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Let be a standard normal random variable and calculate the

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 28E 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 28E

Let ? ?be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate.

Step-by-Step Solution:

Answer : Step 1 of 10 : Let Z be a standard normal random variable The claim is to calculate the probabilities a) We have to find the probability values for P(0 Z 2.17) Let, P(0 Z 2.17) = (2.17) - (0) We have to check standard normal table for the values 2.17 and 0 Therefore, P(0 Z 2.17) = 0.9850 - 0.5000 = 0.4850 Hence, P(0 Z 2.17) = 0.4850. Step 2 of 10 : b) We have to find the probability values for P(0 Z 1) Let, P(0 Z 1) = (1) - (0) We have to check standard normal table for the values 1 and 0 Therefore, P(0 Z 1) = 0.8413 - 0.5000 = 0.3413 Hence, P(0 Z 1) = 0.3413 Step 3 of 10 : c) We have to find the probability values for P(-2.50 Z 0) Let, P(-2.50 Z 0) = (0) - (-2.50) We have to check standard normal table for the values -2.50 and 0 Therefore, P(-2.50 Z 0) = 0.5000 - 0.0062 = 0.4938 Hence, P(-2.50 Z 0) = 0.4938 Step 4 of 10 : d) We have to find the probability values for P(-2.50 Z 2.50) Let, P(-2.50 Z 2.50) = (2.50) - (-2.50) We have to check standard normal table for the values 2.50 and -2.50 Therefore, P(-2.50 Z 2.50) = 0.9938 - 0.0062 = 0.9876 Hence, P(-2.50 Z 2.50) = 0.9876 Step 5 of 10 : e) We have to find the probability values for P( Z 1.37) Let, P( Z 1.37) = (1.37) We have to check standard normal table for the values 1.37 Therefore, P( Z 1.37) = (1.37) = 0.9147 Hence, P( Z 1.37) = 0.9147

Step 6 of 10

Chapter 4, Problem 28E is Solved
Step 7 of 10

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

The full step-by-step solution to problem: 28E from chapter: 4 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. The answer to “Let ? ?be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate.” is broken down into a number of easy to follow steps, and 17 words. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. This full solution covers the following key subjects: appropriate, calculate, drawing, let, normal. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Since the solution to 28E from 4 chapter was answered, more than 303 students have viewed the full step-by-step answer.

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