You test some components with a Weibull distribution at temperatures of 100 Celcius and 60 Celcius. At those temperatures you estimate a characteristic life of 15000 hours and 25000 hours...

You test some components with a Weibull<br>distribution at temperatures of 100 Celcius and<br>60 Celcius. At those temperatures you estimate<br>a characteristic life of 15000 hours and 25000<br>hours respectively. hint: Boltzmann's constant is<br>k = 8.617 x 10-5. Temperature in Celcius +<br>273.15 = temperature in Kelvin. If n is an integer,<br>T(n) = (n – 1)<br>a) Using linear regression, estimate the<br>Arrhenius model parameters A and AH.<br>b) Give the MTTF, median, and 90th percentile<br>at T = 20 Celcius. Assume that the shape<br>parameter is 0.5<br>

Extracted text: You test some components with a Weibull distribution at temperatures of 100 Celcius and 60 Celcius. At those temperatures you estimate a characteristic life of 15000 hours and 25000 hours respectively. hint: Boltzmann's constant is k = 8.617 x 10-5. Temperature in Celcius + 273.15 = temperature in Kelvin. If n is an integer, T(n) = (n – 1) a) Using linear regression, estimate the Arrhenius model parameters A and AH. b) Give the MTTF, median, and 90th percentile at T = 20 Celcius. Assume that the shape parameter is 0.5

Jun 07, 2022

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You test some components with a Weibull distribution at temperatures of 100 Celcius and 60 Celcius. At those temperatures you estimate a characteristic life of 15000 hours and 25000 hours...

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