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
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Transcribed Image Text: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 standard
deviation of the number of cycles is o = 5000. Let X, X21 · ..
specimen. Then the expected value of the sample mean
X25 be a random sample of size 25, where each X; is the number of cycles on a different randomly selected
number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the
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
half of its previous value (the sample size must be quadrupled v
to halve the
standard deviation of X.)
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