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Example 17.2 The deviation of the size of an item from the midpoint of the tolerance field of width 2d equals the sum of two random variables X and Y with probability densities and f(x) p(y): 1 exp x² {-2003) 20² exp {-23). 0,√2T Determine the (conditional) probability density of the random variable X for the nondefective items if the distribution p(y) does not depend on the value assumed by X.

Example 17.2 The deviation of the size of an item from the midpoint of the tolerance field of width 2d equals the sum of two random variables X and Y with probability densities and f(x) p(y): 1 exp x² {-2003) 20² exp {-23). 0,√2T Determine the (conditional) probability density of the random variable X for the nondefective items if the distribution p(y) does not depend on the value assumed by X.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Example 17.2 The deviation of the size of an item from the midpoint of the tolerance field of width 2d equals the sum of two random variables X and Y with probability densities and f(x) ❤(y) = = 1 0x V/Z #7 CXP { - 12/0 exp x² 20² √2TT √/2= exp{-2013) · 20² Determine the (conditional) probability density of the random variable X for the nondefective items if the distribution p(y) does not depend on the value assumed by X.
Transcribed Image Text:Example 17.2 The deviation of the size of an item from the midpoint of the tolerance field of width 2d equals the sum of two random variables X and Y with probability densities and f(x) ❤(y) = = 1 0x V/Z #7 CXP { - 12/0 exp x² 20² √2TT √/2= exp{-2013) · 20² Determine the (conditional) probability density of the random variable X for the nondefective items if the distribution p(y) does not depend on the value assumed by X.
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