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The weekly demand for propane gas (in 1000s of gallons)

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 22E Chapter 4

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 22E

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv ?X ?with pdf a.? ?Compute the cdf of ?X?. b.? ?Obtain an expression for the (100?p?)th percentile. What is the value of ? c.? ?Compute ?E?? ?) and ?? ?? . d.? ?If 1.5 thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the 1.5 thousand gallons is expected to be left at the end of the week? [?Hint?: h(x)Let amount left when demand = ?x.? ]

Step-by-Step Solution:

Solution Step 1: It is given that the weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv X with pdf We have to find the cdf X 100th percentile , median , mean and variance. Step 2 : a) We have to find the cumulative distribution function. x F(x) = P(X x) = f(x)dx 1 x = 2( 1 1/x ) dx 1 = 2 [x+1/x] x 1 = 2[x+1/x - 2] Also note that F(1)=0, F(2)=1. So we can express the cdf of X in the form. b) we have to obtain an expression for 100th percentile. And also want to find the median . In order to find the percentile set F(x)= P , then find the value for x. In 100th percentile the value of P will be = 1 So here it will be F(x)= 2( x+1/x - 2) = 1 2 = (x + 1- 2x) = x/2 2 = (x - 2x-x/2 + 1)=0 = (x-2) (x-½) = 0 = x= 2 or x= ½ So the 100th percentile will be at x= 2 or x=1/2 . Similarly The median is the 50th percentile, which means here P= 0.5 So F(x)= 2( x+1/x - 2) = 0.5 = (x + 1- 2x) = x/4 = (x - 2x-x/4 + 1)=0 2 = (x - 9/4 x +1) = 0 By using quadratic formula b± b 4ac x= 2a X = 9/8 ± 17/8 1

Step 3 of 4

Chapter 4, Problem 22E is Solved
Step 4 of 4

Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

Since the solution to 22E from 4 chapter was answered, more than 442 students have viewed the full step-by-step answer. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. The answer to “The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv ?X ?with pdf a.? ?Compute the cdf of ?X?. b.? ?Obtain an expression for the (100?p?)th percentile. What is the value of ? c.? ?Compute ?E?? ?) and ?? ?? . d.? ?If 1.5 thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the 1.5 thousand gallons is expected to be left at the end of the week? [?Hint?: h(x)Let amount left when demand = ?x.? ]” is broken down into a number of easy to follow steps, and 99 words. This full solution covers the following key subjects: gallons, week, compute, thousand, demand. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. The full step-by-step solution to problem: 22E from chapter: 4 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM.

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