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Let X be a Bernoulli rv with pmf as in Example 3.18. a.

Probability and Statistics for Engineering and the Sciences | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole, Raymond H. Myers

Problem 33E Chapter 3

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Probability and Statistics for Engineering and the Sciences | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole, Raymond H. Myers

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Problem 33E

Let X be a Bernoulli rv with pmf as in Example 3.18. a.? ?Compute E(X2). b.?? how that V(X) = p(1 – p). c.? ?Compute E(X79). Reference example 3.18 Let X = 1 if a randomly selected vehicle passes an emissions test and X = 0 otherwise. Then X is a Bernoulli rv with pmf p(1) = p and p(0) = 1 - p , from which E(X) = 0. P(0) + 1. P(1) = 0(1 – p) + 1(p) = p. That is, the expected value of X is just the probability that X takes on the value 1. If we conceptualize a population consisting of 0s in proportion and 1 – p in proportion p, then the population average is µ = p.

Step-by-Step Solution:

Answer : Step 1 : Let X be a Bernoulli rv with pmf as Let X = 1 if a randomly selected vehicle passes an emissions test and X = 0 otherwise. a). 2 Now we have to calculate E(X ). SoE(X ) = X p(x) E(X ) = X p(x) 2 2 2 E(X ) = 1 × p + 0 × (1 p) E(X ) = p Therefore E(X ) = p. Step 2 : b). We have to show that V(X) = p(1 – p). Then, E(X ) = X p(x) 2 E(X ) = 1 × p + 0 × (1 p) E(X ) = p 2 2 V (X) = E(X ) E(X) V (X) = P (P) 2 V (X) = P(1 P) Therefore V (X) = P(1 P).

Step 2 of 2

Chapter 3, Problem 33E is Solved
Textbook: Probability and Statistics for Engineering and the Sciences
Edition: 9
Author: Ronald E. Walpole, Raymond H. Myers
ISBN: 9780321629111

Probability and Statistics for Engineering and the Sciences was written by and is associated to the ISBN: 9780321629111. This full solution covers the following key subjects: Bernoulli, proportion, compute, population, PMF. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. The answer to “Let X be a Bernoulli rv with pmf as in Example 3.18. a.? ?Compute E(X2). b.?? how that V(X) = p(1 – p). c.? ?Compute E(X79). Reference example 3.18 Let X = 1 if a randomly selected vehicle passes an emissions test and X = 0 otherwise. Then X is a Bernoulli rv with pmf p(1) = p and p(0) = 1 - p , from which E(X) = 0. P(0) + 1. P(1) = 0(1 – p) + 1(p) = p. That is, the expected value of X is just the probability that X takes on the value 1. If we conceptualize a population consisting of 0s in proportion and 1 – p in proportion p, then the population average is µ = p.” is broken down into a number of easy to follow steps, and 125 words. The full step-by-step solution to problem: 33E from chapter: 3 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. Since the solution to 33E from 3 chapter was answered, more than 295 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Probability and Statistics for Engineering and the Sciences, edition: 9.

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