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Get Full Access to Mathematical Statistics With Applications - 7 Edition - Chapter 3 - Problem 3.169
Get Full Access to Mathematical Statistics With Applications - 7 Edition - Chapter 3 - Problem 3.169

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# This exercise demonstrates that, in general, the results provided by Tchebysheffs ISBN: 9780495110811 47

## Solution for problem 3.169 Chapter 3

Mathematical Statistics with Applications | 7th Edition

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Problem 3.169

This exercise demonstrates that, in general, the results provided by Tchebysheffs theorem cannot be improved upon. Let Y be a random variable such that p(1) = 1 18 , p(0) = 16 18 , p(1) = 1 18 .a Show that E(Y ) = 0 and V(Y ) = 1/9. b Use the probability distribution of Y to calculate P(|Y | 3). Compare this exact probability with the upper bound provided by Tchebysheffs theorem to see that the bound provided by Tchebysheffs theorem is actually attained when k = 3. *c In part (b) we guaranteed E(Y ) = 0 by placing all probability mass on the values 1, 0, and 1, with p(1) = p(1). The variance was controlled by the probabilities assigned to p(1) and p(1). Using this same basic idea, construct a probability distribution for a random variable X that will yield P(|X X | 2X ) = 1/4. *d If any k > 1 is specified, how can a random variable W be constructed so that P(|W W | kW ) = 1/k2 ?

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Exam 1: Study Guide Lecture 2: Rafia Chaudhary  Statistics is the art and science of learning from data  Individuals (also known as subjects, instances, and observations) are the things we get our data from. Who/what do we get our data from o These are often people, animals, or other living things but don’t need to be  Variable: any characteristic or piece of information about an individual o Is an answer to a question, like how tall are you What’s your name etc.  Ex: you are working for a drug company and you are developing a new drug to treat the common cold o Individuals: humans o Variables: measure of patient’s symptoms, time it takes f

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##### ISBN: 9780495110811

The answer to “This exercise demonstrates that, in general, the results provided by Tchebysheffs theorem cannot be improved upon. Let Y be a random variable such that p(1) = 1 18 , p(0) = 16 18 , p(1) = 1 18 .a Show that E(Y ) = 0 and V(Y ) = 1/9. b Use the probability distribution of Y to calculate P(|Y | 3). Compare this exact probability with the upper bound provided by Tchebysheffs theorem to see that the bound provided by Tchebysheffs theorem is actually attained when k = 3. *c In part (b) we guaranteed E(Y ) = 0 by placing all probability mass on the values 1, 0, and 1, with p(1) = p(1). The variance was controlled by the probabilities assigned to p(1) and p(1). Using this same basic idea, construct a probability distribution for a random variable X that will yield P(|X X | 2X ) = 1/4. *d If any k > 1 is specified, how can a random variable W be constructed so that P(|W W | kW ) = 1/k2 ?” is broken down into a number of easy to follow steps, and 178 words. This textbook survival guide was created for the textbook: Mathematical Statistics with Applications , edition: 7. Since the solution to 3.169 from 3 chapter was answered, more than 229 students have viewed the full step-by-step answer. Mathematical Statistics with Applications was written by and is associated to the ISBN: 9780495110811. This full solution covers the following key subjects: . This expansive textbook survival guide covers 32 chapters, and 3350 solutions. The full step-by-step solution to problem: 3.169 from chapter: 3 was answered by , our top Statistics solution expert on 07/18/17, 08:07AM.

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