Need some help with this homework question In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate crack initiation) is u = 28,000, and the standarddeviation of the number of cycles is o = 5000. Let X, X21 · ..specimen. Then the expected value of the sample meanX25 be a random sample of size 25, where each X; is the number of cycles on a different randomly selectednumber of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the700000x specimens is E(T,) = nu = 25(28,000) = 700,000. The standard deviations of X and T, areEnter an exact number.5000ox = 0/Vn =1000VnVno = V25(5000)25000If the sample size increases to n = 100, E(X) is unchanged but o y = 500half of its previous value (the sample size must be quadrupled vto halve thestandard deviation of X.)

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In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate crack initiation) is µ = 28,000, and the standard deviation of the number of cycles is o = 5000. Let X,, X2, ... ,X25 be a random sample of size 25, where each X; is the number of cycles on a different randomly selected Vy number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the specimen. Then the expected value of the sample mean 700000 x specimens is E(T,) = nu = 25(28,000) = 700,000. The standard deviations of X and T, are Enter an exact number. 5000 ox = 0/Vn = = 1000 Vn = Vno = V25(5000) 25000 If the sample size increases to n = 100, E(X) is unchanged but o y = 500 standard deviation of Y) , half of its previous value (the sample size must be quadrupled v to halve the

In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate crack initiation) is µ = 28,000, and the standard deviation of the number of cycles is o = 5000. Let X,, X2, ... ,X25 be a random sample of size 25, where each X; is the number of cycles on a different randomly selected Vy number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the specimen. Then the expected value of the sample mean 700000 x specimens is E(T,) = nu = 25(28,000) = 700,000. The standard deviations of X and T, are Enter an exact number. 5000 ox = 0/Vn = = 1000 Vn = Vno = V25(5000) 25000 If the sample size increases to n = 100, E(X) is unchanged but o y = 500 standard deviation of Y) , half of its previous value (the sample size must be quadrupled v to halve the

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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