Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 withcorresponding probabilities 0.15, 0.15, 0.35, and 0.35, respectively.(a) Calculate E(X) and then E(5 – X).E(X)=E(5 - X)(b) Would the repair facility be better off charging a flat fee of $95 or else the amount $150?[(5 - X)]Note: It is not generally true that E.E(Y)150The repair facility-Select--- v be better off charging a flat fee of $95 because E(5 - X)

values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with onding probabilities 0.15, 0.15, 0.35, and 0.35, respectively. alculate E(X) and then E(5 – X). E(X) = 5 - X) - 150 ould the repair facility be better off charging a flat fee of $95 or else the amount $ [(5 – X) Wote: It is not generally true that 150 e repair facility | -Select--- v be better off charging a flat fee of $95 because E (5 - X)]

values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with onding probabilities 0.15, 0.15, 0.35, and 0.35, respectively. alculate E(X) and then E(5 – X). E(X) = 5 - X) - 150 ould the repair facility be better off charging a flat fee of $95 or else the amount $ [(5 – X) Wote: It is not generally true that 150 e repair facility | -Select--- v be better off charging a flat fee of $95 because E (5 - X)]

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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