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The lifetime X of a widget has a Weibull distribution with parameters a = 2 and B = 5. The mean widget lifetime is E[X] = 4.43. The standard deviation of the lifetime is StDev(X) = 2.316. Suppose we order a package of 50 widgets. Let To represent the total lifetime of all the widgets. The mean tota lifetime of all 50 widgets is E(To) = 221.5 and the standard deviation of the total lifetime is StDev(To) = 16.38. W know the mean and standard deviation of To, but we do not know its exact distribution. According to the centra limit theorem, since the sample size n = 50 is considered large, the distribution of To can be considered relatively close to normal. What is the approximate chance the total lifetime of the 50 widgets is more than 240? P(To > 240 )

The lifetime X of a widget has a Weibull distribution with parameters a = 2 and B = 5. The mean widget lifetime is E[X] = 4.43. The standard deviation of the lifetime is StDev(X) = 2.316. Suppose we order a package of 50 widgets. Let To represent the total lifetime of all the widgets. The mean tota lifetime of all 50 widgets is E(To) = 221.5 and the standard deviation of the total lifetime is StDev(To) = 16.38. W know the mean and standard deviation of To, but we do not know its exact distribution. According to the centra limit theorem, since the sample size n = 50 is considered large, the distribution of To can be considered relatively close to normal. What is the approximate chance the total lifetime of the 50 widgets is more than 240? P(To > 240 )

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The lifetime X of a widget has a Weibull distribution with parameters a = 2 and B = 5. The mean widget lifetime is E[X] = 4.43. The standard deviation of the lifetime is StDev(X) = 2.316. Suppose we order a package of 50 widgets. Let To represent the total lifetime of all the widgets. The mean total lifetime of all 50 widgets is E(To) = 221.5 and the standard deviation of the total lifetime is StDev(To) = 16.38. W know the mean and standard deviation of To, but we do not know its exact distribution. According to the central limit theorem, since the sample size n = 50 is considered large, the distribution of To can be considered relatively close to normal. What is the approximate chance the total lifetime of the 50 widgets is more than 240? P( To > 240) Express your answer as a proportion rounded to 4 decimal places.
Transcribed Image Text:The lifetime X of a widget has a Weibull distribution with parameters a = 2 and B = 5. The mean widget lifetime is E[X] = 4.43. The standard deviation of the lifetime is StDev(X) = 2.316. Suppose we order a package of 50 widgets. Let To represent the total lifetime of all the widgets. The mean total lifetime of all 50 widgets is E(To) = 221.5 and the standard deviation of the total lifetime is StDev(To) = 16.38. W know the mean and standard deviation of To, but we do not know its exact distribution. According to the central limit theorem, since the sample size n = 50 is considered large, the distribution of To can be considered relatively close to normal. What is the approximate chance the total lifetime of the 50 widgets is more than 240? P( To > 240) Express your answer as a proportion rounded to 4 decimal places.
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