Probability

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Probability & Statistics for Engineers & Scientists

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Probability & Statistics for Engineers & Scientists
NINTH EDITION

Ronald E. Walpole
Roanoke College

Raymond H. Myers
Virginia Tech

Sharon L. Myers
Radford University

Keying Ye
University of Texas at San Antonio

Prentice Hall

Editor in Chief: Deirdre Lynch Acquisitions Editor: Christopher Cummings Executive Content Editor: Christine O’Brien Associate Editor: Christina Lepre Senior Managing Editor: Karen Wernholm Senior Production Project Manager: Tracy Patruno Design Manager: Andrea Nix Cover Designer: Heather Scott Digital Assets Manager: Marianne Groth Associate Media Producer: Vicki Dreyfus Marketing Manager: Alex Gay Marketing Assistant: Kathleen DeChavez Senior Author Support/Technology Specialist: Joe Vetere Rights and Permissions Advisor: Michael Joyce Senior Manufacturing Buyer: Carol Melville Production Coordination: Lifland et al. Bookmakers Composition: Keying Ye Cover photo: Marjory Dressler/Dressler Photo-Graphics Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and Pearson was aware of a trademark claim, the designations have been printed in initial caps or all caps.

Library of Congress Cataloging-in-Publication Data Probability & statistics for engineers & scientists/Ronald E. Walpole . . . [et al.] — 9th ed. p. cm. ISBN 978-0-321-62911-1 1. Engineering—Statistical methods. 2. Probabilities. I. Walpole, Ronald E. TA340.P738 2011 519.02’462–dc22 2010004857 Copyright c 2012, 2007, 2002 Pearson Education, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the…...

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...Statistics 100A Homework 5 Solutions Ryan Rosario Chapter 5 1. Let X be a random variable with probability density function c(1 − x2 ) −1 < x < 1 0 otherwise ∞ f (x) = (a) What is the value of c? We know that for f (x) to be a probability distribution −∞ f (x)dx = 1. We integrate f (x) with respect to x, set the result equal to 1 and solve for c. 1 1 = −1 c(1 − x2 )dx cx − c x3 3 1 −1 = = = = c = Thus, c = 3 4 c c − −c + c− 3 3 2c −2c − 3 3 4c 3 3 4 . (b) What is the cumulative distribution function of X? We want to find F (x). To do that, integrate f (x) from the lower bound of the domain on which f (x) = 0 to x so we will get an expression in terms of x. x F (x) = −1 c(1 − x2 )dx cx − cx3 3 x −1 = But recall that c = 3 . 4 3 1 3 1 = x− x + 4 4 2 = 3 4 x− x3 3 + 2 3 −1 < x < 1 elsewhere 0 1 4. The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by, 10 x2 f (x) = (a) Find P (X > 20). 0 x > 10 x ≤ 10 There are two ways to solve this problem, and other problems like it. We note that the area we are interested in is bounded below by 20 and unbounded above. Thus, ∞ P (X > c) = c f (x)dx Unlike in the discrete case, there is not really an advantage to using the complement, but you can of course do so. We could consider P (X > c) = 1 − P (X < c), c P (X > c) = 1 − P (X < c) = 1 − −∞ f (x)dx P (X > 20) = 10 dx......

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...Probability, Mean and Median In the last section, we considered (probability) density functions. We went on to discuss their relationship with cumulative distribution functions. The goal of this section is to take a closer look at densities, introduce some common distributions and discuss the mean and median. Recall, we define probabilities as follows: Proportion of population for Area under the graph of   p ( x ) between a and b which x is between a and b  p( x)dx a b The cumulative distribution function gives the proportion of the population that has values below t. That is, t P (t )    p( x)dx  Proportion of population having values of x below t When answering some questions involving probabilities, both the density function and the cumulative distribution can be used, as the next example illustrates. Example 1: Consider the graph of the function p(x). p x  0.2 0.1 2 4 6 8 10 x Figure 1: The graph of the function p(x) a. Explain why the function is a probability density function. b. Use the graph to find P(X < 3) c. Use the graph to find P(3 § X § 8) 1 Solution: a. Recall, a function is a probability density function if the area under the curve is equal to 1 and all of the values of p(x) are non-negative. It is immediately clear that the values of p(x) are non-negative. To verify that the area under the curve is equal to 1, we recognize that the graph above can be viewed as a triangle. Its...

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...Problem 1: Question 1. The probability of a case being appealed for each judge in Common Pleas Court. p(a) | 0.04511031 | 0.03529063 | 0.03497615 | 0.03070624 | 0.04047164 | 0.04019435 | 0.03990765 | 0.04427171 | 0.03883194 | 0.04085893 | 0.04033333 | 0.04344897 | 0.04524181 | 0.06282723 | 0.04043298 | 0.02848818 | Question 2. The probability of a case being reversed for each judge in Common Pleas Court. P® | 0.00395127 | 0.0029656 | 0.0063593 | 0.0035824 | 0.00223072 | 0.00795053 | 0.00725594 | 0.00675904 | 0.00434918 | 0.00477185 | 0.002 | 0.00404176 | 0.00561622 | 0.0104712 | 0.00413881 | 0.00194238 | Question 3. The probability of reversal given an appeal for each judge in Common Pleas Court. p(R/A) | 0.08759124 | 0.08403361 | 0.18181818 | 0.11666667 | 0.05511811 | 0.1978022 | 0.18181818 | 0.15267176 | 0.112 | 0.11678832 | 0.04958678 | 0.09302326 | 0.12413793 | 0.16666667 | 0.1023622 | 0.06818182 | Question 4. The probability of cases being appealed in Common Pleas Court. Probability of cases being appealed in common pleas court | 0.0400956 | Question 5. Identify the best judges in Common Pleas Court according to the three criteria in Questions 1-3: 1) The best judge in Common Pleas Court with the smallest probability in Question 1; Ralph Winkler 2) The best judge in Common Pleas Court with the smallest probability in Question 2; and ......

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...PROBABILITY 1. ACCORDING TO STATISTICAL DEFINITION OF PROBABILITY P(A) = lim FA/n WHERE FA IS THE NUMBER OF TIMES EVENT A OCCUR AND n IS THE NUMBER OF TIMES THE EXPERIMANT IS REPEATED. 2. IF P(A) = 0, A IS KNOWN TO BE AN IMPOSSIBLE EVENT AND IS P(A) = 1, A IS KNOWN TO BE A SURE EVENT. 3. BINOMIAL DISTRIBUTIONS IS BIPARAMETRIC DISTRIBUTION, WHERE AS POISSION DISTRIBUTION IS UNIPARAMETRIC ONE. 4. THE CONDITIONS FOR THE POISSION MODEL ARE : • THE PROBABILIY OF SUCCESS IN A VERY SMALL INTERAVAL IS CONSTANT. • THE PROBABILITY OF HAVING MORE THAN ONE SUCCESS IN THE ABOVE REFERRED SMALL TIME INTERVAL IS VERY LOW. • THE PROBABILITY OF SUCCESS IS INDEPENDENT OF t FOR THE TIME INTERVAL(t ,t+dt) . 5. Expected Value or Mathematical Expectation of a random variable may be defined as the sum of the products of the different values taken by the random variable and the corresponding probabilities. Hence if a random variable X takes n values X1, X2,………… Xn with corresponding probabilities p1, p2, p3, ………. pn, then expected value of X is given by µ = E (x) = Σ pi xi . Expected value of X2 is given by E ( X2 ) = Σ pi xi2 Variance of x, is given by σ2 = E(x- µ)2 = E(x2)- µ2 Expectation of a constant k is k i.e. E(k) = k fo any constant k. Expectation of sum of two random variables is the sum of their expectations i.e. E(x +y) = E(x) + E(y) for any......

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...Odd-Numbered End-of-Chapter Exercises * Chapter 2 Review of Probability 2.1. (a) Probability distribution function for Y Outcome (number of heads) | Y  0 | Y  1 | Y  2 | Probability | 0.25 | 0.50 | 0.25 | (b) Cumulative probability distribution function for Y Outcome (number of heads) | Y  0 | 0 Y  1 | 1 Y  2 | Y 2 | Probability | 0 | 0.25 | 0.75 | 1.0 | (c) . Using Key Concept 2.3: and so that 2.3. For the two new random variables and we have: (a) (b) (c) 2.5. Let X denote temperature in F and Y denote temperature in C. Recall that Y  0 when X  32 and Y 100 when X  212; this implies Using Key Concept 2.3, X  70oF implies that and X  7oF implies 2.7. Using obvious notation, thus and This implies (a) per year. (b) , so that Thus where the units are squared thousands of dollars per year. (c) so that and thousand dollars per year. (d) First you need to look up the current Euro/dollar exchange rate in the Wall Street Journal, the Federal Reserve web page, or other financial data outlet. Suppose that this exchange rate is e (say e  0.80 Euros per dollar); each 1 dollar is therefore with e Euros. The mean is therefore e C (in units of thousands of Euros per year), and the standard deviation is e C (in units of thousands of Euros per year). The correlation is unit-free, and is unchanged. 2.9. | | Value of Y | Probability Distribution of X | | | 14 | 22 | 30 | 40 | 65 | | | Value......

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