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 = 28,000, and the standard deviation of the number of cycles is a = 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 specimen. Then the expected value of the sample a vx number of cycles until first emission is E(X) = u= 28,000, and the expected total number of cycles for the 70000 x specimens is E(T,) = nu = 25(28,000) = 700,000. The standard deviations of X and T, are 5000 = 1000 Vn OT. Vno = V25(5000) - 25000 If the sample size increases ton = 100, E(X) is unchanged but ox = 500 , half of its previous value (the sample size must be quadrupled v to halve the standard deviation of X.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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Author:Carter
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Chapter10: Statistics
Section10.6: Summarizing Categorical Data
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Séléct---
In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate
sample of size 25, where each X, is the number of cycles on a different randomly selected specimen. Then the expected value of th v sample o
sample range
: 28,000, and the standard deviation of the number of cycles is o = 5000. Let X,, X,, . .. , X25 be a random
number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the
70000
x specimens is E(T) = nu = 25(28,000) = 700,000. The standard deviations of X and T, are
sample median
sample mean
5000
ox = 0/Vn =
1000
in
Vno =
OTO
V 25 (5000)
= 25000
If the sample size increases to n =
100, E(X) is unchanged but o x = 500
half of its previous value (the sample size must be quadrupled v
to halve the standard deviation of X.)
Transcribed Image Text:Séléct--- In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate sample of size 25, where each X, is the number of cycles on a different randomly selected specimen. Then the expected value of th v sample o sample range : 28,000, and the standard deviation of the number of cycles is o = 5000. Let X,, X,, . .. , X25 be a random number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the 70000 x specimens is E(T) = nu = 25(28,000) = 700,000. The standard deviations of X and T, are sample median sample mean 5000 ox = 0/Vn = 1000 in Vno = OTO V 25 (5000) = 25000 If the sample size increases to n = 100, E(X) is unchanged but o x = 500 half of its previous value (the sample size must be quadrupled v to halve the standard deviation of X.)
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
sample of size 25, where each X, is the number of cycles on a different randomly selected specimen. Then the expected value of the sample o
28,000, and the standard deviation of the number of cycles is o = 5000. Let X,, X, . . . , X25 be a random
number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the
70000
X specimens is E(T,) = nµ = 25(28,000)
= 700,000. The standard deviations of X and T, are
5000
1000
Vn
OT, = V no = V25(5000)
If the sample size increases to n = 100, E(X) is unchanged but o y = 500
= 25000
half of its previous value (the sample size must be quadrupled v
to halve the standard deviation of X.)
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 sample of size 25, where each X, is the number of cycles on a different randomly selected specimen. Then the expected value of the sample o 28,000, and the standard deviation of the number of cycles is o = 5000. Let X,, X, . . . , X25 be a random number of cycles until first emission is E(X) = µ = 28,000, and the expected total number of cycles for the 70000 X specimens is E(T,) = nµ = 25(28,000) = 700,000. The standard deviations of X and T, are 5000 1000 Vn OT, = V no = V25(5000) If the sample size increases to n = 100, E(X) is unchanged but o y = 500 = 25000 half of its previous value (the sample size must be quadrupled v to halve the standard deviation of X.)
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