referring back to (b), suppose now that the population mean waist size in \(82 \mathrm{~cm}\). Now what is the (approximate) probability that the sample mean will be at least \(86.3 \mathrm{~cm}\) ? In light of this calculation, do you think that \(82 \mathrm{~cm}\) is a reasonable value

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